Class 11 Physics

Statics Short Answer Type Questions

Question 1. Two particles of mass 2 g and 3 g are situated at the locations (3 cm, 5 cm) and (4 cm, 6 cm) respectively. Find the position vector of the centre of mass of the two particles.
Answer:

x = \(\frac{2 \times 3+3 \times 4}{2+3}=3.6 \mathrm{~cm}\)

and y = \(\frac{2 \times 5+3 \times 6}{2+3}=5.6 \mathrm{~cm}\)

Hence, position vector, \(\vec{r}=3.6 \hat{i}+5.6 \hat{j} \mathrm{~cm}\)

Question 2. Explain why the concept of the centre of mass is much more fundamental than the centre of gravity.
Answer:

As the mass of a body remains constant anywhere in the universe, a body always has a definite centre of mass. But at places where there is no gravity (for example, in outer space, inside an artificial satellite rotating around Earth or in case of a free-falling object), the weight of the body becomes zero i.e., there is no centre of gravity.

Hence the concept of the centre of mass is much more fundamental than the centre of gravity.

Question 3.

1. Find the velocity of the centre of mass of two identical particles moving with velocities v₁ and v₂.
2. Show that in the absence of any external force the centre of mass of two moving particles moves with uniform velocity.

Answer:

1. If the mass of each particle is m, the resultant momentum = \(\left(m \vec{v}_1+m \vec{v}_2\right)=m\left(\vec{v}_1+\vec{v}_2\right)\)

If the velocity of the centre of mass of the two particles is \(\vec{v}\), then the momentum of the centre of mass =(m+ m)\(\vec{v}\) = 2m\(\vec{v}\).

∴ 2m\(\vec{v}\) = m(\(\vec{v_1}\)+\(\vec{v_2}\))

Hence, \(\vec{v}=\frac{\vec{v}_1+\vec{v}_2}{2}\)

2. If no external force is applied to the particles, the resultant momentum of the particles remains conserved. That means, on the above example, \(m\left(\vec{v}_1+\vec{v}_2\right)\) = constant. Clearly, both the velocities \(\vec{v}_1\) and \(\vec{v}_2\) can be changed in such a way that the above condition is not violated.

On the other hand, for the centre of mass of the two particles, 2 m\(\vec{v}\) = m(\(\vec{v}_1\)+ \(\vec{v}_2\)) = constant.

As mass m is constant, so \(\vec{v}\) = constant

Hence, in the absence of any external force the centre of mass of two moving particles moves with uniform velocity.

Question 4. Three forces F₁, F₂, F₃ —of which F₂ and F₃ are mutually perpendicular act on a particle of mass m so that the particle is stationary. Find the acceleration of the particle when F₁ is withdrawn.
Answer:

As the particle is stationary, the resultant force \(\vec{F}_1+\vec{F}_2+\vec{F}_3=\overrightarrow{0}\)

Hence, \(\vec{F}_2+\vec{F}_3=-\vec{F}_1\)

∴ \(\vec{F}_2+\vec{F}_3\) is equal and opposite to \(\vec{F}_1\).

∴ \(\vec{F}_2\) and \(\vec{F}_3\) are mutually perpendicular; so the magnitude of \(\left(\vec{F}_2+\vec{F}_3\right)\) is \(\sqrt{F_2^2+F_3^2}\).

When \(\vec{F}_1\) is withdrawn, the resultant force on the particle \(=\vec{F}_2+\vec{F}_3\); in that case the acceleration of the particle will be in the opposite direction of \(\vec{F}_1\).

∴ Magnitude of the acceleration \(=\frac{\sqrt{F_2^2+F_3^2}}{m}\)

WBBSE Class 11 Statics Short Answer Questions

Question 5. Write down its mathematical form. Itoo particles of masses 2g and 3g are situated at the positions (2 cm, 3 cm) and (4 cm, 5 cm) respectively. Find the position vector of the centre of mass of the two particles.
Answer:

x \(=\frac{2 \times 2+3 \times 4}{2+3}=\frac{4+12}{5}=\frac{16}{5}=3.2 \mathrm{~cm}\)

y = \(\frac{2 \times 3+3 \times 5}{2+3}=\frac{6+15}{5}=4.2 \mathrm{~cm}\)

Position vector of the centre of mass, \(\vec{r}=x \hat{i}+y \hat{j}=(3.2 \hat{i}+4.2 \hat{j}) \mathrm{cm}\)

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Question 6. A large number of particles are placed around the origin, each at a distance R from the origin. The distance of the centre of mass of the system from the origin is

1. =R
2. ≤R
3. >R
4. ≥F

Answer:

• If the particles are uniformly placed at a distance R from the origin, then the centre of mass will be situated at the origin.
• Again, when all the particles are placed at a distance R from the origin at one side of it, then the centre of mass will be situated at a distance of R from the origin.
• For any other kind of mass distribution, the centre of mass will be situated between the origin and R.

The option 2 is correct.

Question 7. Two bodies of masses m₁ and m₂ are separated by a distance R. The distance of the centre of mass of the bodies from the mass m₁ is

1. \(\frac{m_2 R}{m_1+m_2}\)
2. \(\frac{m_1 R}{m_1+m_2}\)
3. \(\frac{m_1 m_2}{m_1+m_2} R\)
4. \(\frac{m_1+m_2}{m_1} R\)

Answer:

The coordinates of masses m₁ and m₂ are (0,0) and (F, 0) respectively.

∴ The position of the centre of mass, \(x_{\mathrm{cm}}=\frac{m_1 \times 0+m_2 \times R}{m_1+m_2}=\frac{m_2 R}{m_1+m_2}\)

The option 1 is correct.

Question 8. Two particles A and B (both initially at rest) start moving towards each other under a mutual force of attraction. At the instant when the speed of A is v and the speed of B is 2 v, the speed of the centre of mass is

1. Zero
2. v
3. \(\frac{3 v}{2}\)
4. –\(\frac{3 v}{2}\)

Answer:

As no external force is applied on the two particles, the speed of the centre of mass of these two particles does not change. Therefore, the centre of mass is at rest.

The option 1 is correct.

Question 9. The distance of the centre of mass of a solid uniform cone from its vertex is z₀. If the radius of its base is R and its height is h, then z₀ is equal to

1. \(\frac{h^2}{4 R}\)
2. \(\frac{3 h}{4}\)
3. \(\frac{5 h}{8}\)
4. \(\frac{3 h^2}{8 R}\)

Answer:

The centre of mass of a uniform cone lies at a height of \(\frac{h}{4}\) from the base on the line joining the vertex and the centre of the base.

Hence, it is situated at a depth of \(\frac{3 h}{4}\) from the vertex.

The option 2 is correct.

Equilibrium Conditions in Statics

Question 10. A rod of weight W is supported by two parallel knife edges A and B and is in equilibrium in a horizontal position. The knives are at a distance d from each other. The centre of mass of the rod is at distance x from A. The normal reaction on A is

1. \(\frac{W x}{d}\)
2. \(\frac{W d}{x}\)
3. \(\frac{W(d-x)}{x}\)
4. \(\frac{W(d-x)}{d}\)

Answer:

Let the normal reactions on A and B be N_A and N_B.

At equilibrium, N_A+N_B=W

Again, when the resultant torques with respect to the centre of mass is zero, then \(N_A x+W .0-N_B(d-x)=0 \quad \text { or, } N_A \frac{x}{d-x}-N_B=0\)

So, \(N_A+N_A \frac{x}{d-x}=W\)

or, \(N_A\left(1+\frac{x}{d-x}\right)=W\)

or, \(N_A \frac{d}{d-x}=W\)

∴ \(N_A=\frac{W(d-x)}{d}\)

The option 4 is correct.

Question 11. Two spherical bodies of mass M and 5 M and radii R and 2R are released in free space with initial separation between their centres equal to 12 R. If they attract each other due to gravitational force only, then the distance covered by the smaller body before the collision is

1. 2.5R
2. 4.5R
3. 7.5R
4. 1.5R

Answer:

Their centre of mass will remain stationary due to the absence of any external force.

Hence, if the displacements of the smaller and the larger body are x₁ and x₂ respectively, \(M x_1=5 M x_2 \quad \text { or, } x_2=\frac{x_1}{5}\)

Since, at the time of the collision, the distance between their centres =R + 2R = 3R, so, the net path traversed by the spheres, x₁ + x₂ = 12R- 3R = 9 R

∴ \(x_1+\frac{x_1}{5}=9 R \quad \text { or, } \frac{6}{5} x_1=9 R \quad \text { or, } x_1=7.5 R\)

The option 3 is correct.

Question 12. Four bodies have been arranged at the comers of a rectangle shown. Find the centre of mass of the system.

Answer:

The masses at the corners O, A, B and C of the rectangle are respectively, m₁ = m, m₂ = 2m, m₃ = 3m and m₄ = 2m. the x-axis is chosen along OC and the y-axis along OA.

So the coordinates of the masses are respectively (0,0); (0, b), (a,h) and (a,0).

If (x, y) are the coordinated of the centre of mass, then

x = \(\frac{m_1 x_1+m_2 x_2+m_3 x_3+m_4 x_4}{m_1+m_2+m_3+m_4}\)

= \(\frac{m \cdot 0+2 m \cdot 0+3 m \cdot a+2 m \cdot a}{m+2 m+3 m+2 m}\)

= \(\frac{5 m a}{8 m}=\frac{5}{8} a\)

Similarly, y = \(\frac{m \cdot 0+2 m \cdot b+3 m \cdot b+2 m \cdot 0}{m+2 m+3 m+2 m}\)

= \(\frac{5 m b}{8 m}=\frac{5}{8} b\)

So the centre of mass is at \(\left(\frac{5}{8} a, \frac{5}{8} b\right)\).

Step-by-Step Solutions to Statics Short Questions

Question 13. Why are we not able to rotate a wheel by pulling or pushing along its radius?
Answer:

The centre of mass is at the centre of a wheel and every radial direction passes through this centre. As per definition, any force acting through the centre of mass can produce translation, but no rotation.

Question 14. Why do we use a wrench of a long arm to unscrew a nut tightly fitted to a bolt?
Answer:

Unscrewing a nut means a rotation about the axis of the nut. The application of a higher moment of force will facilitate this rotation. If the force applied remains the same, the moment of force is higher when the arm is longer. So we use a wrench of a long arm to unscrew a nut tightly fitted to a bolt.

Statics Very Short Answer Type Questions

Question 1. A rocket in space is being propelled by the ejection of a gas. What is the velocity of the centre of mass of the rocket-fuel system?
Answer: Zero

Question 2. A rocket in space is being propelled by the ejection of a gas. Will the velocity of the centre of mass of the rocket be zero?
Answer: No

Question 3. Can the centre of mass of a body lie outside the body?
Answer: Yes

Read And Learn More WBCHSE Class 11 Physics Very Short Question And Answers

Question 4. Does the centre of mass of a body necessarily lie inside the body?
Answer: May lie outside the body example, the CM of a circular ring is at the centre of the circle which is outside the ring.

Question 5. Give the location of the centre of mass of a solid cylinder having uniform mass density.
Answer: At the mid-point of the straight line joining the two centres of the circular end faces

Question 6. Two particles, due to their mutual attraction, approach each other. Give the value of the velocity of the centre of mass of the system at the moment when their relative velocity is v.
Answer: Zero

WBBSE Class 11 Statics Very Short Answer Questions

Question 7. A uniform rod of length 1 m is bent at right angles at its mid-point. What is the distance of the centre of mass from the mid-point of the rod?
Answer: 17.7 cm

Question 8. Write down the CGS unit of the moment of force.
Answer: dyn cm

Question 9. Write down the dimension of the moment of force.
Answer: ML²T^-2

Question 10. Write down the name of the anticlockwise moment of a force.
Answer: Positive moment

Question 11. Mention the clockwise moment of a force.
Answer: Neganve moment

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Question 12. The moment of a force about a point is ______ the area of the triangle formed by joining that point, the initial point of application of the force and the final point of application of the force.
Answer: Double

Question 13. ‘Moment of force is a vector’—true or false?
Answer: True

Question 14. Give the vector representation of the moment of force.
Answer: \(\vec{G}=\vec{r} \times \vec{F}\)

Question 15. It is _____ to open a door by pushing it near the outer edge rather than by pushing it near the hinge.
Answer: Easier

Question 16. Some forces act at a point such that their magnitudes and directions can be represented by the sides of a closed polygon taken in order. What is the resultant of the forces?
Answer: Zero

Question 17. Lengths of the arms of a balance are equal but the weights of the pans are different. When an object is put on the left pan, it weighs W₁ and on the right pan, it weighs W₂. What is the actual weight of the object?
Answer: \(\left[\frac{w_1+w_2}{2}\right]\)

Question 18. A man is at the top of a ladder and in another instance he stands on one of the lower steps. In which case does the foot of the ladder have a higher chance of slipping?
Answer: The man at the top of the ladder

Question 19. Can two unequal forces that lie in the same plane produce equilibrium?
Answer: No

Question 20. ‘A gas-filled balloon remains at rest at a certain height. The balloon is in stable equilibrium’—correct or incorrect?
Answer:  Correct

Question 21. ‘When a body revolves in a circular path at a constant speed, it is in an unstable equilibrium’—true or false?
Answer: True

Real-Life Examples of Forces in Statics

Question 22. At one end of a weightless rope of length 10 m, a car of mass M is tied. The other end of the rope is in the hands of a man of mass M. The arrangement stands on a horizontal frictionless surface. The man is at x = 0 and the car is at x = 10 m. As the man pulls the car with the rope, locate the point where they meet.
Answer: x = 5m

Question 23. Under the action of three forces F, F and √2F, a particle remains in equilibrium. Find the angle between the first two forces.
Answer: 90

Question 24. What is the position of the centre of gravity of a uniform rectangular lamina?
Answer: At the point of intersection of the diagonals

Question 25. For a uniform body, if the shape is kept the same but its size is changed uniformly, will the position of the centre of gravity change?
Answer: No

Question 26. When a uniform rod is heated at one of its ends, will the position of the centre of gravity be shifted?
Answer: Yes

Question 27. Does the centre of gravity change it’s position if the shape changes of a body without a change of its weight?
Answer: Yes

Simple Problems in Statics with Answers

Question 28. Is the centre of gravity of a body to be many?
Answer: No

Question 29. Can the centre of gravity of a long glass tube lie at the midpoint of its length?
Answer: Yes

Question 30. A rod weighing W is placed horizontally over knife edges A and B. The Distance between the knife edges is d and the distance of the centre of gravity of the rod from
A is x. What are the reactions at A and B?
Answer: \(\left[W\left(1-\frac{x}{d}\right), W \frac{x}{d}\right]\)

Question 31. ‘Centre of mass and centre of gravity of a body are not the same, in general’—correct or incorrect?
Answer: Correct

WBCHSE Class 11 Physics Statics MCQs

Statics Multiple Choice Questions And Answers

Question 1. Two particles, initially at rest, due to their mutual attraction, move towards each other. When their relative velocity becomes v, the velocity of the centre of mass of the system becomes

1. Zero
2. v
3. 1.5 v
4. 3v

Answer: 1. Zero

Question 2. Position vectors of two equal masses with respect to the origin are \(\vec{a}\) and \(\vec{t}\) respectively. The position vector of the centre of mass of these masses is

1. \(\vec{a}+\vec{b}\)
2. \(\frac{\vec{a}+\vec{b}}{2}\)
3. \(\vec{a} \times \vec{b}\)
4. \(\frac{\vec{b}-\vec{a}}{2}\)

Answer: 2. \(\frac{\vec{a}+\vec{b}}{2}\)

Read And Learn More WBCHSE Class 11 Physics MCQs

Question 3. Position vector of the centre of mass of a system of N particles of total mass M is

1. \(\frac{\sum M \vec{r}_p}{M}\)
2. \(\frac{\sum_{p=1}^N \vec{r}_p}{r}\)
3. \(\frac{\sum_{p=1}^N m_p \vec{p}_p}{\sum_{p=1}^N m_p}\)
4. \(\frac{\sum_{p=1}^N m_p \vec{r}_p}{\sum_{p=1}^N r_p}\)

Answer: 3. \(\frac{\sum_{p=1}^N m_p \vec{p}_p}{\sum_{p=1}^N m_p}\)

WBCHSE Class 11 Physics Statics MCQs

Question 4. A stick is thrown in air. It lands a little away from the thrower. The locus of the path of the centre of mass of the stick will be a parabola

1. In all cases
2. Only when the stick is uniform
3. If the stick had only linear motion and no rotational motion
4. If the stick is of such shape that its centre of mass is on the stick itself and not outside

Answer: 1. In all cases

Question 5. A man is hanging from a rope attached to a balloon containing hot air. The system is at rest in the air. If the man climbs up the rope to the balloon, then the centre of mass of the system

1. Remains at rest
2. Moves upwards
3. Moves downwards
4. First moves up and then moves back to the initial position

Answer: 1. Remains at rest

WBBSE Class 11 Statics MCQs

Question 6. Four masses m, m, 2m and 2m  are kept at four vertices of a square of side a. The coordinates of the centre of mass of the system are

1. \(\left(\frac{a}{2}, 2 a\right)\)
2. \(\left(\frac{a}{2}, a\right)\)
3. \(\left(\frac{a}{2}, \frac{2 a}{3}\right)\)
4. \(\left(a, \frac{a}{3}\right)\)

Answer: 3. \(\left(\frac{a}{2}, \frac{2 a}{3}\right)\)

Question 7. All the particles of a body are situated at a distance R from the origin. The distance of the centre of mass of the body from the origin is

1. =R
2. ≤R
3. >R
4. ≥R

Answer: 2. ≤R

West Bengal Class 11 Physics Multiple Choice Questions 

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Question 8. Consider a system of two identical particles. One of the particles is at rest and the other has an acceleration \(\vec{a}\). The centre of mass has an acceleration

1. Zero
2. \(\frac{1}{2} \vec{a}\)
3. \(\vec{a}\)
4. \(2 \vec{a}\)

Answer: 2. \(\frac{1}{2} \vec{a}\)

Question 9. A body falling vertically downwards under gravity breaks in two parts of unequal masses. The centre of mass of the two parts taken together shifts horizontally towards

1. Heavier piece
2. Lighter piece
3. Does not shift horizontally
4. Depends on the vertical velocity at the time of breaking

Answer: 3. Does not shift horizontally

Question 10. A uniform metal disc of radius R is taken and out of it, a disc of diameter R is cut off from the end. The centre of mass of the remaining part will be

1. \(\frac{R}{4}\) from the centre
2. \(\frac{R}{3}\) from the centre
3. \(\frac{R}{5}\) from the centre
4. \(\frac{R}{6}\) from the centre

Answer: 4. \(\frac{R}{6}\) from the centre

Question 11. The centre of mass of a solid cone along the line from the centre of the base to the vertex is at

1. One-fourth of the height
2. One-third of the height
3. One-fifth of the height
4. None of these

Answer: 1. One-fourth of the height

Equilibrium and Forces MCQs

Question 12. A pulley fixed to the ceiling carries a string with blocks of masses m and 3 m attached to its ends. The masses of string and pulley are negligible. When the system is released, the acceleration of the centre of mass will be

1. Zero
2. \(\frac{g}{4}\)
3. \(\frac{g}{2}\)
4. –\(\frac{g}{2}\)

Answer: 3. \(\frac{g}{2}\)

Question 13. Two particles each of mass 1 g are placed at distances of 1m and 3m respectively from the origin along x -the axis. The centre of mass of the system from the origin is

1. 1 m
2. 2 m
3. 2.5 m
4. 3.5 m

Answer: 2. 2 m

Question 14. A uniform metre scale of mass m is suspended horizontally by two vertical ropes fitted at its two ends. An object of mass 2 m is placed at the 75 cm mark on the scale. The ratio of tension in the two strings is

1. 1:2
2. 1:3
3. 2:3
4. 3:4

Answer: 1. 1:2

WBCHSE Physics Chapter-wise MCQs 

Question 15. The height of a solid cone is h and the radius of its circular base is r. The cone has been placed on its base on an inclined plane. The maximum angle of inclination for which the cone will not topple over is

1. \(\cos ^{-1} \frac{2 r}{h}\)
2. \(\tan ^{-1} \frac{4 r}{h}\)
3. \(\tan ^{-1} \frac{3 r}{h}\)
4. \(\sin ^{-1} \frac{4 r}{h}\)

Answer: 3. \(\tan ^{-1} \frac{3 r}{h}\)

West Bengal Class 11 Physics Multiple Choice Questions 

In this type of question, more than one option is correct.

Question 16. Four forces act on a point object. The object will be in equilibrium, if

1. All of them are in the same plane
2. They are opposite to each other in pairs
3. The sum of x, y and z components of forces is zero separately
4. They form a closed figure of 4 sides when added as per polygon law.

Answer: 

2. They are opposite to each other in pairs

3. The sum of x, y and z components of forces is zero separately

4. They form a closed figure of 4 sides when added as per polygon law.

Class 11 Physics Statics Questions WBCHSE 

Mathematical Problems in Statics MCQs

Question 17. The surfaces shown are smooth. The system is released from rest, x-and y-components of acceleration of the centre of mass are

1. \(\left(a_{\mathrm{cm}}\right)_x=\frac{m_1 m_2 g}{m_1+m_2}\)
2. \(\left(a_{c m}\right)_x=\frac{m_1 m_2 g}{\left(m_1+m_2\right)^2}\)
3. \(\left(a_{\mathrm{cm}}\right)_y=\left(\frac{m_2}{m_1+m_2}\right)^2 g\)
4. \(\left(a_{\mathrm{cm}}\right)_y=\left(\frac{m_2}{m_1+m_2}\right) g\)

Answer:

2. \(\left(a_{c m}\right)_x=\frac{m_1 m_2 g}{\left(m_1+m_2\right)^2}\)

3. \(\left(a_{\mathrm{cm}}\right)_y=\left(\frac{m_2}{m_1+m_2}\right)^2 g\)

Class 11 Physics Statics Questions WBCHSE 

Question 18. A block of mass m is placed at rest on a smooth wedge of mass M placed at rest on a smooth horizontal surface. As the system is released

1. The centre of mass of the system remains stationary
2. The centre of mass of the system has an acceleration g vertically downward
3. The momentum of the system is conserved along the horizontal direction
4. Acceleration of centre of mass is vertically downward (a < g)

Answer:

3. The momentum of the system is conserved along the horizontal direction

4. Acceleration of centre of mass is vertically downward (a < g)

Statics Long Answer Type Questions

Question 1. Describe the moment of a force about a point as a vector quantity.
Answer:

O is the point of rotation, P is the point of application of a force \(\vec{F}\), \(\vec{r}=\overrightarrow{O P}\), position vector of the point \(\vec{P}\), with respect to the point O.

Then, the moment of the force \(\vec{F}\) about the point O, is \(\vec{G}\) = \(\vec{r}\) x \(\vec{F}\)

According to the properties of the cross product, the direction of the vector \(\vec{G}\) is along the direction of advance of a righthanded screw, when it is rotated from \(\vec{r}\) towards \(\vec{F}\). The direction of \(\vec{G}\), as shown in the figure, is perpendicular to the plane of rotation, i.e., always along the axis of rotation.

Read and Learn More Class 11 Physics Long Answer Questions

Question 2. What is the significance of a moment o fa force about a point?
Answer:

The moment of a force on a body about a point is an external influence that can rotate the body about that point. Higher torque would produce more rotation if the body remains the same.

Question 3. It is easier to open or close a door by pushing it at the edges than by pushing it closer to the hinges. Explain.
Answer:

A door is fitted to the door frame using hinges. Opening or closing of a door means a rotation about these hinges. The application of a higher moment of force will facilitate this rotation. If the force applied remains the same, the moment of force is higher when the arm is longer. So the door is usually pushed farthest from the hinges, i.e., at the edges.

WBBSE Class 11 Statics Long Answer Questions

Question 4. State whether three forces of magnitude 1 dyn, 2 dyn and 3 dyn acting simultaneously on a body can keep it In equilibrium.
Answer:

If 1 dyn and 2 dyn forces act on a body in parallel, then the resultant force on the body becomes (1+2)dyn or 3 dyn. Now if the given 3 dyn force acts on the body along the same line as the resultant force but in the opposite direction to, then the resultant force on the body becomes zero. Hence the body will be in equilibrium.

Question 5. A bomb Is thrown vertically upwards. At its topmost position, it explodes into a number of fragments. What would be the locus of the centre of gravity of the bomb?
Answer:

Forces developed due to an explosion are internal forces and therefore the motion of the centre of gravity does not get affected. Hence, the centre of gravity would come down vertically like a freely falling body.

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Question 6. A ping pong ball is floating at the vertex of a vertical fountain. State whether the equilibrium of the ball is vertically stable, unstable or neutral.
Answer:

The equilibrium of the ball, in the vertical direction, is stable. At the top of the fountainhead, the thrust of water acts vertically upwards and the weight of the ball acts downwards. The two forces acting on the ball, being equal and opposite, keep the ball in equilibrium at A.

• If the ball is lifted upward a little to B, the thrust is less than the weight and so the ball returns to the equilibrium position A.
• On the other hand, if the ball is taken down to position C, the thrust being greater than the weight of the ball, pushes it up to the equilibrium position A . Hence, the pingpong ball is in a stable equilibrium.

Step-by-Step Solutions to Statics Problems

Question 7. A rectangular parallelepiped of mass m has three sides of length l, 2l and 3l respectively. State with reason, in which position the block should be kept at rest on a horizontal floor so as to attain the highest stability of equilibrium.
Answer:

The longer sides 2l and 3l should form the horizontal base of the parallelepiped. In this position, maximum stability of equilibrium can be achieved. Centre of gravity is at its lowest possible position, i.e., at a height of \(\frac{l}{2}\) from the base and the base area is maximum.

Question 8. Two particles, initially at rest, are approaching each other due to their mutual force of attraction. What is the velocity of the centre of mass of the system when the particles have a relative velocity v?
Answer:

For a two-particle system, the mutual attraction is an internal force and has no effect on the motion of the centre of mass. Hence, the velocity of the centre of mass is zero.

Question 9. A piece of stone thrown vertically upwards comes to rest momentarily at its maximum height. Is the stone in equilibrium at this position?
Answer:

The stone, though momentarily at rest at the maximum height, is not in equilibrium there. It has a downward acceleration, due to the earth’s gravitational force.

Question 10. Should there be any change in the position of the centre of gravity of a hollow sphere when it is

1. Half filled,
2. Completely filled with water?

Answer:

The centre of gravity of an empty hollow sphere is at its centre.

1. When it is half filled with water, the centre of gravity is lowered.
2. When it is completely filled with water the centre of gravity again shifts to the centre.

Question 11. It is easier to stand on two legs than to stand on a single leg. Why?
Answer:

A body remains in stable equilibrium when the vertical line through its centre of gravity remains well within its base. Equilibrium is disturbed when this vertical line falls outside the base. It is difficult to maintain the vertical line passing through one foot only.

This reduces the stability while standing on one leg. Standing on two legs increases the base area through which the vertical line is more likely to pass and hence, a higher stability of equilibrium is achieved.

Question 12. Passengers on a boat should not be allowed to stand up on the boat while crossing a river. Why?
Answer:

The centre of gravity of a boat with its passengers should be as low as possible, so that it may always be in a stable equilibrium. But if the passengers stand up on the boat, the centre of gravity of the system also moves up. The system then attains an unstable condition. Then, even for a very small tilting, the boat may capsize. So, the passengers on a boat are not allowed to stand up while crossing a river.

Question 13. Two arms of a common balance are unequal. How can the balance be used to find the correct mass of an object?
Answer:

Suppose the correct mass of the object is m and the lengths of the left and right arms of the balance are x and y respectively. At first, the object is kept on the left pan and the counterpoising weight m₁ on the right pan.

Hence, \(m x=m_1 y \text { or, } \frac{m}{m_1}=\frac{y}{x}\)…(1)

Next, the object is kept on the right pan and the counterpoising weight m₂ on the left pan.

Hence, \(m_2 x=m y \text { or, } \frac{m}{m_2}=\frac{y}{x}\)…(2)

From equations (1) and (2) we get, \(\frac{m}{m_1}=\frac{m_2}{m} \text { or, } m^2=m_1 m_2 \text { or, } m=\sqrt{m_1 m_2}\)

Hence, the true mass m of the object can be calculated.

Real-Life Examples of Static Equilibrium

Question 14. A businessman uses a faulty balance of unequal arms of lengths x and y respectively. He sells W kg of tea to each customer. While selling to the 1st and the 2nd customers, he keeps the counterpoising weight on the left pan and on the right pan respectively. Does he gain or lose?
Answer:

Suppose the lengths of the left and the right arms of the balance are x and y respectively. For the first customer, he placed the counterpoising weight in the left pan. In this case, the tea is placed in the other pan. Suppose W₁ kg of tea balances W kg weight.

Hence, W x x = W₁ x y

or, \(W_1=\frac{W x}{y}\)

For the second customer, he placed the W kg counterpoising weight in the right pan. Suppose, in this case, W₂ kg tea is placed in the other pan.

∴ W₂ x x = W x y

or, \(W_2=\frac{W y}{x}\)

The customers pay for 2 Wkg but get (W₁ + W₂) kg of tea.

Now \(W_1+W_2-2 W=\frac{W x}{y}+\frac{W y}{x}-2 W\)

= \(W\left(\frac{x}{y}+\frac{y}{x}-2\right)=\frac{W}{x y}(x-y)^2\) which is a positive quantity.

∴ (W₁ + W₂) > 2 W. Hence, there is a loss for the businessman.

Question 15. A businessman uses a faulty balance of unequal arms. He buys some old papers from a person and for this, he uses a \(\frac{1}{2}\) kg counterpoising weight. He then readily agrees to weigh the papers alternately by changing the pans of the balance during successive weighings. Show that he gains in every 1 kg of purchase. [Upthrust due to air is neglected.)
Answer:

Let the length of the left and the right arms of the faulty balance be x and y (x>y). Suppose Wy kg of paper on the right pan balances the \(\frac{1}{2}\) kg counterpoising weight on the left pan.

∴ \(\frac{1}{2} \cdot x=W_1 \cdot y \text { or, } W_1=\frac{1}{2} \cdot \frac{x}{y}\)

Suppose W₂ kg of paper is required on the left pan to balance \(\frac{1}{2}\) kg counterpoising weight on the right pan.

∴ \(W_2 x=\frac{1}{2} y \text { or, } W_2=\frac{1}{2} \cdot \frac{y}{x}\)

Paper received by the businessman, \(W_1+W_2=\frac{1}{2}\left(\frac{x}{y}+\frac{y}{x}\right)=\frac{1}{2} \cdot\left(\frac{x^2+y^2}{x y}\right)\)

= \(\frac{1}{2} \cdot\left[2+\frac{(x-y)^2}{x y}\right]=1+\frac{1}{2} \frac{(x-y)^2}{x y}\)

Hence the businessman gains in every kilogram of purchase.

Mathematical Problems in Statics

Question 16. A person may stand on the lower steps or at the top of a ladder. In which of these cases does the possibility of sliding of the ladder become maximum?
Answer:

The possibility of sliding becomes maximum when the person stands at the top of the ladder because the combined centre of gravity of the system is very high and the system becomes unstable.

Question 17. Why is it easier to walk with two equal loads in both hands than with the entire load in one hand?
Answer:

While carrying a load in one hand the centre of gravity comes down to some extent, but it also shifts towards the side of the load. As a result, the line of action of the weight tends to shift away from the base. In this state, to maintain equilibrium, the body has to be tilted towards the other side.

It is quite difficult to walk in such a manner. While carrying two equal loads in both hands, the centre of gravity of the system is not displaced sideways, but it only comes’ clown. Hence, the body with two loads becomes more stable and? walking becomes easier.

Question 18. Babies fall down frequently while trying to walk. Why?
Answer:

The centre of gravity of a baby, while crawling, remains quite low and the line of action of its body weight always passes through the base. As the baby tries to stand, the centre of gravity of the body goes up and so the body becomes unstable.

To keep the body stable, the baby should keep its body erect, so that the line of action of the body weight passes through the base. Hence, due to a lack of sufficient practice, babies fall down frequently.

Statics – Equilibrium Of Body

The Conditions Of Equilibrium

WBBSE Class 11 Conditions of Equilibrium Notes

A body is said to be in equilibrium when

1. The linear acceleration of the body is zero and
2. The angular acceleration of any axis is zero.

If the linear and angular accelerations are zero, the body is not necessarily at rest. A body having uniform velocity or uniform angular velocity also has zero linear acceleration and zero angular acceleration, respectively.

Equilibrium Of Body Definition: A body is in equilibrium if it is either at rest or in motion with uniform linear or angular velocity, or with a combination of both uniform linear and angular velocities.

Read and Learn More: Class 11 Physics Notes

• From Newton’s second law of motion, we know that the acceleration of a body is related to the applied external force. Whether any external force is acting on the body cannot be determined from the observation of the equilibrium of the body.
• There may or may not be any external force acting on the body when the body is in equilibrium.
• No object can remain in equilibrium under the action of a single force but a number of forces together may keep the object in equilibrium. Again, only external forces can change the state of equilibrium of a body.
• Internal forces like interatomic forces cannot shift a body from its equilibrium position as internal forces occur in pairs and balance each other. Hence, internal forces play no part in determining the equilibrium state of a body.

From the above discussions, it is evident that there are two conditions under which a body can be kept in equilibrium when forces are applied at different points on it. We consider only the action of coplanar forces in our discussion.

Equilibrium Of Body 1st Condition: For equilibrium, the vector sum (i.e., resultant) of all coplanar forces acting on a body must be zero.

First Condition of Equilibrium Explained

Mathematically, we can express this condition as \(\vec{R}=\vec{F}_1+\vec{F}_2+\vec{F}_3+\cdots=\sum_i \vec{F}_i=\overrightarrow{0}\)…(1)

Resolving the forces into mutually perpendicular components along suitably chosen x and y-axes, we get,

⇒ \(R_x =F_{1 x}+F_{2 x}+F_{3 x}+\cdots=\sum_i F_{i x}=0\)….(2)

and \(R_y=F_{1 y}+F_{2 y}+F_{3 y}+\cdots=\sum_i F_{i y}=0\)….(3)

Hence, The Condition Can Be Expressed In Another Form: For coplanar forces acting on a body, the sum of components along a given direction and along the direction perpendicular to it should separately be zero.

When this condition is fulfilled, the body will not have a linear acceleration.

Equilibrium Of Body 1st Condition Vector Form: Equation (1) is the vector form of the 1st condition.

Alternative Description Of The 1st Condition: Applying the law of polygon of vectors it can be stated that if the magnitude and direction of forces (F₁, F₂, F₃, etc.), acting on a body, can represent the sides of a closed polygon taken in order, then the body remains in equilibrium under the action of the forces.

Alternative Description Of The 2nd Condition: The algebraic sum of moments of the coplanar forces acting on a body, taken about any point, should be zero.

• The mathematical expression for the above condition may be written by taking moments about any algebraic point O. The sum of the moments of the forces is, τ = F₁r₁ + F₂r₂ + F₃r₃+…. = 0
• If the algebraic sum of the moments of the coplanar forces is equal to zero about a point on that plane, the sum will also be zero about all other points on that plane. For example, the sum will be zero about the point O’ also.
• When this condition is fulfilled, the angular acceleration of the body will be zero. A body will remain in equilibrium when both conditions are fulfilled together.
• Shows an incomplete polygon formed by the vectors. Clearly, in this case, the first condition has not been fulfilled.

Alternative Description Vector form: Vector form of the 2nd condition,

∴ \(\vec{\tau}=\overrightarrow{r_1} \times \vec{F}_1+\vec{r}_2 \times \vec{F}_2+\vec{r}_3 \times \vec{F}_3+\cdots\)

= \(\sum_i \overrightarrow{r_i} \times \vec{F}_i=\overrightarrow{0}\)….(4)

Equilibrium Under Two Forces: When two equal and oppositely directed forces act along the same straight line, the resultant force becomes zero. The condition for equilibrium is that both forces acting on the body should lie on the same straight line, and be equal in magnitude but directed opposite to each other.

Second Condition of Equilibrium Overview

Equilibrium Under Two Forces Example: A pendulum is suspended from a rigid support using a string. When the string is vertical, the pendulum remains in equilibrium. In this case, the weight (W) of the pendulum and the tension (T) in the string are equal and opposite.

But if the bob is displaced from the equilibrium position, the weight W and the tension T are neither equal nor opposite. Hence, the system cannot be in equilibrium.

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Equilibrium Under Three Parallel Forces: Conditions for the equilibrium of a body under the action of three parallel forces are

1. They must be coplanar,
2. The resultant of any two forces must be equal and opposite to the third force and
3. The resultant of two forces and the third force should have the same line of action.

A body in equilibrium under the action of three coplanar forces \(\vec{F}_1\), \(\vec{F}_2\), and \(\vec{F}_3\) is shown. The line of action of the resultant \(\vec{R}\), of \(\vec{F}_1\) and \(\vec{F}_2\), is the same as that of \(\vec{F}_3\).

∴ \(\vec{R}\) = –\(\vec{F}_3\)

The conditions for equilibrium under three non-parallel forces can be expressed in any of the following four ways:

1. Law Of Resultant: The three forces should be coplanar and the resultant of any two of the forces must be equal and opposite to the third one and the lines of action of all the three forces should pass through the same point

2. Law Of Triangle Of Forces: For three non-parallel, coplanar forces acting on a body, if the lines of action of the forces pass through the same point and the magnitudes and directions of these forces can be represented by the three sides of a triangle taken in order, then the body remains in equilibrium.

Suppose the forces P, Q, R act on a body at a single point and the magnitude and direction of the forces \(\vec{P}\) and \(\vec{Q}\) are represented by the two sides AB and BC of a triangle ABC,

i.e., \(\vec{A B}\) = \(\vec{P}\) and \(\vec{B C}\) = \(\vec{Q}\)

From the triangle law of vector addition \(\overrightarrow{A B}+\overrightarrow{B C}=\overrightarrow{A C}\)

or, \(\vec{P}+\vec{Q}=\overrightarrow{A C} \text { or, } \vec{P}+\vec{Q}+\vec{R}=\overrightarrow{A C}+\vec{R}\)

The first condition of equilibrium requires that the resultant, \(\vec{P}\) + \(\vec{Q}\) + \(\vec{R}\) = \(\vec{0}\)

∴ \(\overrightarrow{A C}+\vec{R}=\overrightarrow{0} \text { or, } \vec{R}=-\overrightarrow{A C}=\overrightarrow{C A}\)

Hence, the magnitude and direction of the third force \(\vec{R}\), is represented by the third side \(\overrightarrow{C A}\), of triangle ABC.

3. Law Of Orthogonal Components: A body under the action of three non-parallel, coplanar forces remains in equilibrium if

1. The lines of action of the forces pass through the same point and
2. The algebraic sum of the components of the forces taken along two mutually perpendicular axes is zero separately.

Three coplanar forces \(\vec{P}\), \(\vec{Q}\), \(\vec{R}\) act at the point O. x and y-axes are chosen along two mutually perpendicular lines OX and OY. Suppose P, Q, and R, the three forces, make angles α, β and γ respectively with the x-axis.

Now, the components of the three forces along the x and y-axes are,

⇒ \(P_x=P \cos \alpha, Q_x=Q \cos \beta, R_x=R \cos \gamma\)

⇒ \(P_y=P \sin \alpha, Q_y=Q \sin \beta, R_y=R \sin \gamma\)

Hence, the sum of the components along the x-axis, \(S_x=P_x+Q_x+R_x=P \cos \alpha+Q \cos \beta+R \cos \gamma\)

and some of the components along y-axis \(S_y=P_y+Q_y+R_y=P \sin \alpha+\mathrm{Q} \sin \beta+\mathrm{R} \sin \gamma\)

According to the condition for equilibrium of an object under the action of three forces, the resultant S of S_x and S_y should be zero,

i.e„ S = \(\sqrt{s_x^2+s_y^2}=0\)

This is true only when S_x = 0 and S_y = 0.

∴ S_x = 0 and S_y = 0 is the required condition.

Static vs Dynamic Equilibrium

4. Larni’s Theorem: Anybody, under the action of three coplanar forces acting at a point, remains in equilibrium when each force is directly proportional to the sine of the angle between the other two forces.

OX, OY, OZ are the lines of action of three forces \(\vec{P}\), \(\vec{Q}\), \(\vec{R}\) acting at a point O of the body.

Let \(\overrightarrow{O A}=\vec{P}\) and \(\overrightarrow{O B}=\vec{Q}\). So, the diagonal \(\overrightarrow{O D}\) of the parallelogram OADB represents the resultant of forces \(\vec{P}\) and \(\vec{Q}\). In other words, \(\vec{P}\) + \(\vec{Q}\) = \(\overrightarrow{O D}\).

∴ \(\vec{P}\) + \(\vec{Q}\) = \(\overrightarrow{O D}\) + \(\vec{R}\)

When the body is in equilibrium, \(\vec{P}+\vec{Q}+\vec{R}=\overrightarrow{0} \text { or, } \overrightarrow{O D}+\vec{R}=\overrightarrow{0} \text { or, } \vec{R}=-\overrightarrow{O D}=\overrightarrow{D O}\)

Clearly, \(\overrightarrow{O A}, \overrightarrow{A D}, \overrightarrow{D O}\) form the arms of the triangle OAD, which, taken in order, represent the three forces \(\vec{P}, \vec{Q}, \vec{R}\)

From the properties of triangle, \(\frac{P}{\sin \angle O D A}=\frac{Q}{\sin \angle A O D}=\frac{R}{\sin \angle O A D}\)

Here, sin∠ODA = sin∠BOD =sin( 180° – ∠YOZ) = sin ∠YOZ = sin(Q, R), where (Q, R) is the angle between Q and R.

Similarly, sin ∠AOD = sin(R, P) and sin ∠OAD = sin (P, Q)

∴ \(\frac{P}{\sin (Q, R)}=\frac{Q}{\sin (R, P)}=\frac{R}{\sin (P, Q)}\)

This is the mathematical representation of Larni’s theorem for the equilibrium of a body under the action of three coplanar forces.

Equilibrant: The single force that balances the resultant of all other forces, is called the equilibrant. F₃ is the equilibrant of F₁ and F₂. It can also be said that F₁ is the equilibrant of F₂ and F₃, or F₂ is the equilibrant of F₁ and F₃.

Statics – Equilibrium Numerical Examples

Applications of Equilibrium in Physics

Example 1. Three forces F, F and √2F acting at a point are in equilibrium. Find the angles between the forces.
Solution:

Given

Three forces F, F and √2F acting at a point are in equilibrium.

Show the three forces. Let the angles between the forces be α, β and γ, as shown.

∴ α + β + γ = 360°

or, β + γ = 360°- α

∴ sin(α) = -sinα

By Lami’s theorem \(\frac{\sqrt{2} F}{\sin \alpha}=\frac{F}{\sin \beta}=\frac{F}{\sin \gamma}\)

∴ β = γ

Also, \(\frac{F}{\sin \beta}=\frac{\sqrt{2 F}}{\sin \alpha}\) or, \(\sin \alpha=\sqrt{2} \sin \beta\).

Now, \(\sin \alpha=-\sin (\beta+\gamma)=-\sin 2 \beta=-2 \sin \beta \cos \beta\)

Hence, \(-2 \sin \beta \cos \beta=\sqrt{2} \sin \beta\)

or, \(\cos \beta=-\frac{1}{\sqrt{2}}=\cos 135^{\circ}\)

∴ \(\beta=135^{\circ}=\gamma\)

and \(\alpha=360^{\circ}-\left(135^{\circ}+135^{\circ}\right)=90^{\circ}\).

Example 2. One end of a light string is fixed to the ceiling and the other end is tied to the walk A body of mass 500 g is suspended from that string in such a way that the part of the string towards the wall remains horizontal and the portion towards the ceiling, makes an angle of 30° with the ceiling. Find the tension on the two parts of the string.
Solution:

Given

One end of a light string is fixed to the ceiling and the other end is tied to the walk A body of mass 500 g is suspended from that string in such a way that the part of the string towards the wall remains horizontal and the portion towards the ceiling, makes an angle of 30° with the ceiling.

Let the positive y-axis be vertically upwards and the x-axis be horizontal. Let T₁ = tension in the horizontal part of the string, T₂ = tension in the other part of the string, W = weight of the suspended body = 500 x 981 dyn.

Sum of the components of the forces along the y-axis = T₂ sin30°- W

Sum of the components of the forces along the x-axis = T₁ – T₂cos30°

As the system is in equilibrium T₂ sin30°- W= 0….(1)

T₁ – 2 cos30° = 0…(2)

From (1), \(\frac{1}{2} T_2=W\)

or, \(T_2=2 W\)

= \(2 \times 500 \times 981 \mathrm{dyn}=2 \times \frac{500 \times 981}{10^5} \mathrm{~N}=9.81 \mathrm{~N}\)

From (2), \(T_1=T_2 \cos 30^{\circ}=9.81 \times \cos 30^{\circ}=8.5 \mathrm{~N} \text {. }\)

Short Answer Questions on Equilibrium Conditions

Example 3. A weight W is suspended by using two strings. One of the strings makes an angle of 30° with the vertical. What should be the direction of the other string; so that the tension in it becomes minimum? Find the tension in each string at this position.
Solution:

Given

A weight W is suspended by using two strings. One of the strings makes an angle of 30° with the vertical.

Let the angle between the two strings be α.

Given that the angle between the weight W and the first string = 180° -30° = 150°

Hence, the angle between the weight and the second string

=360° – 150°- α

= 210°- α

For equilibrium, using Lami’s theorem, \(\frac{T_2}{\sin 150^{\circ}}=\frac{W}{\sin \alpha} \text { or, } \frac{T_2}{0.5}=\frac{W}{\sin \alpha} \text { or, } T_2=\frac{W}{2 \sin \alpha}\)

For T₂ to be minimum, sina should be maximum i.e., sinα = 1 or α = 90°. Hence, the tension in the second string will be minimal when it is at right angles to the first string.

At this setting, \(T_2=\frac{W}{2}\)

Also, \(\frac{W}{\sin 90^{\circ}}=\frac{T_1}{\sin \left(210-90^{\circ}\right)}=\frac{T_1}{\sin 120^{\circ}}\)

or, W = \(\frac{T_1}{\sin 60^{\circ}}\)

∴ \(T_1=W \sin 60^{\circ}=W \times \frac{\sqrt{3}}{2}\)

∴ \(T_1=\frac{\sqrt{3} W}{2} \text { and } T_2=\frac{W}{2} .\)

Example 4. From the markings, 20 cm, 40 cm, 60 cm, 80 cm and 100 cm on a light metre ruler, masses of 1 g, 2 g, 3 g, 4 g, and 5 g respectively are suspended. At which mark should the ruler be pivoted so that it remains horizontal?
Solution:

Given

From the markings, 20 cm, 40 cm, 60 cm, 80 cm and 100 cm on a light metre ruler, masses of 1 g, 2 g, 3 g, 4 g, and 5 g respectively are suspended.

The gravitational forces on 1g, 2g,3g,4g, and 5g masses are downward parallel forces. Suppose their resultant acts at the x cm mark on the scale. Hence,

x = \(\frac{\sum_{i=1}^5 F_i x_i}{\sum_{i=1}^5 F_i}\)

= \(\frac{1 \times 20+2 \times 40+3 \times 60+4 \times 80+5 \times 100}{1+2+3+4+5} \)

= \(\frac{1100}{15}=73.3\)

When the downward resultant and the upward normal force of the support, act at the same point, the metre ruler remains in equilibrium. Hence, the ruler should be pivoted at the 73.3 cm mark.

Example 5. Two parallel forces P and Q (P> Q) are acting at two points, A and B, of a body. If the forces Interchange their positions, by how much will the point of action of the resultant of the two forces shift along the line AB?
Solution:

Given

Two parallel forces P and Q (P> Q) are acting at two points, A and B, of a body. If the forces Interchange their positions,

Suppose C is the point of action of the resultant in the first case.

∴ P x AC = Q x BC…(1)

If the new position of the point of action of the resultant is then Q x AC’ = P x BC’ …(2)

From (1) we get, P x AC = Q(AB – AC) or,(P + Q) x AC = Q x AB

∴ Ac = \(\frac{Q}{P+Q} \times A B\)…(3)

Similarly from (2) we get, \(A C^{\prime}=\frac{P}{P+Q} \times A B\)…(4)

∴ Subtracting equation (3) from equation (4) we get, \(A C^{\prime}-A C=\frac{A B}{(P+Q)}(P-Q) \text { or, } C C^{\prime}=\frac{P-Q}{P+Q} \times A B \text {. }\)

Example 6. The result of two parallel forces P and Q is P.If the force P shifts by a distance x, parallel to Itself, prove that R will shift by \(\frac{P x}{P+Q}\).
Solution:

Given

The result of two parallel forces P and Q is P.If the force P shifts by a distance x, parallel to Itself,

Suppose the parallel forces P and Q act at points A and B and the resultant R acts at C.

∴ P x AC = Q x BC

or, \(\frac{P}{Q}=\frac{B C}{A C} \text { or, } \frac{P+Q}{Q}=\frac{B C+A C}{A C}=\frac{A B}{A C}\)

∴ AC = \(A B \times \frac{Q}{P+Q}\)

Now if the force P acts at point D such that AD = x, then the resultant also shifts and acts at E instead of C.

Hence, Px DE = Qx BE

or, \(\frac{P}{Q}=\frac{B E}{D E} or, \frac{P+Q}{Q}=\frac{B E+D E}{D E}=\frac{B D}{D E}=\frac{A B-x}{A E-x}\)

∴ \((A E-x)(P+Q)=(A B-x) Q\)

AE = \(\frac{Q \cdot A B+P \cdot x}{P+Q}\)

∴ Displacement of the line of action of the resultant, CE = \(A E-A C=\frac{Q \cdot A B+P \cdot x}{P+Q}-\frac{Q \cdot A B}{P+Q}=\frac{P x}{P+Q}\)

Example 7. The mass of a uniform rod of length 10 m is 10 kg. The rod is placed over knife edges A and B. One end of the rod is resting on the knife edge A and the other end is 2 m outside the knife edge B. A 30-kg weight is now suspended 2 m away from end A. Find the magnitude of the normal forces on the knife edges.
Solution:

Given

The mass of a uniform rod of length 10 m is 10 kg. The rod is placed over knife edges A and B. One end of the rod is resting on the knife edge A and the other end is 2 m outside the knife edge B. A 30-kg weight is now suspended 2 m away from end A.

Let R₁ and R₂ be the normal forces at knife edges A and B respectively. The weight of the rod, 10 kg x 9.8 m · s^-2 = 98 N, is acting at O, the mid-point of the rod.

From the conditions of equilibrium, R₁ + R₂ = (30 + 10) x 9.8 = 392 N

Taking moments about the point A, R₂ x AB = 30 x 9.8 x AD+ 10 x 9.8 x AO

or, R₂ x 8 = 294 x 2 + 98 x 5

∴ R₂ = 134.75 N

∴ R₁ = 392-134.75 = 257.25 N

Free Body Diagrams and Equilibrium

Example 8. A ladder weighing W rests with one end against a rough vertical wall and the other end on a rough floor. It is inclined at an angle of 45° with the horizontal. The coefficient of friction of the ladder with the floor and the wall are μ and μ’ respectively. Show that the minimum horizontal force that can move the base of the ladder towards the wall is \(\frac{W\left(1+2 \mu-\mu \mu^{\prime}\right)}{2\left(1-\mu^{\prime}\right)}\)
Solution:

Given

A ladder weighing W rests with one end against a rough vertical wall and the other end on a rough floor. It is inclined at an angle of 45° with the horizontal. The coefficient of friction of the ladder with the floor and the wall are μ and μ’ respectively.

A horizontal force F is applied at point B of the ladder AB, such that the ladder is at the limiting stage of equilibrium. The figure shows the forces acting at this stage.

At equilibrium, the horizontal components give, F= R’ + μR…(1)

and the vertical components give, W+μ’R’ = R…(2)

∴ ∠ABO = 45° , we have OA = OB = 2L (say). At equilibrium, the moments of the forces about B give \(W L+\mu^{\prime} R^{\prime} \cdot 2 L=R^{\prime} \cdot 2 L\)

∴ \(W=2 R^{\prime}\left(1-\mu^{\prime}\right)\)

or, \(R^{\prime}=\frac{W}{2\left(1-\mu^{\prime}\right)}\)….(3)

From (2) we get, R = \(W\left[1+\frac{\mu^{\prime}}{2\left(1-\mu^{\prime}\right)}\right]\)

and from (1) we get, F = \(W\left[\frac{1}{2\left(1-\mu^{\prime}\right)}+\mu+\frac{\mu^{\prime} \mu}{2\left(1-\mu^{\prime}\right)}\right]\)

= \(W \frac{\left(1+2 \mu-\mu \mu^{\prime}\right)}{2\left(1-\mu^{\prime}\right)}\)

Example 9. A uniform ladder of length L and mass M leans against a frictionless vertical wall. The ladder makes an angle θ with the ground and the coefficient of the static friction between the ground and the foot of the ladder is μ. What maximum height a man of mass m can climb up the ladder so that the ladder will not skid?
Solution:

Given

A uniform ladder of length L and mass M leans against a frictionless vertical wall. The ladder makes an angle θ with the ground and the coefficient of the static friction between the ground and the foot of the ladder is μ.

The value of limiting friction between the ground and the ladder is μn.

Let the man climb a maximum height x up the ladder. The point P in the figure denotes the position of the man.

Here frictional force between the ladder and ground, f = μn

where n = contact or normal force on the ladder by ground = (Mg + mg) and the contact force on the ladder by the wall is n’.

At equilibrium, net force on the ladder is zero, i.e., n’ – f = 0 or, n’ = μ(M+ m)g

Also, the moment about the point B is zero.

i.e., \(n^{\prime} L \sin \theta-M g \frac{L}{2} \cos \theta-m g \cdot x \cos \theta=0\)

or, \( m x \cos \theta=\mu(M+m) L \sin \theta-\frac{M L}{2} \cos \theta\)

∴ x = \(\frac{\mu(M+m) L \sin \theta-\frac{M L}{2} \cos \theta}{m \cos \theta}\)

WBCHSE Class 11 Physics Moment Of A Force Notes

Statics – Moment Of A Force

Let the handle of a tube-well with O as the fulcrum be AB. Generally, the portion OA is used as the handle, i.e., the handle can turn about O, in a vertical plane. Obviously, the axis of rotation is at right angles to the plane. When part OA is pushed down, OB moves up and water rises through a piston connected at point B.

• From experience, we know that, if a force F is applied at O, the handle does not turn. It is also hard to lift water by applying force F at a little distance from O. But by applying force F at the endpoint A, the handle can easily be turned. Hence, it is not true that a body will always rotate on the application of a force on it.
• Rotation is possible only when the line of action of the force passes through a point which is at some distance from the axis of rotation. The rotational tendency increases with the increase in the perpendicular distance of the line of action of the force from the centre of rotation O.

The perpendicular distance of the line of action of the force, from the axis of rotation is called the arm of that force. In the figure shown, the length of the arms of the force F are 0, OL and OM. Hence, the rotation of a body depends on two factors:

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1. Magnitude (F) and
2. Arm (d) of the applied force.

They together generate a physical quantity known as the moment of a force.

WBCHSE Class 11 Physics Moment of a Force Notes 

Moment Of A Force Definition: The force applied on a body and the perpendicular distance of the line of action of the force from the axis of rotation together, set up a rotational tendency of the body and is called the moment of the force about that axis of rotation.

• The product of the magnitude of the applied force (F) and the perpendicular distance (d) of the line of action of the force from the axis of rotation gives the magnitude of the moment of force, G = Fd.
• As shown in the figure, for different lines of action of F, the generated moments are 0, Fx OL and Fx OM respectively. As OM > OL > 0, the generated moment is the largest, when the line of action of F passes through A.
• The handle can easily turn when F acts at A. But the generated moment is zero when F acts at O.
• It is important to note that, a moment of force always refers to an axis of rotation. As a special case, when a particle rotates in a plane perpendicular to the axis of rotation, the point of intersection of the axis with that plane is sometimes called the point of rotation.
• Then the term, ‘moment of a force about a point of rotation’, may be used.
• A hinged door or window opens easily when pushed at a point farthest from the hinges. A door cannot be opened easily when we push on the hinges.

Unit And Dimension Of Moment: As defined, moment of a force = magnitude of the force x perpendicular distance of the line of action of the force from the axis of rotation. Hence, unit of moment = unit of force x unit of length.

Dimension of moment = dimension of force x dimension of length = MLT^-2 · L = ML²T^-2

Algebraic Notation Of Moment Of A Force: A number of forces acting on the same plane are called coplanar forces. Rotation, set up by the coplanar forces, also remains confined in that plane only. Suppose a body AB is able to turn about point O on a fixed plane (on the plane of the paper.

1. The moment of the force F₁ is zero.
2. For F₂, the rotation of the body AB is anticlockwise. As per convention, for anticlockwise rotation, the moment of a force is taken as positive.
3. For F₃, the rotation is clockwise. Here, the moment of the force F₃ is taken as negative.

Hence, the moments of coplanar forces can be expressed as ordinary positive or negative algebraical quantities.

Algebraic Sum Of The Moments Of Force: Let F₁, F₂ and F₃ be coplanar forces that act simultaneously on the body AB. Algebraic sum of moments on the body G = 0 + F₂ x OM + (-F₃ x ON) = F₂ X OM – F₃ X ON

If the sum

1. Is zero, the body will not rotate,
2. Is positive, the body turns anticlockwise
3. Is negative, it turns clockwise.

If the resultant of F₁, F₂ and F₃ is R and s is the arm of R, then, G = R x s.

WBCHSE Class 11 Physics Moment of a Force Notes 

Understanding Torque and Moment of Force

Geometrical Representation Of Moments: Let AB represent the magnitude, direction and line of action of the force F.

Moment of the force F about O = magnitude of F x perpendicular distance of O from the line of action

= AB x OC = 2 x (1/2 AB X OC) = 2 X area of ΔOAB

Hence, a moment of a force about a point is twice the area of the triangle formed by the endpoints of the force vector and the point about which the moment is taken.

Geometrical Representation Of Moments Vector Form: The moment of a force (G) is a vector quantity. Moment is represented in vector form as \(\vec{G}=\vec{r} \times \vec{F}\)

• The vector \(\vec{G}\) is illustrated. O is the point of rotation, i.e., the point of intersection of the axis of rotation, with the plane of rotation (the plane of the paper).
• P is the point of application of the force \(\vec{F}, \vec{r}=\overrightarrow{O P}\) = position vector of the point of application of the force, with respect to the point of rotation.
• By the definition of a cross product, the vector \(\vec{G}=\vec{r} \times \vec{F},\), is directed upwards from the plane of the paper. Hence, the torque always acts along the direction of the axis of rotation.

Class 11 Physics Moment of a Force Study Material WBCHSE 

Condition For Not Toppling Of A Moving Train Or Car On A Horizontal Circular Track: Suppose a train or a car takes a turn along a circular horizontal path about the point O. The effective centripetal force acting on the car is \(\frac{m v^2}{r}\), where m = mass of the car, v = speed of the car, r = OD = radius of the circular path.

Since the frictional forces of the wheels A and B with the road (f₁ and f₂ respectively) supply the necessary centripetal force, \(f_1+f_2=\frac{m v^2}{r}\)….(1)

Again the sum of the normal reactions on the two wheels by the ground is equal to the weight of the car, i.e., R₁ + R₂ = mg….(2)

Suppose the distance between the two wheels of the car = 2 d, i.e., AD = DB = d and the centre of mass (G) of the car is at a height h above the ground, i.e., GD = h.

So, the condition of not toppling of the car is attained if the algebraic sum of the moments of the forces about G is zero,

i.e., \(R_2 \times B D-R_1 \times A D+f_1 \times D G+f_2 \times D G=0\)

or, \(R_2 d-R_1 d+\left(f_1+f_2\right) h=0\)

or, R₂d-R₁d+(f₁+f₂)h = 0 …(3)

From equations (1) and (3) we get, \(R_1-R_2=\frac{m v^2 h}{r d}\)…(4)

Solving equations (2) and (4) we get, \(R_1=\frac{m}{2}\left(g+\frac{v^2 h}{r d}\right)\)…(5)

and \(R_2=\frac{m}{2}\left(g-\frac{v^2 h}{r d}\right)\)…(6)

It is clear that the value of R₁ can never be zero. But with the increase in speed of the car, the value of R₂ gradually decreases. For a speed v₀, if the value of R₂ is zero then,

g = \(\frac{v_0^2 h}{r d} \text { or, } v_0^2=\frac{d r g}{h} \text { or, } v_0=\sqrt{\frac{d r g}{h}}\)…(7)

In this situation, as reaction R₂ is zero, wheel B remains floating above the path. If the value of the speed exceeds v₀, the car will be overturned outwards with wheel A as the centre, i.e., the car will be overturned in the opposite direction of the centre of rotation O.

Class 11 Physics Moment of a Force Study Material WBCHSE 

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Moment of Force Formula and Derivation

Special Note: Circular Motion we observe that \(v_m=\sqrt{\mu r g}\)

Comparing this equation with equation (7) from above, it can be inferred that

if \(\mu>\frac{d}{h} \text {, then } v_m>v_0\)

i.e., the car will topple before skidding.

Again, \(\text { if } \mu<\frac{d}{h} \text {, then } v_m<v_0\)

i.e., the car will skid before it topples.

Now, normally the coefficient of friction between the wheels of a car and the road or the wheels of a train and the railway tracks is around 0.8. If the magnitude of \(\frac{d}{h}\) is less, then even vehicles moving with a small value of speed along a horizontal circular track would show a tendency to topple.

This is the reason why the ratio of d and h is kept high, while designing vehicles. It is to be noted that d is half the distance between the two wheels (either front or rear) and h is the height of the centre of mass from the surface on which the vehicle is moving.

As \(\frac{d}{h}\) is built into the design of a vehicle, it generally does not overturn. In reality, vehicles do not topple before skidding in most of the cases.

WBCHSE Physics Class 11 Moment of Force Lecture Notes 

Statics – Moment Of A Force Numerical Examples

Short Answer Questions on Moment of Force

Example 1. Masses of 2 kg and 5 kg are suspended at the two ends of a light rod of length 1.4 m. At which point should the rod be supported at to keep the system horizontal?
Solution:

Given

Masses of 2 kg and 5 kg are suspended at the two ends of a light rod of length 1.4 m.

The rod is kept horizontal, and so does not have any rotational motion. Suppose the rod is supported at O.

Taking moments about the point O, 2 x OA – 5 x OB = 0

Let OA = x

∴ OB = (1.4-x)

Then, 2 x = 5(1.4 -x) or, 7x = 7 or, x = 1 m.

Hence, the rod is supported 1 m away from the 2 kg mass.

Example 2. One end of a rope of length L is tied to a vertical pole. A man is pulling the rope with a constant force at its other end. Which point on the pole should the rope be tied at to uproot the pole most easily?
Solution

Given

One end of a rope of length L is tied to a vertical pole. A man is pulling the rope with a constant force at its other end.

Suppose the rope is tied at point B of the pole OA and is pulled by a force F from the other end C. In order to uproot the pole, a rotational motion has to be set up about the base O of the pole. The moment of the force F about O sets up the rotation.

The moment = F x OD

= F x OB sinθ

= F x BC cosθ sinθ

= \(\frac{1}{2}\) F- L sin2θ (BC = L)

The moment will be maximum when sin 2θ = 1 or, θ = 45°.

Hence, OB = BCcosθ = L cos45° = \(\frac{L}{\sqrt{2}}\)

Hence, the rope should be tied at a height of \(\frac{L}{\sqrt{2}}\) from the ground.

WBCHSE Physics Class 11 Moment of Force Lecture Notes 

Applications of Moment of Force in Everyday Life

Example 3. There is an obstruction of height h in front of a wheel of radius r weighing W. What is the minimum horizontal force that is to be applied at the centre O of the wheel to overcome the obstruction? Given, h< r.
Solution:

Given

There is an obstruction of height h in front of a wheel of radius r weighing W.

The wheel touches the obstacle at point B. Suppose a force F attempts to turn the wheel about point B. The weight of wheel W, on the other hand, resists this rotation. Hence to overcome the obstacle, the moment of the force F about B must be greater than the moment of W about B.

For F to be the minimum, F x BD = W x BC

Here, BD = OC = OA – AC = r-h

BC = \(\sqrt{O B^2-O C^2}=\sqrt{r^2-(r-h)^2}=\sqrt{2 r h-h^2}\)

Hence, F(r-h) = \(W \sqrt{2 r h-h^2}\)

or, F = \(\frac{W \sqrt{2 r h-h^2}}{r-h}\).

Moment of a Force Class 11 Physics Notes WBCHSE 

Example 4. A horizontal force F is applied on a uniform sphere of radius r and weight W, and as a result, the sphere slides over the horizontal table. If the coefficient of friction between the sphere and the table is p > then show that h = \(r\left(1-\frac{\mu W}{F}\right)\)
Solution:

Given

A horizontal force F is applied on a uniform sphere of radius r and weight W, and as a result, the sphere slides over the horizontal table. If the coefficient of friction between the sphere and the table is p >

The point of contact between the table and the sphere is Q. Here, the force of friction f acts along the surface of the table through the point Q opposite to the force F. The weight of the sphere W acts vertically downwards through the centre of the sphere. Since the sphere slides over the horizontal table, it has no rotational motion.

So, with respect to the centre O, f x r = F(r- h) or, μWx r = F(r-h)

h = \(r\left(1-\frac{\mu W}{F}\right)\)

WBCHSE Class 11 Physics Torque and Moment of Force Notes 

Real-Life Examples Illustrating Moment of Force

Example 5. A uniform iron rod of length 50 cm is bent, at right angles at 30 cm from one end, giving it an L-shape. The L-shaped rod is suspended freely from its point of bending. Find the angle that the 30 cm arm makes with the vertical at equilibrium.
Solution:

Given

A uniform iron rod of length 50 cm is bent, at right angles at 30 cm from one end, giving it an L-shape. The L-shaped rod is suspended freely from its point of bending.

Suppose the rod is bent at point B, AB = 20 cm, BC = 30 cm, and B is the point of suspension. Suppose BC, in equilibrium, makes an angle θ with the vertical.

Let the mass per unit length of the rod be w. So the mass of AB is 20 w and the mass of BC is 30 w. At equilibrium, the moments about point B due to the two masses, should be equal.

Let x and y be the perpendicular distances of the centres of mass of the arms AB and BC respectively, from the vertical line through B. The centre of mass is at the midpoint of each arm.

∴ 20 wx = 30 wy

or, 20 x 10 cosθ = 30 x 15 sinθ or, tanθ = \(\frac{4}{9}\)

or, θ = 24°.

The angle that the 30 cm arm makes with the vertical at equilibrium is 24°.

Centre Of Mass Or C M

WBBSE Class 11 Centre of Mass Notes

An extended body is made up of a large number of particles. If the extended body is rigid, the shape and size of the body remain unaltered because the relative distances between the particles do not change.

Similarly, if a group of isolated particles are arranged in such a way that their relative distances are fixed, then the system of particles can also be considered as an extended rigid body.

When a force is applied to a particle in a fixed direction, the particle undergoes only a linear motion. Rotational motion cannot be set up.

• On the other hand, if an external force is applied to a system of particles (which may be an extended rigid body in real life), it may result in a pure linear motion, a pure rotational motion, or a combination of the two. The type of motion depends on the line of action of the applied force.
• We define the centre of mass of the system of particles as a point such that, if the net resultant of the applied forces acts along a line passing through this point, the body undergoes only linear motion.
• This means that we can consider the entire mass of the system of particles, or the extended body, to be concentrated at the centre of mass. The kinematics of an extended body becomes considerably easier when we study the motion of its centre of mass.
• The centre of mass need not be located within the body. We have to be cautious when we work out the effects of forces on an extended body.

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If the line of action of the resultant of all applied forces pass through the centre of mass, then the net torque vanishes. This gives rise to translational motion only. But, if the line of action of the resultant does not pass through the centre of mass, then there will be both translational and rotational motion.

Consider a disc kept on a table. A radial force F₁, applied at point A on the disc, produces only a rectilinear motion. Similarly, by applying a force F₂ at point B, or force F₃ at D, we get rectilinear motion only.

Key Concepts of Centre of Mass in Physics

• Hence, it is seen that the point of intersection C of the lines of action of F₁, F₂, and F₃ is such a point that any force applied through it generates linear motion only. The point C is the centre of mass of the disc.
• It is clear from the above discussion that, for an annular disc the centre of mass C is still at its centre. It highlights the fact that the centre of mass need not be located within the body.
• When the line of action of the applied force does not pass through the centre of mass, there will be a mixed motion of the body. The line of action of F₄, the force applied at A, does not pass through the centre of mass and so it produces both rotational and translational motions.

Center Of Mass Definition: The centre of mass of an extended body or a system of particles is such a unique point that, any force applied through that point produces only translational motion of the body, but no rotational motion.

While studying the linear motion of a body, the motion of the entire body need not be considered. Instead, it is sufficient to analyse the motion of the centre of mass of the body. This centre of mass effectively acts as a particle, where the entire mass of the body may be assumed to be concentrated.

Centre Of Mass Of A Two-Particle System: Let us consider a system of two point masses m₁ and m₂ connected by a light inextensible rod. This system is kept along the x-axis of a two-dimensional reference frame.

The position of the centre of mass can be considered to be the average position of the total mass of the system. The equation which describes the position of the centre of mass of the two-particle system, can be understood with the help of the following example.

Centre of Mass Formula and Derivation

Let us consider that in a shop two types of pens are available at a price of c₁ and c₂ respectively. If n₁ and n₂ are the numbers of these two types of pens bought from the shop, the average price for each of the (n₁ + n₂) pens is

∴ c = \(\frac{n_1 c_1+n_2 c_2}{n_1+n_2}\)

It can be shown that the value of c is between c₁ and c₂. Similarly, it can be stated that the average position of the total mass of the two-point masses or the position of the centre of mass of the two-particle system is

∴ \(x_{\mathrm{cm}}=\frac{m_1 x_1+m_2 x_2}{m_1+m_2}\)…(1)

where x₁ and x₂ are the position coordinates of m₁ and m₂ respectively.

Centre Of Mass Of Infinite Number Of Point Mourn In A Three-Dimensional Coordinate System: Let us consider that the position coordinates of the masses \(m_1, m_2, \cdots \text { etc. are }\left(x_1, y_1, z_1\right),\left(x_2, y_2, z_2\right), \cdots\) etc. and the coordinates of the centre of mass is (x_cm, y_cm, z_cm). Following equation (1), we can write

⇒ \(\left.\begin{array}{l}
x_{\mathrm{cm}}=\frac{m_1 x_1+m_2 x_2+\cdots}{m_1+m_2+\cdots}=\frac{\sum_i m_i x_i}{\sum_i m_i} \\
y_{\mathrm{cm}}=\frac{\sum_i m_i y_i}{\sum_i m_i}, z_{\mathrm{cm}}=\frac{\sum_i m_i z_i}{\sum_i m_i}
\end{array}\right\}\)….(2)

Centre Of Mass Of Continuous Bodies: if we consider the body to have a continuous distribution of matter, the summation in the expression of the centre of mass should be replaced by integration. So, instead of talking about the i-th particle, we may consider a small element of the body, having a mass dm. If x, y, z are the coordinates of this small mass dm, then in terms of integrals, equation (2) can be written as

⇒ \(\left.\begin{array}{l}
x_{\mathrm{cm}}=\frac{1}{M} \int x d m, y_{\mathrm{cm}}=\frac{1}{M} \int y d m \\
z_{\mathrm{cm}}=\frac{1}{M} \int z d m
\end{array}\right\}\)…(3)

The integration should be performed under proper limits so that as the integration variable goes through the limits, the elements cover the entire body.

Example Centre Of Mass Of A Uniform Straight Rod: Let M be the mass and L be the length of a rod. Let us take the left end of the rod as the origin and the x-axis along the length of the rod. The rod may be considered to be an aggregate of small elements. Let us consider a small element AB of length dx.

Since the rod is uniform, mass per its unit length = \(\frac{M}{L}\)

So, the mass of the element AB is dm = \(\frac{M}{L}\)dx

The coordinates of the element are (x, 0,0).

Therefore, the x-coordinate of the centre of mass of the rod is

⇒ \(x_{\mathrm{cm}}=\frac{1}{M} \int x d m=\frac{1}{M} \int_0^L \frac{M}{L} x d x=\frac{1}{L}\left[\frac{x^2}{2}\right]_0^L=\frac{L}{2}\)

The y-coordinate is \(y_{\mathrm{cm}}=\frac{1}{M} \int y d m=0\)

Similarly, z_cm = 0

So the centre of mass is at (\(\frac{L}{2}\)0, 0), i.e., at the mid-point of the rod.

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Some Important Information About The Centre Of Mass:

1. The centre of mass is a unique point of a body, i.e., a body cannot have more than one centre of mass.
2. A body with a regular geometrical shape and of uniform density has its centre of mass at its geometrical centre.
3. If the density of a body is non-uniform and varies from one end to the other, then the centre of mass shifts towards the heavier end.
   • For example, the centre of mass of a tumbler of muddy water is lower than that of a tumbler of pure water. This is because the suspended clay particles settle downwards making the density at the bottom higher.
4. In a system of bodies, if the mass of one body is much higher than that of the others, then the centre of mass of the system almost coincides with that of the heavy body.
   • For example, the centre of mass of the solar system is at the centre of the sun, because the masses of the other celestial bodies in the solar system are negligible compared to that of the sun.
5. The Centre of mass of a body may lie outside the body, for example, the centre of mass of a ring, or a bangle, or a door knocker is at its centre.
6. The position of the centre of mass changes with the change in the shape of a body. For example, the centre of mass of a metal rod is at the mid-point of the rod, but if the rod is bent into a circular ring, the centre of mass shifts to the centre of the circle.

Momentum Conservation Motion Of Centre Of Mass:

Relation Between Momentum Of The Centre Of Mass Of An Extended Body Or A System Of Particles And Momenta Of Constituent Parties Of The Body: Let m₁, m₂, m_g,…. be the masses of the constituent particles of an extended object or a system of particles. The position vectors of these particles are \(\vec{r}_1, \vec{r}_2, \vec{r}_3, \ldots\)respectively. If \(\vec{r}_{\mathrm{cm}}\) is the position vector of the centre of mass, then,

⇒ \(\vec{r}_{\mathrm{cm}}=x_{\mathrm{cm}} \hat{i}+y_{\mathrm{cm}} \hat{j}+z_{\mathrm{cm}} \hat{k}\)

= \(\frac{\sum_i m_i x_i}{\sum_i m_i} \hat{i}+\frac{\sum_i m_i y_i}{\sum_i m_i} \hat{j}+\frac{\sum_i m_i z_i}{\sum_i m_i} \hat{k}\)

∴ \(\vec{r}_{\mathrm{cm}} \sum_i m_i=\sum_i\left\{m_i\left(x_i \hat{i}+y_i \hat{j}+z_i \hat{k}\right)\right\}=\sum_i m_i \vec{r}_i\)

∴ \(M \vec{r}_{\mathrm{cm}}=\sum_i m_i \vec{r}_i\)

(Mass of the body or the system of particles, \(M=\sum_i m_i\))

Therefore, the momentum of the centre of mass of the body or the system of particles,

⇒ \(M \frac{d \vec{r}_{\mathrm{cm}}}{d t}=\sum_i m_i \frac{d \vec{r}_i}{d t}\)

or, \(M \vec{v}_{\mathrm{cm}}=\sum_i m_i \vec{v}_i\)…(4)

Hence, momentum of the centre of mass = sum of momenta of constituent particles of the body

Relation Between The Force On The Centre Of Mass And The Forces Upon The Particles: Differentiating equation (4) with respect to time we get,

⇒ \(M \frac{d}{d t}\left(\vec{v}_{\mathrm{cm}}\right)=\sum_i m_i \frac{d \vec{\nu}_i}{d t}\)

or, \(M \vec{a}_{\mathrm{cm}}=\sum_i m_i \vec{a}_i\)

∴ \(\vec{F}_{\mathrm{cm}}=\sum_i \vec{F}_i=\vec{F}_1+\vec{F}_2+\vec{F}_3+\cdots\)…(5)

Now, \(\vec{F}_1\) the force on the first particle is not a single force, but it is the vector sum of all the forces acting on it. Similar arguments can be made for the second particle, the third panicle and so on.

Understanding Centre of Mass in Systems of Particles

These forces on each particle can be classified as external force (exerted by bodies outside the system) and internal force (exerted by the particles on one another).

From Newton’s laws of motion, we know that these internal forces occur in equal and opposite pairs. So in equation (5) their contribution is zero. Hence only the external forces contribute to the equation (5). Rewriting equation (5) as,

⇒ \(\vec{F}_{\mathrm{cm}}=\vec{F}_{\mathrm{ext}}\)

or, \(M \vec{a}_{\mathrm{cm}}=\vec{F}_{\text {ext }}\)…(6)

where \(\vec{F}_{\mathrm{ext}}\) is the sum of all external forces acting on the particles of the system.

Conservation Of Momentum Of Centre Of Mass: From Newton’s laws of motion we know that for zero external force, the momentum of a body or momenta of particles remains unaltered,

i.e., \(\text { if } F_{\mathrm{ext}}=0, \sum_i m_i \overrightarrow{v_i}=\text { constant }.\)

Thus, from equation (4), \(M \vec{v}_{\mathrm{cm}}\) = constant, i.e., the momentum of the centre of mass is constant.

Hence, for zero external force, the momentum of the centre of mass is conserved. As mass is a constant, the conservation of momentum implies the conservation of velocity.

It is to say that, for zero external force, the centre of mass of a body remains stationary or moves with uniform velocity.

It is important to note that the velocity of the centre of mass of an extended body or a system of particles remains unaltered even under the action of internal forces acting among the constituent particles of the body or the system, provided there is no external force acting on the body.

Applications of Centre of Mass in Real Life

Illustrative Examples:

1. Change Of Position Of A Passenger On A Boat: Consider a boat with a passenger at position A. The centre of mass of the boat-passenger system shifts to C from the mid-point of the boat due to the mass of the passenger.
   • When the passenger moves from end A to end B, the boat displaces towards the opposite direction, so that the centre of mass of the system is now located near end B.
   • In this case, no external force acts on the boat-passenger system. Hence, the position of the centre of mass with respect to water remains the same, i.e., at C. When the passenger moves in one direction, the boat also shifts in the opposite direction so that the position of the centre of mass of the system does not change.
2. Explosion Of An Object: Suppose a cannonball moving with velocity v along OP explodes at P. After the explosion, the ball breaks up into a large number of fragments and the fragments scatter in different directions. Only two fragments and their paths after the explosion are shown.
   • When the centre of mass of the two fragments is determined, it is seen that the centre of mass is at A, B, etc.
   • Hence the centre of mass of the two fragments continues moving with the same velocity v and in the same direction as before. In this case, the internal forces developed due to the explosion did not influence the motion of the centre of mass.

Statics – Centre Of Mass Numerical Examples

Short Answer Questions on Centre of Mass

Example 1. Two particles of masses 1 kg and 2 kg are positioned on x-and y-axes at distances of 1 m and 2 m from the origin respectively. Find the position of the centre of mass of the system.
Solution:

Given

Two particles of masses 1 kg and 2 kg are positioned on x-and y-axes at distances of 1 m and 2 m from the origin respectively.

The mass of the first particle, m₁ = 1 kg and its position is (1, 0).

The mass of the second particle is m₂ = 2 kg and it is positioned at (0,2).

Let the coordinates of the centre of mass of the system be (x_cm, y_cm)

∴ \(x_{\mathrm{cm}}=\frac{m_1 x_1+m_2 x_2}{m_1+m_2}=\frac{1 \times 1+2 \times 0}{1+2}=\frac{1}{3} \mathrm{~m}\)

and \(y_{\mathrm{cm}}=\frac{m_1 y_1+m_2 y_2}{m_1+m_2}=\frac{1 \times 0+2 \times 2}{1+2}=\frac{4}{3} \mathrm{~m}\)

Hence, the distance of the centre of mass from the 1 kg mass = \(\sqrt{\left(1-\frac{1}{3}\right)^2+\left(0-\frac{4}{3}\right)^2}=\frac{2 \sqrt{5}}{3} \mathrm{~m}\) and that from the 2 kg mass = \(\sqrt{\left(0-\frac{1}{3}\right)^2+\left(2-\frac{4}{3}\right)^2}=\frac{\sqrt{5}}{3} \mathrm{~m}.\)

Example 2. Two masses, initially at rest, attract each other with a constant force. If there is no external force acting on the masses, prove that the centre of mass of the system remains stationary.
Solution:

Given

Two masses, initially at rest, attract each other with a constant force. If there is no external force acting on the masses,

Let the line joining the two masses be taken as the x-axis. The initial position of mass m₁ is x₁ and that of mass m₂ is x₂ (x₂ > x₁).

Hence, the initial position of the centre of mass is \(x_{\mathrm{cm}}=\frac{m_1 x_1+m_2 x_2}{m_1+m_2}\)

Suppose a force F acts on the first mass along the positive x direction. Hence, a force F will act on the second mass along the negative direction of the x-axis as per Newton’s third law of motion.

Hence, acceleration of the first mass, \(a_1=\frac{F}{m_1}\) and displacement in time t = \(\frac{1}{2} a_1 t^2\).

So, the position of the first one after a time t, \(x_1{ }^{\prime}=x_1+\frac{1}{2} a_1 t^2=x_1+\frac{1}{2} \frac{F}{m_1} t^2\)

Similarly, the position of the second one after a time t, \(x_2{ }^{\prime}=x_2+\frac{1}{2} \cdot \frac{(-F)}{m_2} t^2=x_2-\frac{1}{2} \frac{F}{m_2} t^2\)

Hence, the position of the centre of mass after time t \(x_{\mathrm{cm}}^{\prime}=\frac{m_1 x_1{ }^{\prime}+m_2 x_2{ }^{\prime}}{m_1+m_2}=\frac{m_1 x_1+\frac{F t^2}{2}+m_2 x_2-\frac{F t^2}{2}}{m_1+m_2}\)

or, \(x_{\mathrm{cm}}^{\prime}=\frac{m_1 x_1+m_2 x_2}{m_1+m_2}=x_{\mathrm{cm}}\)

Hence, the centre of mass remains stationary.

Example 3. A small sphere of radius R is kept attached to the smooth inner wall of a hollow sphere of radius 6 R. The system is kept on a frictionless horizontal table. On releasing the small sphere, as it reaches the other end of the diameter, what will be the coordinates of the centre of the large sphere? Masses of the small and large spheres are M and 4M respectively.
Solution:

Given

A small sphere of radius R is kept attached to the smooth inner wall of a hollow sphere of radius 6 R. The system is kept on a frictionless horizontal table. On releasing the small sphere, as it reaches the other end of the diameter,

The coordinates of the centre of the large sphere is (L,0). Hence, the coordinates of the centre of the small sphere is (L+6R-R,0) or (L + 5R, 0). So, if x_cm are the coordinates of the centre of mass, then,

⇒ \(x_{\mathrm{cm}}=\frac{m_1 x_1+m_2 x_2}{m_1+m_2}\)

= \(\frac{4 M L+M(L+5 R)}{5 M}=\frac{5 M(L+R)}{5 M}=L+R\)

The coordinates of the centre of mass remain unchanged in the absence of any external force.

Let the coordinates of the centre of the large sphere be (a, 0) when the small sphere reaches the opposite end A.

The coordinates of the centre of the small sphere become (a – 5R, 0). As the position coordinates of the centre of mass remain unchanged,

L + R = \(\frac{4 M a+M(a-5 R)}{4 M+M}=\frac{5 M a-5 M R}{5 M}=a-R\)

∴ a = L + R + R = L + 2R

Hence, the new coordinates of the centre of the large sphere will be (L+2R, 0).

Real-Life Examples Illustrating Centre of Mass

Example 4. The position vectors of two masses of 6 and 2 units are \(6 \hat{i}-7 \hat{j} \text { and } 2 \hat{i}+5 \hat{j}-8 \hat{k}\) respectively. Deduce the position of their centre of mass.
Solution:

Given

The position vectors of two masses of 6 and 2 units are \(6 \hat{i}-7 \hat{j} \text { and } 2 \hat{i}+5 \hat{j}-8 \hat{k}\) respectively.

Here m₁ = 6 units; its position vector \(\vec{r}_1=6 \hat{i}-7 \hat{j}\); m₂ = 2 units; its position vector \(\vec{r}_2=2 \hat{i}+5 \hat{j}-8 \hat{k}\)

The position vector of the centre of mass of the two masses \(\vec{r}_{\mathrm{cm}}=\frac{m_1 \vec{r}_1+m_2 \vec{r}_2}{m_1+m_2}\)

= \(\frac{6(6 \hat{i}-7 \hat{j})+2(2 \hat{i}+5 \hat{j}-8 \hat{k})}{6+2}=5 \hat{i}-4 \hat{j}-2 \hat{k}\)

So, the coordinates of the centre of mass are (5, -4, -2).

Example 5. Locate the centre of mass of a triangular lamina.
Solution:

The centre of mass of a triangular lamina

ABD is a triangular lamina. It is divided into a few narrow strips parallel to the base BD.

• Due to symmetry, the centre of mass of each strip lies at its mid-point. The midpoints of the strips lie on the median AE of the triangular lamina. This means that the centre of mass of the triangular lamina must lie on the median AE.
• By the same reasoning, the centre of mass of the lamina must lie on the medians BF and DG. Obviously, the centre of mass of the triangular lamina lies at the intersection of the three medians, i.e., at the centroid C of the triangle ABD.

Example 6. Three particles of masses 1 g, 2 g, and 3 g are placed at the vertices of an equilateral triangle of side 1 m. j Locate the centre of mass of the system.
Solution:

Given

Three particles of masses 1 g, 2 g, and 3 g are placed at the vertices of an equilateral triangle of side 1 m. j

Suppose, the triangle lies on the XY- plane. The particle of mass m₁ = 1 g is placed at the origin O which is one of the vertices of the triangle. m₁ and m₂ lie on the x-axis. Let the coordinates of the centre of mass of the system be (x_cm, y_cm).

∴ \(x_{\mathrm{cm}}=\frac{m_1 x_1+m_2 x_2+m_3 x_3}{m_1+m_2+m_3}=\frac{1 \times 0+2 \times 1+3 \times \frac{1}{2}}{1+2+3}\)

= \(\frac{7}{12} \mathrm{~m}\)

and \(y_{\mathrm{cm}}=\frac{m_1 y_1+m_2 y_2+m_3 y_3}{m_1+m_2+m_3}=\frac{1 \times 0+2 \times 0+3 \times \frac{\sqrt{3}}{2}}{1+2+3}\)

= \(\frac{\sqrt{3}}{4} \mathrm{~m}\)

∴ The coordinates of the centre of mass are \(\left(\frac{7}{12}, \frac{\sqrt{3}}{4}\right)\)

Example 7. In the HCl molecule, the separation between the nuclei of the two atoms is about 1.27 x 10^-10 m. Find the approximate location of the Centre of mass of the molecule. Given that a chlorine atom is about 35.5 times as massive as a hydrogen atom and nearly all the mass of an atom is concentrated in its nucleus.
Solution:

Given

In the HCl molecule, the separation between the nuclei of the two atoms is about 1.27 x 10^-10 m.

Let the mass of the H-atom be m.

So the mass of the Cl-atom is 35.5 m.

Let the coordinate of the H-atom, x₁ = 0; then the coordinate of the Cl-atom, x₂ = r.

So, the coordinate of the centre of mass,

x = \(\frac{m_1 x_1+m_2 x_2}{m_1+m_2}=\frac{m \cdot 0+35.5 m \cdot r}{m+35.5 m}\)

= \(\frac{35.5}{36.5} r=\frac{35.5}{36.5} \times 1.27 \times 10^{-10}=1.235 \times 10^{-10} \mathrm{~m} .\)

Example 8. Two objects of mass 10kg and 2kg are moving with velocities \((2 \hat{i}-7 \hat{j}+3 \hat{k}) \mathrm{m} / \mathrm{s}\) and \((-10 \hat{i}+35 \hat{j}-3 \hat{k}) m/s\) respectively. Calculate the velocity of the centre of mass of the system.
Solution:

Given

Two objects of mass 10kg and 2kg are moving with velocities \((2 \hat{i}-7 \hat{j}+3 \hat{k}) \mathrm{m} / \mathrm{s}\) and \((-10 \hat{i}+35 \hat{j}-3 \hat{k}) m/s\) respectively.

Centre of mass of the two objects \(\vec{r}_{\mathrm{cm}}=\frac{m_1 \vec{r}_1+m_2 \vec{r}_2}{m_1+m_2}\)

Therefore, the velocity of the centre of mass \(\vec{v}_{\mathrm{cm}}=\frac{m_1 \vec{v}_1+m_2 \vec{v}_2}{m_1+m_2}\)

= \(\frac{10(2 \hat{i}-7 \hat{j}+3 \hat{k})+2(-10 \hat{i}+35 \hat{j}-3 \hat{k})}{10+2}=2 \hat{k} \mathrm{~m} / \mathrm{s}\)

Example 9. A stone is dropped from the top of a tower of a height of 60 m. At the same time, another stone is thrown up from the foot of the tower with a velocity of 20 m/s. Calculate the displacement of the centre of mass of the two stones at the time of collision, (g = 10 m/s²)
Solution:

Given

A stone is dropped from the top of a tower of a height of 60 m. At the same time, another stone is thrown up from the foot of the tower with a velocity of 20 m/s.

Initially, the centre of mass of the two stones is at a height of 30 m from the ground.

Acceleration of the centre of mass = \(\frac{m \times g+m \times g}{m+m}=g \text { (downward) }\)

Let the stones collide over time t.

At that time distance travelled by the first stone = \(\frac{1}{2} g t^2\)

and distance travelled by the second stone = ut-\(\frac{1}{2} g t^2\)

Now, \(\frac{1}{2} g t^2+u t-\frac{1}{2} g t^2=60\)

∴ t = \(\frac{60}{u}=\frac{60}{20}=3 \mathrm{~s}\)

The initial speed of the centre of mass, \(u_{\mathrm{cm}}=\frac{m \times 0+m \times 20}{m+m}=10 \mathrm{~m} / \mathrm{s} \text { (upward) }\)

∴ Displacement of the centre of mass

= \(u_{\mathrm{cm}} \times t-\frac{1}{2} g t^2=10 \times 3-\frac{1}{2} \times 10 \times 9=-15 \mathrm{~m}\)

Therefore, the centre of mass will move downward by 15 m.

Newtonian Gravitation And Planetary Motion

WBBSE Class 11 Newtonian Gravitation Notes

Newtonian Gravitation And Planetary Motion Introduction: Since ancient times, scientists have been extremely curious to learn more about the stars and planets.

• Since the time of Copernicus, it is known that planets move around the sun. But to investigate the cause and nature of this motion, it was necessary to know the exact positions of the planets in the sky at different times.
• Astrophysicist Tycho Brahe, for many years, observed the positions of planets without a telescope (Galileo discovered the telescope after the death of Brahe) and published a lot of information about this.
• Kepler analysed the observational findings of Tycho Brahe and arrived at three laws about planetary motion. These laws are known as Kepler’s laws of planetary motion.

Newton’s Law of Universal Gravitation Explained

Read and Learn More: Class 11 Physics Notes

 Newtonian Gravitation And Planetary Motion – Kepler’s Laws

Understanding Kepler’s Laws of Planetary Motion

Kepler’s Laws First Law: Every planet moves in an elliptical orbit with the sun at one of its foci.

Kepler’s Laws Second Law: The line joining the sun and a planet sweep out equal areas in equal intervals of time, i.e., the areal velocity of a planet is constant.

Kepler’s Laws Third Law: The square of the time period of revolution of a planet is directly proportional to the cube of the length of the semi-major axis of its elliptical orbit.

The orbits of planets are known as Keplerian orbits while their motions are known as Keplerian motions.

Proof Of Kepler’s Second Law: Let us consider a planet of mass m moving in an elliptical orbit with the sun at focus S. Also, let \(\vec{r}\) be the position vector of the planet with respect to the sun and \(\vec{F}\) be the required centripetal force for the planet.

The torque exerted on this planet by this force about the sun, \(\)

(\(\vec{r}\) and \(\vec{F}\) are oppositely directed)

But \(\) (\(\vec{L}\) = angular momentum of the planet)

∴ \(\frac{d \vec{L}}{d t}=0 \quad \text { or, } \vec{L}=\text { constant }\)

Now, if the planet moves from position P to P’ in very small time A t, then the area swept out by the radius vector \(\vec{r}\) is,

⇒ \(\Delta \vec{A}=\text { area of triangular region } S P P^{\prime}\)

= \(\frac{1}{2} \vec{r} \times \overrightarrow{P P^{\prime}}\)

Now, \(\quad \overrightarrow{P P^{\prime}} =\Delta \vec{r}=\vec{v} \Delta t=\frac{\vec{p}}{m} \Delta t\)

∴ \(\quad \Delta \vec{A}=\frac{1}{2} \vec{r} \times \frac{\vec{p}}{m} \Delta t\)

or, \(\quad \frac{\Delta \vec{A}}{\Delta t}=\frac{1}{2 m}(\vec{r} \times \vec{p})=\frac{\vec{L}}{2 m}\) (because \(\vec{L}=\vec{r} \times \vec{p}\))

Hence, \(\quad \frac{\Delta \vec{A}}{\Delta t}=\text { constant }\) (because \(\vec{L} \text { and } m\)) are constant

Thus, the areal velocity of the planet remains constant, i.e., the radius vector joining the sun and a planet sweeps out equal areas in equal intervals of time.

Gravitational Attraction Of Extended Bodies

Gravitational Force and Planetary Orbits

Any extended body comprises a large number of particles. Thus, to find the mutual force of attraction between two extended bodies, the force of attraction between each particle of the first body and that of the second body is to be determined individually.

• The resultant of these forces, given by the vector addition method, gives the magnitude and direction of the force of attraction between the two extended bodies. Clearly, this method is complicated and unwieldy. This method can only be used in the case of regular-shaped bodies.
• But in the case of two extended bodies, however large they may be, if the distance separating them is much greater than their sizes, it can be assumed that the entire mass of each body is concentrated at its centre of mass. As a result, two extended bodies can be considered as two particles at their respective centres of mass.
• In this way, the complication involved in applying Newton’s law of gravitation may be removed. This approach can be applied to celestial bodies, like the earth and the moon, even if they are large in size and cannot be considered as particles.
• The distance between them is so large in comparison to their sizes that they may be considered as particles situated at their respective centres of mass for the purpose of measuring the gravitational force between them.

Now taking M = mass of the earth, m = mass of the moon and r = the distance between their centres of mass, the force of gravitation between them = \(\frac{G M m}{r^2}\).

When the distance between the two bodies is small compared to their sizes, this method of calculation fails.

When the gravitational attraction between a homogeneous spherical body and a particle situated outside the body is considered, it is assumed that the entire mass of the spherical body is concentrated at its centre of mass.

Hence, the earth, the sun, the moon and other planets are taken as spherical and the law is applied. A substance is called homogeneous when the physical properties (like density) of its different parts are identical.

Newtonian Gravitation And Planetary Motion – Sun Earth Moon Numerical Examples

Short Answer Questions on Planetary Motion

Example 1. Assuming that the moon moves around the earth in a circular orbit of radius 3.8 x 10⁵ km in 27 days and the earth moves around the sun in a circular orbit of radius 1.5 x 10⁸ km in 365 days, find the ratio of the masses of the sun and the earth.
Solution:

Given

Assuming that the moon moves around the earth in a circular orbit of radius 3.8 x 10⁵ km in 27 days and the earth moves around the sun in a circular orbit of radius 1.5 x 10⁸ km in 365 days,

From Kepler’s law, \(\frac{T^2}{r^3}\) = constant

where T = time period, r = radius of the orbit.

If the mass of the sun is M₀ and the radius of the earth’s orbit around the sun is r₀, the orbital speed,

⇒ \(v_0=\sqrt{\frac{G M_0}{r_0}}\)

∴ Time period T₀ of the earth = \(\frac{2 \pi r_0}{v_0}=2 \pi r_0 \sqrt{\frac{r_0}{G M_0}}=2 \pi \sqrt{\frac{r_0^3}{G M_0}}\)

Similarly, if the mass of the earth is M, the radius of the moon’s orbit is r, time period,

T = \(2 \pi \sqrt{\frac{r^3}{G M}}\)

∴ \(\frac{T}{T_0}=\sqrt{\left(\frac{r}{r_0}\right)^3 \frac{M_0}{M}} \text { or, }\left(\frac{T}{T_0}\right)^2=\left(\frac{r}{r_0}\right)^3 \cdot \frac{M_0}{M}\)

or, \(\frac{M_0}{M}=\left(\frac{T}{T_0}\right)^2 \cdot\left(\frac{r_0}{r}\right)^3=\left(\frac{27}{365}\right)^2 \cdot\left(\frac{1.5 \times 10^8}{3.8 \times 10^5}\right)^3\)

= 3.37 x 10⁵ = 337000

Hence, the sun is 337000 times more massive than the Earth.

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Example 2. Find the mass of the sun considering the orbit of the earth to be circular. Given, a distance of the earth from the sun = 1.49 x 10¹³ cm and G = 6.66 x 10^-8 CGS unit.
Solution:

Given

Given, a distance of the earth from the sun = 1.49 x 10¹³ cm and G = 6.66 x 10^-8 CGS unit.

Let the mass of the sun be M₀, the distance of the earth from the sun r, and the time period of the earth around the sun T.

∴ Mass of the Sun, \(M_0=\frac{4 \pi^2 r^3}{G T^2}\)

= \(\frac{4 \times \pi^2 \times\left(1.49 \times 10^{13}\right)^3}{6.67 \times 10^{-8} \times(365 \times 24 \times 60 \times 60)^2}\)

= \(2 \times 10^{33} \mathrm{~g}=2 \times 10^{30} \mathrm{~kg}\)

Newtonian Gravitation And Planetary Motion – Mass And Average Density Of The Earth Numerical Examples

Example 1. The average density of the earth is 5500 kg · m^-3, the gravitational constant is 6.7 x 10^-11 N · m² · kg^-2 and the radius of the earth is 6400 km. Using the given values, find the magnitude of the acceleration due to gravity on the surface of the earth.
Solution:

Given

The average density of the earth is 5500 kg · m^-3, the gravitational constant is 6.7 x 10^-11 N · m² · kg^-2 and the radius of the earth is 6400 km. Using the given values,

It is known \(\rho=\frac{3 g}{4 \pi R G}\)

or, \(g=\frac{4 \pi \rho R G}{3}\)

g =\(\frac{4 \times 22 \times 5500 \times 6400 \times 10^3 \times 6.7 \times 10^{-11}}{7 \times 3}\)

= \(9.88 \mathrm{~m} \cdot \mathrm{s}^{-2} .\)

Example 2. If the earth is considered as a solid sphere of iron of; radius 6.37 x 10⁶ m and of density 7.86 g · cm^-3, what will be the magnitude of the acceleration due to gravity on the earth’s surface? Gravitational constant = 6.58 x 10^-8 CGS unit.
Solution:

Given

If the earth is considered as a solid sphere of iron of; radius 6.37 x 10⁶ m and of density 7.86 g · cm^-3,

It is known \(\rho=\frac{3 g}{4 \pi R G}\)

∴ g = \(\frac{4 \pi \rho R G}{3}=\frac{4 \times 22 \times 7.86 \times 6.37 \times 10^8 \times 6.58 \times 10^{-8}}{7 \times 3}\)

= \(1380.55 \mathrm{~cm} \cdot \mathrm{s}^{-2}\)

Newtonian Gravitation And Planetary Motion Conclusion

The force with which any two material particles in the universe attract each other is called gravitation.

Newton’s Law Of Gravitation: Any two material particles in the universe attract each other along their line of joining. This force of attraction is directly proportional to the product of the masses of the two particles and inversely proportional to the square of the distance separating them.

Applications of Newtonian Gravitation in Space

• The amount of force with which any two material particles of unit mass kept at a unit distance apart attract each other is called the gravitational constant.
• If a body of unit mass is kept at a point in any gravita¬tional field, the gravitational force acting on that body is called gravitational intensity at that point.
• The amount of work done to bring a body of unit mass from infinity to any point in a gravitational field is called the gravitational potential at that point.
• The force with which anybody on or near the earth’s surface is attracted by the earth is called gravity.
• The acceleration produced in a body falling under the influence of the force of gravity is called the acceleration due to gravity.

Kepler’s laws related to the motion of planets and satellites:

1. First Law: Keeping the sun at one of the foci, each planet revolves in an elliptical path around the sun.
2. Second Law: The line joining the sun and a planet covers equal areas in equal intervals of time.
3. Third Law: The square of the time period of revolution of a planet around the sun is directly proportional to the cube of the semimajor axis of its orbit.

The minimum velocity that should be given to a body so that it can escape from the earth or other planets or sat¬ellites is called its escape velocity.

Some uses of an artificial satellite:

1. Weather information,
2. Communication,
3. Military and defence surveillance,
4. Remote sensing.

If the relative angular velocity of an artificial satellite revolving in the equatorial plane is zero with respect to the diurnal motion of the earth, then, from the earth’s surface, the satellite appears to be at rest at the same place. This kind of satellite is called a geostationary satellite.

Newtonian Gravitation And Planetary Motion Useful Relations For Solving Numerical Problems

If the distance between two particles of masses m₁ and m₂ is r and if the mutual force of attraction between the two particles is F, then according to Newton’s law of gravitation,

F = \(\) where G is the gravitational constant whose value in SI is 6.67 x 10^-11 N · m² · kg^-2.

Gravitational intensity at a point situated at a distance r from the centre of a body of mass M is E = \(\frac{G M}{r^2}\)

Work done in bringing a body of unit mass from infinity to the position r, i.e., the gravitational potential at a distance r from the centre of a body of mass M is V = –\(\frac{G M}{r}\)

The relation between the gravitational field intensity and potential is E = \(\frac{d V}{d r}\)

Acceleration due to gravity at any point on the earth’s surface, g = \(\frac{G M}{R^2}\), where M is the mass and R is the radius of the earth, and G is the gravitational constant.

The relation between the gravitational constant G and the mean density (ρ) of the earth is \(\frac{3 g}{4 \pi R G}\).

1. The relation between the acceleration due to gravity at an altitude h above the surface of the earth (g’) and the acceleration due to gravity on the earth’s surface (g) is g’ = g(1-\(\frac{2h}{R}\)) (this equation is valid for h<<R)
2. The relation between the acceleration due to gravity at a depth h below the earth’s surface (g’) with the acceleration due to gravity on the earth’s surface (g) is g’ = g = (1-\(\frac{h}{R}\)
3. The value of the acceleration due to gravity at θ – latitude (g’) of the earth is \(g^{\prime}=g\left(1-\frac{\omega^2 R}{g} \cos ^2 \theta\right)\)
   1. At the poles, θ = 90° ; so, g’ = g.
   2. At the equator, θ = 0°; so, \(g^{\prime}=g\left(1-\frac{\omega^2 R}{g}\right)\)

If a planet of mass m revolves around the sun of mass M₀ along a circular path of radius r, then the orbital speed of that planet, v = \(v=\sqrt{\frac{G M_0}{r}}.\).

Orbital angular velocity, \(\omega=\sqrt{\frac{G M_0}{r^3}}\)

Time period of revolution, T = \(T=2 \pi \sqrt{\frac{r^3}{G M_0}}\)

The value of the escape velocity of a body from the surface of the earth, \(v_e=\sqrt{2 g R}.\).

If the mass of the earth = M, radius of the earth = R, mass of an artificial satellite = m, the orbital speed of the satellite = v and the height of the orbit above the surface of the earth = h, then the orbital speed of the artificial satellite, v \(=\sqrt{\frac{G M}{R+h}}\) and the time period of revolution of the satellite, T = \(2 \pi \sqrt{\frac{r^3}{8}}\), where r = R+h.

In the case of a satellite revolving very close to the surface of the earth, its orbital speed, \(v=\sqrt{g R}\)

and the time period of revolution, T = \(2 \pi \sqrt{\frac{R}{g}}\)

If the mass of the earth = M, the radius of the earth = R, the mass of a geostationary satellite = m, the distance of the satellite from the centre of the earth = r and its time period of revolution = T, then the orbital speed of the geostationary satellite,

v = \(\sqrt{\frac{g R^2}{r}}\)

and the height of the orbit of the geostationary satellite, r \(=\left(\frac{g R^2 T^2}{4 \pi^2}\right)^{1 / 3}\)

If an artificial satellite of mass m revolves along a circular path of radius r, the kinetic energy of the satellite,

K = \(\frac{G M M}{2 r}\) [where M = mass of the earth]

The potential energy is U = –\(\frac{G M m}{r}\)

Total energy of the satellite, E = –\(\frac{G M m}{2r}\)

Newtonian Gravitation And Planetary Motion Assertion Reason Type Questions And Answers

Direction: These questions have statement 1 and statement 2 of the four choices given below, choose the one that best describes the two statements.

1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.
2. Statement 1 is true, statement 1 is true; statement 2 is not a correct explanation for statement 1.
3. Statement 1 is true, statement 2 is false.
4. Statement 1 is true; statement 2 is false.

Question 1.

Statement 1: The force of gravitation between a sphere of mass M₁ and a rod of mass M₂ is = \(\frac{G M_1 M_2}{r} \text {. }\)

Statement 2: Newton’s law of gravitation holds correct for point masses.

Answer: 4. Statement 1 is true; statement 2 is false.

Question 2.

Statement 1: An astronaut in an orbiting space station above the earth experiences weightlessness

Statement 2: An object moving around the earth under the influence of the earth’s gravitational force is in a state of free fall.

Answer: 1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.

Question 3.

Statement 1: If time period of a satellite revolving in a circular orbit in the equatorial plane is 24 h, then it must be a geostationary satellite.

Statement 2: Time period of a geostationary satellite is 24 hours.

Answer: 4. Statement 1 is true; statement 2 is false.

Question 4.

Statement 1: Gravitational force between two masses in the air is F. If they are immersed in the water force will remain F.

Statement 2: Gravitational force does not depend on the medium between the masses.

Answer: 1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.

Question 5.

Statement 1: The binding energy of a satellite does not depend upon the mass of the satellite.

Statement 2: Binding energy is the negative value of the total energy of the satellite.

Answer: 4. Statement 1 is true; statement 2 is false.

Question 6.

Statement 1: Kepler’s laws for planetary motion are a consequence of Newton’s laws.

Statement 2: Kepler’s laws can be derived by using Newton’s laws.

Answer: 4. Statement 1 is true; statement 2 is false.

Real-Life Examples of Gravitational Effects

Newtonian Gravitation And Planetary Motion Match Column A With Column B.

Question 1.

Answer: 1. A, C, 2. A, B, C, D, 3. A, B, C, D, 4. A, C, D

Question 2. In Column A, four artificial satellites of Earth are shown. In Column B, some statements are given related to motion or other facts about the satellites.

Answer: 1. A, D, 2. B, D, 3. B, D, 4. C

Question 3.

Answer: 1. C, 2. B, 3. A

Question 4. In the elliptical orbit of a planet, as the planet moves from the apogee position to the perigee position, match the following columns.

Answer: 1. C, 2. B, 3. B, 4. A

Newtonian Gravitation And Planetary Motion Comprehension Type Questions And Answers

Read the following passages carefully and answer the questions at the end of them.

Question 1. The PE of the earth-satellite system is shown by a solid line as a function of distance r (the separation between the earth’s centre and satellite). The total energy of two objects which may or may not be bound to earth are shown in the graph by dotted lines,

1. Mark the correct statement(s).

1. The object having total energy E₁ is the bounded one
2. The object having total energy E₂ is the bounded one
3. Both objects are bounded

Answer: 1. The object having total energy E₁ is the bounded one

2. If the object having total energy E₁ is having the same PE curve as shown, then

1. r₀ is the maximum distance of the object from the earth’s centre
2. This object and earth system is a bounded one
3. The KE of the object is zero when r = r₀
4. All of the above

Answer: 4. All of the above

3. If both the objects gave the same PE curve as shown, then

1. For the object having total energy E₂, all values of r are possible
2. For the object having total energy E₂ only values of r < r₀ are possible
3. For the object having total energy E₁ all values of r are possible
4. None of the above

Answer: 1. For the object having total energy E₂, all values of r are possible

Question 2. A uniform metal sphere of radius a and mass M is surrounded by a thin uniform spherical shell of equal mass and radius 4a. The centre of the shell falls on the surface of the inner sphere.

1. Find the gravitational field intensity at point P₁.

1. \(\frac{G M}{16 a^2}\)
2. \(\frac{G M}{8 a^2}\)
3. \(\frac{G M}{2 a^2}\)
4. \(\frac{G M}{4 a^2}\)

Answer: 1. \(\frac{G M}{16 a^2}\)

2. Find the gravitational field intensity at point P₂.

1. \(\frac{21 G M}{900 a^2}\)
2. \(\frac{61 G M}{450 a^2}\)
3. \(\frac{61 G M}{900 a^2}\)
4. \(\frac{61 G M}{1800 a^2}\)

Answer: 3. \(\frac{61 G M}{900 a^2}\)

Question 3. The gravitational field in a region is given by \(\vec{E}=\left(5 \mathrm{~N} \cdot \mathrm{kg}^{-1}\right) \hat{i}+\left(12 \mathrm{~N} \cdot \mathrm{kg}^{-1}\right) \hat{j}\)

1. Find the magnitude of the gravitational force acting on a particle of mass 2 kg placed at the origin.

1. 26 N
2. 30 N
3. 20 N
4. 35 N

Answer: 1. 26 N

2. Find the potential at the points (12 m, 0) and (0,5 m) if the potential at the origin is taken to be zero.

1. -30J · kg^-1_, -30 J · kg^-1
2. -40J · kg^-1_, -30 J · kg^-1
3. -60J · kg^-1_, -60 J · kg^-1
4. -40J · kg^-1_, -50 J · kg^-1

Answer: 3. -60J · kg^-1_, -60 J · kg^-1

3. Find the change in gravitational potential energy if a particle of mass 2 kg is taken from the origin to the point (12 m, 5 m).

1. -225 J
2. -240 J
3. -245 J
4. -250 J

Answer: 2. -240 J

4. Find the change in potential energy if the particle is taken from (12 m, 0) to (0,5 m).

1. -10 J
2. -50 J
3. Zero
4. -60 J

Answer: 3. Zero

Newtonian Gravitation And Planetary Motion Integer Answer Type Questions

In this type, the answer to each of the questions Is a single-digit integer ranging from 0 to 9.

Question 1. Gravitational acceleration on the surface of a planet is (√6/11)g, where g is the gravitational acceleration on the surface of the earth. The average mass density of the planet is \(\frac{2}{3}\)times that of the earth. If the escape speed on the surface of the earth is taken to be 11 km · s^-1, the escape speed on the surface of the planet in km · s^-1 will be
Answer: 3

Question 2. A binary star consists of two stars A (mass 2.2 M_s) and B (mass 11 M_s), where M_s is the mass of the sun. They are separated by distance d and are rotating about their centre of mass, which is stationary. The ratio of the total angular momentum of the binary star to the angular momentum of star B about the centre of mass is
Answer: 6

Question 3. A particle is projected vertically upwards from the surface of the earth of radius R, with a kinetic energy equal to half of the minimum value needed for it to escape. The maximum height to which it rises above the surface of the earth is nR. What should be the value of n?
Answer: 1

Question 4. The time period of a satellite A, whose orbital radius is r₀, is T₀ and the time period of a satellite B having radius 4r₀ is T_B. Find the ratio of T_B and T₀.
Answer: 8

Question 5. Two particles of masses m₁ and m₂ are kept at a separation of r. When a third particle is kept at a distance of \(\frac{d}{3}\) from m, then it does not experience any net force. Determine the ratio \(\frac{m_2}{m_1}\).
Answer: 4

 

Newtonian Gravitation And Planetary Motion Long Answer Type Questions

WBBSE Class 11 Long Answer Questions on Gravitation

Question 1. How will the value of the acceleration due to gravity be affected if

1. The earth stops rotating
2. Does the Earth rotate faster?

Answer:

Due to the rotation of the earth, the outward centrifugal force acting on bodies on the earth’s surface and directed away from the centre of the earth decreases the weight of these bodies slightly. Hence, there is a small apparent change in g.

Apparent weight = mg’ = mg- mω² Rcos²θ; g’ = effective acceleration due to gravity, g = actual value of the acceleration due to gravity, ω = angular velocity of the earth, R = radius of the earth, θ = latitude of the place.

∴ g’ = g – ω²Rcos²θ.

1. If the earth stops rotating, ω = 0 and g’ = g. Thus the effective value of g increases by ω²Rcos²θ.
2. If the earth rotates faster, the value of co will increase. This will cause a further decrease in the apparent value of g. It is to be noted that changes in the speed of rotation of the earth about its axis do not influence the value of g at the poles.

Read and Learn More Class 11 Physics Long Answer Questions

Example 2. What will be the effect on the value of the acceleration due to gravity on the earth’s surface, if the radius of the earth suddenly reduces to half its present value and the mass remains constant?
Answer:

Let the mass of the earth be M and its radius be R.

Acceleration due to gravity on the earth’s surface, g = \(\frac{G M}{R^2}\)

When the radius changes to latex]\frac{R}{2}[/latex] and the mass remains constant, acceleration due to gravity

g’ = \(\frac{G M}{\left(\frac{R}{2}\right)^2}=\frac{4 G M}{R^2}\)

∴ g’ = 4g.

Thus, the acceleration due to gravity would be 4 times the present value.

Question 3. If the mass and the radius of the earth suddenly reduce to half their present values,

1. How many hours will be there in a day and
2. How many days will constitute a year?

Answer:

1. The time taken by the earth to spin once about its axis is the measure of a day. If the earth is taken as a uniform sphere, its moment of inertia about the axis of spinning, I = \(\frac{2}{5}\)MR² (M = mass of the earth and R = radius of the earth). Let the angular velocity of the earth be ω. When both the mass and radius of the earth become half the present values, a moment of inertia and angular velocity change to I’ and ω’.

∴ I’ = \(I^{\prime}=\frac{2}{5}\left(\frac{M}{2}\right)\left(\frac{R}{2}\right)^2=\frac{1}{8} \times \frac{2}{5} M R^2=\frac{1}{8} I\)

From the law of conservation of angular momentum \(I \omega=I^{\prime} \omega^{\prime} \text { or, } I \omega=\frac{1}{8} I \omega^{\prime} \text { or, } \omega^{\prime}=8 \omega\)

As angular velocity increases to 8 times the present value, the length of a day will be \(\frac{24}{8}\) or 3 h.

2. The time taken by the earth to revolve once around the sun (1 year) does not depend on the mass or radius of the earth. But as each day is now of 3 h duration and the total time in a year remains unchanged, it will now contain 365 x 8 = 2920 days.

Newton’s Law of Gravitation Detailed Explanations

Example 4. Explain why a person in an artificial satellite, orbiting the earth, feels himself to be weightless.
Answer:

Let the artificial satellite be in an orbit of radius r with respect to the centre of the earth. The necessary centripetal force for keeping the man of mass m moving in the orbit is supplied by the resultant of the gravitational force of the earth mg’ on the person and the upward reaction R of the plane of the satellite on the person.

That is mg’ – R = \(\frac{m v^2}{r}\)….(1)

The necessary centripetal force, to keep the satellite of mass M moving, is supplied by the gravitational attraction of the earth on the satellite.

Hence, Mg’ = \(\frac{m v^2}{r}\) or, g’ = \(\frac{v^2}{r}\)…..2)

From (1) and (2) we have, mg’ – R = mg’ or, R = 0

As the upward reaction force is zero, the person feels weightless.

Question 5. The sun and the earth both exert a gravitational force of attraction on any object on the earth’s surface. The two gravitational forces are in opposite directions at noon and in the same direction at midnight. Will an object weigh more at midnight?
Answer:

Given

The sun and the earth both exert a gravitational force of attraction on any object on the earth’s surface. The two gravitational forces are in opposite directions at noon and in the same direction at midnight.

The Earth is a planet in the solar system. We know that there is no effect of the gravitational pull of the earth on any object in an artificial satellite, i.e., the object becomes weightless.

Similarly, any object on the surface of the earth is weightless with respect to the sun, i.e., there is no effect of the sun’s gravitational force on the object. As a result, the weight of an object (which is equal to the gravitational pull of the earth experienced by the body) remains the same both at noon and at midnight.

Question 6. Does the moon have any weight?
Answer:

Any planet or satellite orbiting due to the effect of the gravitational force of the respective star or planet is weightless. Hence, the moon does not have any weight with respect to the Earth.

WBBSE Class 11 Sample Questions on Orbital Motion

Question 7. If the value of the universal gravitational constant G starts decreasing very slowly with time, what will be the effect on the motion of the moon around the Earth? Explain clearly.
Answer:

Given

If the value of the universal gravitational constant G starts decreasing very slowly with time,

Let the earth be at O and the moon, while revolving around the earth, reach point A. When the value of G remains unchanged, the moon will continue its motion along path AB, that is to say, its orbit will remain unchanged.

On the other hand, if suddenly the value of G falls to zero, the earth will have no attraction on the moon and the moon will fly off along AC, tangential to the orbit. If the value of G decreases slowly with time, the path of the moon will be towards AD, i.e., along some direction between AB and AC. As a result, the moon will gradually move away from the Earth and will take a spiral path as shown.

Question 8. Select the correct option stating the reason for your choice. A satellite is revolving around the earth in a circular path. It has

1. Constant velocity
2. Constant acceleration
3. Variable acceleration.

Answer:

An object continuously changes the direction of its velocity while moving in a circular path. Hence, the velocity is not a constant. Also, during circular motion, there is an acceleration along the radius of the orbit directed towards the centre.

The magnitude of this acceleration is constant, but its direction changes continuously. Hence, the acceleration is not constant. The acceleration of the satellite keeps changing, and therefore, option 3 is correct.

Question 9. Select the correct option stating the reason for your choice. If the mass of the earth remains the same, but its diameter is decreased by 1%, the value of the acceleration due to gravity on the surface of the earth

1. Remains the same
2. Decreases
3. Increases.

Answer:

The value of acceleration due to gravity on the earth’s surface is given by g = \(\frac{G M}{R^2}\). Thus, if M remains constant and diameter, i.e., radius R decreases, the value of g will increase. Hence, option 3 is correct.

Gravitational Potential Energy: Long Answer Questions

Question 10. Between the earth and the moon, which one has a greater escape velocity? Give reason. Or, Is the value of the escape velocity from the earth’s surface the same as that from the surface of the moon?
Answer:

Escape velocity from a planet or satellite \(v_e=\sqrt{2 g R}\)  where g = acceleration due to gravity on the surface of that planet or satellite and R = radius of that planet or satellite. Values of g and R for the earth are much greater than that for the moon. Hence, the escape velocity from the earth’s surface is greater.

Question 11. Two satellites move around the earth at the same height. The mass of one satellite Is twice that of the other. Which satellite has a greater velocity?
Answer:

Given

Two satellites move around the earth at the same height. The mass of one satellite Is twice that of the other.

The orbital speed of a satellite in a fixed orbit at a height h from the earth’s surface is \(v=\sqrt{\frac{G M}{R+h}}\), where M = mass of the earth, R = radius of the earth. Clearly, orbital speed does not depend on the mass of the orbiting body. Hence, both move with the same velocity.

Question 12. Acceleration due to gravity on the surface of a planet is 196 cm · s^-2. If It Is safe to jump from a height of 2 m on the Earth, what is the safe height for taking a jump on that planet?
Answer:

Given

Acceleration due to gravity on the surface of a planet is 196 cm · s^-2. If It Is safe to jump from a height of 2 m on the Earth,

If a man jumps from a height h and v is the velocity attained on reaching the earth’s surface, then v = \(\sqrt{2 g h}\)

Let the safe height for taking a jump on the other planet be h’. In that case, the velocity attained on reaching the planet’s surface is,

v’ = \(\sqrt{2 g^{\prime} h^{\prime}} .\)

For the problem v’ = v

∴ gh = g’h’ or, h’ = \(\frac{g}{g \prime}\) h

Given g = 9.8 m · s^-2 , h = 2m and g’ = 196 cm · s^-2 = 1.96 m · s^-2

∴ h’ = \(\frac{9.8 \times 2}{1.96}=10 \mathrm{~m}\)

Hence, it is quite safe to take a jump from a maximum height of 10 m from the surface of the planet.

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Long Answer Questions on Kepler’s Laws of Planetary Motion

Question 13. A freely falling body has no weight explains.
Answer:

Suppose, a body of mass m resting on an imaginary plane is falling down with an acceleration due to gravitational attraction. If the force of reaction of the plane on the body be R (upwards), mg – R = ma or, R = m(g-a)

For a freely falling body a = g and thus R – m(g- g) = 0.

∴ The reaction force acting on the body is zero and thus its apparent weight is zero.

Question 14. A body is released in space from an artificial satellite moving in a fixed orbit. What happens to the body?
Answer:

A body is released in space from an artificial satellite moving in a fixed orbit.

The released body will revolve around the earth in the same orbit and with the same velocity as that of the It should be noted that if a small body is projected with an initial velocity from an orbiting satellite out into space, there will be a change in the orbit of the body. Using the principle of conservation of angular momentum, in this case, it can be shown that there will be a small change in the orbit of the satellite.

Question 15. Why are tidal waves not observed in a large lake?
Answer:

There is no change in the total volume of water in a lake or sea, due to the gravitational attraction of the moon or the sun. In that portion of the sea surface, where the gravitational attraction is high, sea water accumulates causing high tide. Sea water flows from the area of minimum attraction to the area of maximum attraction and low tide occurs at the place where the attraction is minimum.

However, irrespective of the size, no lake is very large; the gravitational pull of the moon practically remains the same at any two places on the lake’s surface at a time. As there is no scope for the flow of water from one place to another in a lake, tidal waves are not observed even on the surfaces of large lakes.

Question 16. What is the condition for setting up an artificial satellite at a height h from the earth’s surface? (Radius of the earth = R, mass of the earth = M)
Answer:

Suppose height of the artificial satellite from the earth’s surface is h.

Hence, the radius of the orbit is (R+ h). If v is the orbital speed, the centripetal force necessary to keep the satellite in that orbit is \(\frac{m v^2}{(R+h)}\) (m = mass of the satellite).

If the mutual force of gravitation between the earth and the satellite can supply this centripetal force, the satellite would be in the right Gravitational force between the satellite and the earth
is \(\frac{G M m}{(R+h)^2}\)

Hence, the necessary condition is \(\frac{m v^2}{R+h}=\frac{G M m}{(R+h)^2} \text { or, } v=\sqrt{\frac{G M}{R+h}}\)

Question 17. At what velocity will a rocket escape the or, gravitational pull of the Earth?
Answer:

To be free from the gravitational pull of the earth, a rocket need not have any minimum velocity. When the fuel in the rocket bums, a reaction force acts on the rocket. When the average reaction force exceeds the average gravitational attraction of the earth on the rocket, it escapes the earth’s gravitational field.

Question 18. What will be the feeling of an astronaut inside a satellite about his weight when the satellite is in the process of being launched by a rocket?
Answer:

The velocity of the satellite is increased rapidly with the help of the rocket as the satellite is launched from the earth’s surface. The upward acceleration of the satellite sometimes increases up to 15 times the value of the acceleration due to gravity.

If the mass of the astronaut = m, the upward reaction of the plane of the satellite on the astronaut = R and acceleration of the satellite a = 15g, then R- mg = ma = m – 15g or, R = 16mg This is the apparent weight of the astronaut. He feels 16 times heavier than his actual weight.

Question 19. A piece of matter of mass m is thrown up vertically from the earth’s surface and it rises up to a height R. (Radius of the earth = R.) What is the initial velocity of the piece of matter? Show that the increase in potential energy of the piece of matter = \(\frac{1}{2}\)mgR.
Answer:

Given

A piece of matter of mass m is thrown up vertically from the earth’s surface and it rises up to a height R. (Radius of the earth = R.)

The potential energy of a body of mass m on the surface of the earth, i.e., at a distance R from the centre of the earth = –\(\frac{G M m}{R}\).

If the initial upward velocity is u, its kinetic energy = \(\frac{1}{2}\)mu². Thus, the total energy of the body on the surface of the earth = \(\frac{1}{2}\)mu² – \(\frac{G M m}{R}\).

At a height R from the earth’s surface, which is at a distance 2R from the centre of the earth, potential energy = –\(\frac{G M m}{2R}\). As per the given condition, kinetic energy at that height = 0. Hence, the total energy = –\(\frac{G M m}{2R}\)

From the law of conservation of energy \(\frac{1}{2} m u^2-\frac{G M m}{R}=-\frac{G M m}{2 R}\)

or, \(\frac{1}{2} m u^2=\frac{G M m}{2 R}\)

or, initial velocity \(u=\sqrt{\frac{G M}{R}}\)

Also, as per the law of conservation of energy, an increase in potential energy = a decrease in kinetic energy

= \(\frac{1}{2} m u^2=\frac{G M m}{2 R}=\frac{1}{2} m \frac{G M}{R^2} \cdot R=\frac{1}{2} m g R .\)

Understanding Planetary Motion: Long Answer Format

Question 20. The force of gravitation on all objects Is directly proportional to their masses, yet a heavy body does not fall faster than a light body. Why?
Answer:

Given

The force of gravitation on all objects Is directly proportional to their masses, yet a heavy body does not fall faster than a light body.

If the force of gravity on a body is F and the mass of the body is m, then F ∝ m

or, F = km [ k = proportionality constant]

or, \(\frac{F}{m}\) = k ….(1)

Suppose mass m gains an acceleration of an under the action of force F. Hence, from Newton’s second law of motion,

F = ma or, \(\frac{F}{m}\) = a…….(2)

From (1) and (2), a = k

Hence, acceleration produced on any mass is a constant, so all bodies fall to the earth with equal speed.

Question 21. What will be the change In the mutual force of gravitation between two bodies when the mass of each body as well as the distance separating them is doubled?
Answer:

In the first case, the force of gravitation \(F_1=\frac{G m_1 m_2}{r^2}\) where m₁ and m₂ are masses of the two bodies and r is the distance separating them.

In the second case, the force of gravitation \(F_2=G \cdot \frac{2 m_1 \cdot 2 m_2}{(2 r)^2}=G \cdot \frac{m_1 m_2}{r^2}\)

∴ F₂ = F₁, i.e., there will be no change in the gravitational force.

Question 22. Show that the orbital speed of an artificial satellite increases as the energy of the satellite decreases.
Answer:

Total energy ofan artificial satellite = –\(\frac{G M m}{2r}\) and its kinetic energy = +\(\frac{G M m}{2r}\) . In this case, M = mass of the earth, m = mass of the artificial satellite and r = distance of the satellite from the centre of the earth.

When the total energy of the satellite starts decreasing, i.e., the value of –\(\frac{G M m}{2r}\) starts falling, the value of the kinetic energy +\(\frac{G M m}{2r}\)starts increasing. As the orbital speed is directly proportional to the square root of the kinetic energy, it also increases.

Question 23. The value of the acceleration due to gravity at a height h from the surface of the earth is g₁ and that at a depth h below the earth’s surface is g₂. Show that(radius of earth R>>h) \(\frac{g_2}{g_1}=1+\frac{h}{R}\).
Answer:

Given

The value of the acceleration due to gravity at a height h from the surface of the earth is g₁ and that at a depth h below the earth’s surface is g₂.

It is known \(g_1=1-\frac{2 h}{R}\) and \(g_2=1-\frac{h}{R}\)

∴ \(\frac{g_2}{g_1}=\frac{1-\frac{h}{R}}{1-\frac{2 h}{R}}=\left(1-\frac{h}{R}\right)\left(1-\frac{2 h}{R}\right)^{-1}\)

= \(\left(1-\frac{h}{R}\right)\left(1+\frac{2 h}{R}\right)[because R \gg h]\)

= \(1-\frac{h}{R}+\frac{2 h}{R}[because R \gg h]\)

= \(1+\frac{h}{R}\)

Question 24. If the acceleration due to gravity at a height h from the surface of the earth is the same as that at a depth h below the surface of the earth, prove that h = \(\frac{\sqrt{5}-1}{2} R\) (R = radius of the earth).
Answer:

Given

If the acceleration due to gravity at a height h from the surface of the earth is the same as that at a depth h below the surface of the earth

Acceleration due to gravity on the earth’s surface, g = \(\frac{G M}{R^2}\)

Acceleration due to gravity at a height h from the earth’s surface \(g_1=\frac{G M}{(R+h)^2}\)

Hence, \(\frac{g_1}{g}=\frac{R^2}{(R+h)^2}\)

Also, taking the earth as a uniform sphere of radius R, acceleration due to gravity at a depth h

⇒ \(g_2=g \frac{(R-h)}{R}\)

∴ \(\frac{g_1}{g_2}=\frac{R^3}{(R+h)^2(R-h)}\)

If \(g_1=g_2\)

⇒ \(R^3=\left(R^2-h^2\right)(R+h)=R^3+R^2 h-h^2 R-h^3 \)

or, \(h^2+R h-R^2=0\)

∴ h = \(\frac{-R \pm \sqrt{R^2+4 R^2}}{2}=\frac{(-1 \pm \sqrt{5}) R}{2}\)

As a negative value of h is not acceptable, h= \(\frac{\sqrt{5}-1}{2} R\)

Applications of Newtonian Gravitation: Long Answers

Question 25. The force of attraction of the earth on an apple is of the same magnitude as the force of attraction of the apple on the earth, yet the apple moves towards the earth, but the earth does not move towards the apple. Why?
Answer:

Given

The force of attraction of the earth on an apple is of the same magnitude as the force of attraction of the apple on the earth, yet the apple moves towards the earth, but the earth does not move towards the apple.

Suppose the earth of mass M attracts an apple of mass m with a force F towards it. As per the law of gravitation, the apple also attracts the earth with the same force F towards it.

Acceleration of the apple due to the earth’s pull

= \(\frac{\text { force on apple }}{\text { mass of apple }}=\frac{F}{m}\)

Acceleration of the earth due to the attraction of the apple = \(\frac{\text { force on the earth }}{\text { mass of the earth }}=\frac{P}{M}\)

Hence, \(\frac{\text { acceleration of the apple }}{\text { acceleration of the earth }}=\frac{\frac{F}{m}}{\frac{F}{M}}=\frac{M}{m}\)

As the mass of the apple is negligible compared to that of the earth, \(\frac{M}{m}\)>>1.

That is to say, acceleration of the apple >> acceleration of the earth. Practically, the apple moves towards the earth. The motion of the earth towards the apple is too small to be noticed.

Question 26. In which case will the decrease in the value of the acceleration due to gravity be greater (with respect to that on the earth’s surface)—at 1 km above the earth’s surface or 1 km below the earth’s surface?
Answer:

Let the radius of the earth = R

Value of acceleration due to gravity on the earth = g

Decrease in g at a height h = \(\frac{2 h g}{R}\)…..(1)

and decrease in g at a depth h = \(\frac{h g}{R}\)….(2)

Comparing the two equations, it can be concluded that the decrease in the acceleration due to gravity at a height of 1 km above the earth’s surface will be greater.

Question 27. Due to some reason, if the average distance of the earth from the sun decreases, will the length of a year Increase or decrease? Give reasons for your answer.
Answer:

If the average distance of the earth from the sun is r and the time period of revolution of the earth around the sun is T, then from Kepler’s law T² ∝ r³. Hence, with the decrease in r, the span of a year (T) would decrease.

Question 28. What happens to a satellite if its orbital speed is greater than the escape velocity corresponding to that orbit?
Answer:

In any orbit, the orbital speed of an artificial satellite is less than the escape velocity from the planet. Now, for some reason, if the orbital speed exceeds the escape velocity, the gravitational attraction of the planet will not be able to keep the satellite in its orbit. The satellite will move out of the gravitational field of the planet following a spiral path.

Question 29. Why is the span of a year smaller for a planet closer: to the sun?
Answer:

As per Kepler’s law (time period of revolution)² ∝ (radius of the orbit)³ Radius of the orbit for a planet closer to the sun is smaller. Hence, the time period, i.e., the span of a year is also smaller.

Question 30. Suppose there existed a planet that went around the sun eight times as fast as the Earth. What would be its orbital size as compared to that of the Earth?
Answer:

If the period of revolution of the earth around the sun is T, the period of the planet is \(\frac{T}{8}\). If the semi-major axes of their elliptical orbits are r₁ and r₂ respectively, then from Kepler’s law,

⇒ \(\left(\frac{T}{T / 8}\right)^2=\left(\frac{r_1}{r_2}\right)^3\)

or, \(\frac{r_1}{r_2}=8^{2 / 3}=4, \text { or, } r_2=\frac{r_1}{4}\)

So, the orbital size of the planet is 1/4 th that of the earth.

Question 31. The force of gravitation between two spherical masses M and m, placed in air, is F. The space between the two spheres is filled with a liquid of relative density 3. What will be the change in the value of the gravitational force acting between the spheres?
Answer:

Given

The force of gravitation between two spherical masses M and m, placed in air, is F. The space between the two spheres is filled with a liquid of relative density 3.

The force of gravitation does not depend on the properties of the medium in which the two bodies are placed. Hence, the force in the second case is also F.

Question 32. The conservation of what physical quantity does Kepler’s second law refer to?
Answer:

Kepler’s second law refers to the law of conservation of angular momentum.

According to Kepler’s second law, the area swept out by the line joining the Earth and the sun in unit time is a constant.

Area swept out in unit time by the line

= \(\frac{1}{2}\)rω x r =  \(\frac{1}{2}\)r²ω = k (constant)

In this case,

r = distance of the earth from the sun

ω = angular velocity of the earth around the sun

If the linear speed of the earth is v and its mass is m, the angular momentum of the earth

= mv x r = mω² = mx 2k = constant

Thus, the angular momentum is conserved.

Question 33. What is the appearance of a communicating satellite from its plane of projection?
Answer:

Communication satellites are actually geostationary satellites. This artificial satellite revolves around the earth with an angular velocity equal to that of the diurnal motion of the earth (and follows the same spinning direction), i.e., it revolves around the earth in 24 hours.

Hence, the angular velocity of the satellite becomes zero with respect to the angular velocity of the earth. Hence, when observed from the plane of projection, i.e., from the earth’s surface, the satellite will appear to be stationary in the sky.

Question 34. Is it possible to place an artificial satellite in an orbit such that it is always visible in the sky of Lucknow or New Delhi? Give the reason.
Answer:

Only a geostationary satellite, which revolves around the earth on the equatorial plane, may always be visible directly overhead from someplace on the equator. Lucknow and New Delhi are not on the equator (nearly at 25°N latitude).

So, if a geostationary satellite is on this side of the earth above the equator, it would always be visible from Lucknow or New Delhi in an oblique direction towards the south, not directly overhead.

Question 35. The gravitational potential energy of a body on the surface of the earth is -6.4 x 10⁶ joule. Explain the statement.
Answer:

The gravitational potential energy of a body on the surface of the earth is -6.4 x 10⁶ joule.

The meaning of the statement is that 6.4 x 10⁶ joule of energy would be required to send away the body outside the gravitational field of the earth.

Question 36. If T is the period of revolution of an artificial satellite orbiting very close to the earth’s surface and ρ is the density of the earth, then show that ρT² is a universal constant.
Answer:

Given

If T is the period of revolution of an artificial satellite orbiting very close to the earth’s surface and ρ is the density of the earth,

The period of revolution of an artificial satellite orbiting very close to the Earth’s surface is given by

T =  \(2 \pi \sqrt{\frac{R}{g}}\); R = radius of the earth

g = acceleration due to gravity on the earth’s surface

We know that the average density of the earth is given by \(\rho=\frac{3 g}{4 \pi G R} \text { or, } \frac{R}{g}=\frac{3}{4 \pi G \rho}\)

∴ T \(=2 \pi \sqrt{\frac{3}{4 \pi G \rho}}\)

or, \(T^2=\frac{4 \pi^2 \times 3}{4 \pi G \rho} or, \rho T^2=\frac{3 \pi}{G}\) (a universal constant)

Question 37. Three particles, each of mass m, are placed at the vertices of an equilateral triangle. What is the force acting on a particle of mass 2m placed at the I centroid D of the triangle?
Answer:

Given

Three particles, each of mass m, are placed at the vertices of an equilateral triangle.

The positions of the particles are shown.

Let DA = DB = DC = x.

The gravitational forces on the particle at D, due to the particles at A, B, and C are equal, with value \(\frac{G(2 m)(m)}{t^2}\) directed as shown.

By symmetry the resultant force on 2m =0.

Question 38. Acceleration due to gravity decreases as we go below the surface of the earth. What will be the nature of the graph between the change of acceleration due to gravity and the depth below the surface of the earth?
Answer:

Given

Acceleration due to gravity decreases as we go below the surface of the earth.

The acceleration due to gravity at a depth d below the surface of the earth is given by, \(g^{\prime}=g\left(1-\frac{d}{R}\right)=g-g \frac{d}{R} \text { or, } g-g^{\prime}=g \frac{d}{R}\)

Since g and R are constants, g-g’ ∝ d

So the graph between (g-g’) and d will be a straight line.

Newtonian Gravitation And Planetary Motion Very Short Answer Type Questions

WBBSE Class 11 Very Short Answer Questions on Gravitation

Question 1. What is the name of the mutual force acting between any two bodies in the universe?
Answer: Gravitation

Question 2. What is the value of G in the CGS system?
Answer:

The value of G in the CGS system

6.67 x 10^-8 dyn · cm² · g^-2

Question 3. What is the unit of G in SI?
Answer:

The unit of G in SI

N · m² · kg^-2

Read And Learn More WBCHSE Class 11 Physics Very Short Question And Answers

Question 4. What is the unit of gravity in SI?
Answer:

The unit of gravity in SI

Newton or N

Question 5. State whether the value of g increases or decreases due to an increase in depth below the earth’s surface.
Answer: Decreases

Question 6. Does the friction arise due to gravitation?
Answer: No

Question 7. What will be the force of attraction of 1 kg lead on the earth and the force of attraction of the earth on 1 kg lead?
Answer: Equal

Question 8. What is the value of G in SI?
Answer:

The value of G in SI

6.67 x 10^-11 N · m² · kg^-2

Question 9. What is the unit of G in the CGS system?
Answer:

The unit of G in the CGS system

dyn · cm²  · g^-2

Question 10. What is the dimension of G?
Answer:

The dimension of G

M^-1L³T^-2

Question 11. ‘The gravitational force is a conservative force’ true or false?
Answer: False

Question 12. Is the gravitational force between two bodies in air equal to that inside water?
Answer: Yes

Newton’s Law of Gravitation VSA Questions

Question 13. Does gravitational force depend on temperature?
Answer: No

Question 14. State whether your weight will increase or decrease if the daily rotation of the earth ceases.
Answer: Increase

Question 15. What will be the gravitational potential on the surface of a planet of mass M and radius R?
Answer: –\(\frac{GM}{R}\)

Question 16. If the potential and kinetic energies of an artificial satellite revolving around the earth are U and K respectively, what will be the relation between U and K?
Answer: U = -2K

Question 17. If a body of mass m is at a distance r in the gravitational field of a body of mass M, what will be the gravitational potential energy of the body?
Answer: –\(\frac{G M m}{r}\)

Question 18. What is the weight of a body at the centre of the earth?
Answer: Zero

Question 19. What is the value of the acceleration due to gravity at the centre of the earth?
Answer: Zero

Question 20. Where does a body weigh more at the pole or at the equator?
Answer: At the pole

Question 21. What is the value of the acceleration due to gravity of all bodies (heavy or light) at a particular place?
Answer: Same

Question 22. Write down the relation between g and G.
Answer: g = \(\frac{G M}{r^2}\)

Question 23. What is the relation between the mean density of the earth and G?
Answer: \(\rho=\frac{3 g}{4 \pi G R}t\)

Question 24. Write down the variation of the acceleration due to gravity at a point above the surface of the earth to the square of the distance of that point from the centre of the earth.
Answer: Inversely proportional

Question 25. ‘The value of g at the poles is greater than its value at the equator.’—true or false?
Answer: True

Question 26. What is the value of the acceleration due to gravity on the surface of the earth?
Answer: Maximum

Kepler’s Laws of Planetary Motion Short Answers

Question 27. The decrease in the acceleration due to gravity at a height above the earth’s surface is _______ the decrease in the acceleration due to gravity at the same depth below the surface of the earth.
Answer: Double

Question 28. The acceleration due to gravity at a height above the earth’s surface is ______ than that at the same depth below the surface of the earth.
Answer: Less

Question 29. Give the value of the angular velocity of the earth for which the acceleration due to gravity at the equator would be zero.
Answer: 1.24 x 10^-3 rad · s^-1

Question 30. If the diurnal motion of the earth ceases, write the variation of g at all places except at the poles.
Answer: Increase

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Question 31. If the angular velocity of the earth is increased, write the variation of g at all places except at the poles.
Answer: Decrease

Question 32. If the radius of the earth is halved keeping its mass unchanged, then the acceleration due to gravity would be _____ times its previous value.
Answer: 4

Question 33. During free fall, the weight of a body becomes _______
Answer: Zero

Question 34. What will be the value of acceleration due to gravity on the earth’s surface if its radius is decreased by 1%, keeping the mass unchanged?
Answer: Increase by 2%

Question 35. Where is the escape velocity of a body greater from the earth or from the moon?
Answer: From the earth

Question 36. How does the escape velocity of a body from the earth depend on the mass of the body thrown?
Answer: Does not depend on the mass of the body

Question 37. From the surface of a planet, the escape velocity of all bodies— heavy or light—is ________
Answer: Equal

Question 38. From the surface of the moon, give the value of the escape velocity.
Answer: 2.4 km · s^-1

Question 39. What is the relation between orbital and escape velocity?
Answer: \(v_e=\sqrt{2 v_0}\)

Question 40. When a body coming from infinity falls on the earth’s surface, give its velocity.
Answer: \(\sqrt{2} g R\)

Question 41. The orbital speed of a planet around the sun depends on the mass of the planet—is this statement true or false?
Answer: False

Gravitational Potential Energy VSA Questions

Question 42. A comet orbits the sun in a highly elliptical orbit. Does the comet have a constant?

1. Linear speed,
2. Angular speed,
3. Angular momentum,
4. Kinetic energy,
5. Potential energy,
6. Total energy throughout its orbit?

Neglect any mass loss of the comet when it comes very close to the sun.

Answer:

1. Not a constant,
2. Not a constant,
3. Constant,
4. Not a constant,
5. Not a constant,
6. Constant.

Question 43. The time period of revolution of an artificial satellite in an orbit just outside the equatorial plane depends only on the mean density of the earth. Is this statement true or false?
Answer: True

Question 44. An artificial satellite appears to be stationary with respect to an observer on the earth. At what height (approx.) above the earth’s surface is it placed?
Answer: 36000 km

Question 45. What is the name of the orbit where an artificial satellite appears to be at rest from the Earth?
Answer: Parking orbit

Question 46. What time will a polar satellite take to revolve around the earth once?
Answer: 2h

Question 47. In which plane of the earth, does the orbit of a geostationary satellite always lie?
Answer: Equatorial plane

Question 48. Write down one most important applications of geostationary satellites.
Answer: As a communication satellite

Question 49. Assuming that the earth revolves around the sun in a circular orbit, the line joining the earth and the centre of its orbit covers ______ area.
Answer: Constant

Question 50. If the distance between the sun and the earth were half the present value, the span of a year would be __________ days.
Answer: 129

Question 51. If two satellites revolve in two different circular orbits, the orbital speed of the external satellite will be _______ than that of the inner satellite.
Answer: Less

Acceleration Due to Gravity Short Answer Questions

Question 52. ‘The orbital speed of a satellite of the earth does not depend on the mass of the satellite’— true or false
Answer: True

Question 53. Two satellites A and B are revolving around the Earth at the same height. The mass of the satellite A is twice that of B. What will be the speed of A that of B?
Answer: Equal

Question 54. If the orbital speed of an artificial satellite revolving very close to the earth is v and the escape velocity from the surface of the earth is v_e, then v_e = _________ x v.
Answer: √2

Question 55. A geostationary satellite orbits the earth once in 24 h in the __________ of the diurnal motion of the earth.
Answer: Direction

Question 56. If the radius of the earth is R, what will be the distance of a geostationary satellite from the centre of the earth?
Answer: Approximately 7R

Question 57. A person inside an artificial satellite revolving around the earth feels weightless’—true or false?
Answer: True

Question 58. The orbit of a polar satellite lies in the __________ plane.
Answer: Polar meridian

Question 59. The total energy of an artificial satellite of mass m revolving around the earth of mass M is _________. The radius of the orbit of the satellite = r.
Answer: –\(\frac{G M M}{2 r}\)

Question 60. What will be the length of a year if the distance between the earth and the sun is increased to twice the present value?
Answer: 1032 days

Question 61. What is the dimension of gravitational intensity?
Answer:

Short Answers on Orbital Motion for Class 11

The dimension of gravitational intensity

LT^-2

Question 62. What is the unit of the intensity of the gravitational field?
Answer:

The unit of the intensity of the gravitational field

N · kg^-1

Question 63. Where is the weight of a body greater—at the equator or at the poles?
Answer: At the poles

Question 64. State whether your weight will increase or decrease if the angular velocity of the earth increases.
Answer: Decrease

Escape Velocity Very Short Answer Questions

Question 65. State whether the value of g increases or decreases if one moves from the equator to the pole.
Answer: Increases

Question 66. Where on the surface of the earth is the value of g least due to the diurnal motion of the earth?
Answer: At the equator

Question 67. If the force of gravity ceases to act suddenly, then all bodies would become weightless’—true or false?
Answer: True