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Circular Motion Long Answer Type Questions

Long Answer Questions on Circular Motion for Class 11

Question 1. Why is a centripetal force necessary for a uniform circular motion?
Answer:

A centripetal force necessary for a uniform circular motion because

According to Newton’s first law of motion, if there is no external force acting on a body, the body remains at rest or moves with uniform velocity. So, to rotate a body along a circular path, its inertia has to be overcome.

For this reason, an external force has to be applied to the body. This external force acts towards the centre of the circular path radially and is called the centripetal force. In the absence of this force, uniform circular motion is not possible.

Read and Learn More Class 11 Physics Long Answer Questions

Question 2. Centripetal force is called a real force, but centrifugal force is called a pseudo force. Give reasons in support of this statement.
Answer:

Centripetal force is called a real force, but centrifugal force is called a pseudo force.

The force that arises due to a mutual action-reaction process is called a real force. We know that in the absence of any external force, a moving body continues to move with uniform velocity.

• So, with respect to any inertial frame of reference, when a body moves along a circular path, at every moment its direction of motion changes.
• This change in direction of motion is possible when an external force acts on the body. This external force is the centripetal force.
• As centripetal force is an external force acting on a rotating body, it is a real force.
• On the other hand, in an accelerated or rotating frame of reference, i.e., in a non-inertial frame of reference, a body is acted upon by a force whose direction is opposite to the direction of the centripetal acceleration of the frame.
• Since this force is not generated due to a mutual action-reaction process between different bodies, it is a pseudo force.
• This force has no existence in any inertial frame of reference. This force, which arises due to an acceleration of the reference frame, can be felt only in the non-inertial frame of reference.
• In a rotating frame of reference, the pseudo force, which acts on a body and is equal but opposite to the centripetal force is called centrifugal force.

Question 3. What should be the length of a day on the earth when a body has no apparent weight at the equatorial region? (Radius of the earth = 6400 km) Or, What should be the time period of rotation of the earth about its own axis so that a person on the equator feels weightless? (Equatorial radius = 6400 km ,g = 9.8m · s^-2)
Answer:

Weight mg of a body acts towards the centre of the earth. If the radius of the earth is R, at the equatorial region, the centrifugal force experienced by the body is mω²R, where, ω is the angular velocity of the earth about its own axis.

When the centrifugal force acting on the body becomes equal and opposite to the weight of the body, the apparent weight of the body becomes zero. In that case,

mg = \(m \omega^2 R\)

or, \(\omega=\sqrt{\frac{g}{R}}\) = time period of rotation of the earth (T)

= \(\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{R}{g}}=2 \pi \times \sqrt{\frac{6400 \times 10^3}{9.8}}=5077.6 \mathrm{~s}\)

= \(1 \mathrm{~h} 24.6 \mathrm{~min}\).

Question 4. What is the maximum speed with which a car can move over a convex bridge?
Answer:

To move over a convex bridge, the car requires a centripetal force. This centripetal force is provided by the weight of the car.

Let us consider that when the car moves with a certain velocity v, the weight of the car is just sufficient to provide that necessary centripetal force; but if the car moves with a velocity greater than v, the car loses its contact with the bridge.

Let the mass of the car be m and the radius of the circular path be r. So, the condition for the car to not lose contact with the surface of the bridge is \(\frac{m v^2}{r}\) ≤ m g or,  v ≤ \(\sqrt{g r}\)

So, a car can move with a maximum speed of √gr over a convex bridge of radius of curvature r such that it does not lose contact with the bridge.

Question 5. Radius of a curved path is r and the coefficient of friction between the wheel of a car and that path is μ. What should be the maximum speed with which the car can take a turn without skidding in that curved path?
Answer:

Weight of the car = mg; so normal force offered by the path = mg.

Hence, limiting frictional force = μmg.

[μ = coefficient of friction between the wheel of the car and the path]

The limiting frictional force provides the centripetal force necessary to prevent skidding.

So, if the maximum speed of the car is v, then the condition for no skidding in that path is, \(\frac{m v^2}{r}=\mu m g \quad \text { or, } \quad v=\sqrt{\mu r g} \text {. }\)

Question 6. While taking a turn, is it possible for a cyclist to lean | at an angle of 45° with the vertical?
Answer:

We know that when a cyclist takes a turn along a circular path of radius r with velocity v, he leans at an angle of θ with the vertical.

In this case, \(\tan \theta=\frac{v^2}{r g}\)

Again, normal force on the cycle = mg, where m is the mass of the cycle along with the cyclist. So, if the coefficient of friction is μ, then limiting frictional force = μmg. This limiting frictional force provides the necessary centripetal force.

Hence, \(\frac{m v^2}{r}=\mu m g \quad \text { or, } \quad \mu=\frac{v^2}{r g}\)…(2)

From the equations (1) and (2) we get, tanθ = μ.

It is given that θ = 45°; so, μ = tan45° = 1.

But the coefficient of friction between the wheel and the road can never be 1; actually, it is less than 1.

Hence, the cycle will skid when the cyclist leans at an angle of 45° with the vertical.

Examples of Long Answer Questions in Circular Motion

Question 7. A stone is tied to the end of a string. When the string is whirled in a circular path, the string can never be kept horizontal. Explain the reason.
Answer:

When a stone tied to the end of a string is whirled along a horizontal circular path, the horizontal component of the tension in the string provides the necessary centripetal force and its vertical component balances the weight of the stone.

If the stone is rotated along a horizontal circular path by keeping the string horizontal, there would be no vertical component of the tension (T) in the string. In that case, the weight of the body (mg) cannot be balanced.

As a result, the stone would dip downwards leaving the horizontal circular path until the vertical component of T would be just sufficient to balance the weight of the stone. Hence, during rotation of the stone along a horizontal circular path, the string can never be kept horizontal.

Question 8. When a body is hung from a string, it does not snap, But when the same mass is set into rotation along a horizontal path at high speed holding the other end of the string, the string snaps. What is the reason?
Answer:

Given

When a body is hung from a string, it does not snap, But when the same mass is set into rotation along a horizontal path at high speed holding the other end of the string, the string snaps.

If the mass of the body is m, then in the first case, tension in the string, T’ = mg ….(1)

Let the length of the thread be l, linear velocity of the body be v, and tension in the thread be T, when the body is rotated in the horizontal plane.

Required centripetal force for revolution, \(\frac{m v^2}{l \sin \theta}=T \sin \theta\)…(2)

Also, \(m g=T \cos \theta\)…(3)

From equations (1) and (3) we get, \(T^{\prime}=T \cos \theta\)

∴ 0 ≤ θ <\(\frac{\pi}{2}\)

∴ 0< cosθ ≤1

∴ \(T^{\prime}<T\)

So tension in the thread is greater in the second case than that in the first case, and hence, in the first case though the thread does not snap, in the second case the thread may snap.

Question 9. Two identical trains are running in opposite directions over two tracks along the equator at equal speed. Will both the trains exert the same force on the tracks?
Answer:

Both trains will not exert the same force on the tracks. The train, running towards the east from the west, possesses a greater angular velocity than that of the earth with respect to the axis of rotation of the earth. This is because the earth is also rotating from the west to the east with respect to the same axis.

• As a result, the outward centrifugal force, which is directly proportional to aω², will increase. Hence, the apparent weight of the train will be reduced. The apparent weight is the difference of the real weight and the centrifugal force on the train.
• Consequently, the train will exert comparatively less force on the tracks. On the other hand, the train running towards the west from the east will have an increase in its apparent weight, and hence, will exert a comparatively greater force on the tracks.

Question 10. A hollow cylinder of radius r is rotating about its own vertical axis. Both ends of the cylinder are open. A piece of stone of mass m remains fixed on the inner side of the cylinder. What should be the minimum velocity of rotation of the cylinder so that the stone remains fixed on the surface of the cylinder without falling down? Coefficient of static friction between the wall of the cylinder and the stone = μ, acceleration due to gravity = g. Also, prove that this velocity is independent of the mass of the stone.
Answer:

Given

A hollow cylinder of radius r is rotating about its own vertical axis. Both ends of the cylinder are open. A piece of stone of mass m remains fixed on the inner side of the cylinder.

The forces acting on the piece of stone are

1. Weight of the stone, W = mg
2. Limiting frictional force, F = μR

[ft = coefficient of friction, R = normal force of the wall,]

Minimum velocity of rotation = v;

The radius of the cylinder = r.

In equilibrium (i.e., when the stone remains fixed to the wall),

W = mg = F or, mg = μR

Again, R = \(\frac{m v^2}{r}\)

∴ mg = \(\mu \cdot \frac{m v^2}{r}\text { or, } v^2=\frac{g r}{\mu}\text { or, } v=\sqrt{\frac{g r}{\mu}}\)

The above expression is independent of mass.

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Practice Long Answer Questions on Uniform Circular Motion

Question 11. When a moving bus takes a right turn, the passengers inside the bus seem to lean towards the left Explain why.
Answer:

When a bus takes a turn, the centrifugal force felt by the passengers is directed opposite to the direction of the turn. Hence, when a bus takes a right turn, the passengers inside the bus seem to lean towards the left.

Question 12. A small glass marble is put on a smooth gramophone disc. When the disc starts rotating, the marble flies off the disc. Explain why.
Answer:

When the disc starts rotating, the marble flies off because the frictional force between the surface of the gramophone disc and the marble cannot provide the necessary centripetal force required by the marble for its revolution.

Question 13. In a circus show, a motorcycle rider inside a ‘death- -well’ can revolve on the erect wall without falling down. What is the reason behind it?
Answer:

While revolving on the erect wall inside the well, the rider derives the necessary centripetal force for revolving in a circular path from the horizontal component of the normal force. Due to the presence of this normal force, an upward frictional force is generated which balances the combined weight of the motorcycle and the cyclist. For this reason, the rider does not fall down.

Question 14. Why should a signboard mentioning the safe maximum speed be cited before the bend on a horizontal road or railway track?
Answer:

If the speed of a vehicle is more than the maximum safe limit written on the board, the frictional force acting on the tyres of the vehicle by the road cannot provide the necessary centripetal force for its turning, and hence, there is a chance of skidding.

To avoid an accident, the driver of the vehicle should know the maximum speed before he reaches the bend. The signboard provides him with that knowledge.

Question 15. When a motor car travels on a convex road the passengers inside feel lighter. Why?
Answer:

When a motor car travels on a convex road, the resultant of the weight (W) of any passenger inside it and the upward normal force (R) on the surface of the car supplies the necessary centripetal force to the passenger to travel along the convex road (i.e., circular path),

i.e., \(W-R=\frac{m v^2}{r}\),

where, m = mass of the passenger,

v = linear velocity of the car

and r = radius of the circular path

or, R = \(W-\frac{m v^2}{r}\)…(1)

Due to this normal force R, the passenger feels this weight. M=Now, according to equation (1), since R < W, the passenger feels lighter.

Question 16. A funnel is rotating around its vertical axis with a constant frequency ν rev/s. A small cube of mass m is placed on the inside wall of the funnel carefully. The wall of the funnel makes an angle θ with the horizontal. If μ is the coefficient of friction between the funnel and the cube, r is the distance between the centre of mass of the cube and the rotational axis, then find the maximum and minimum value of v for which the cube remains static with respect to the funnel.

Answer:

Given

A funnel is rotating around its vertical axis with a constant frequency ν rev/s. A small cube of mass m is placed on the inside wall of the funnel carefully. The wall of the funnel makes an angle θ with the horizontal. If μ is the coefficient of friction between the funnel and the cube, r is the distance between the centre of mass of the cube and the rotational axis

Frequency of rotation of the funnel = ν rev/s.

The centripetal force required for the cube of mass m to rotate around the circular path of radius r is,

F_r = mω²r = m(2πv)²r = 4π²ν²mr

Let the block slide down the wall of the funnel when the funnel is at rest. Suppose, v₁ is the minimum value of frequency for which the block remains static. At that time, the direction of the frictional force will be upward along the walls of the funnel.

Balancing the forces on the cube, we get

mg =(ncosθ + fsinθ) [n = normal reaction, f = limiting value of the static friction]

= \(n \cos \theta+\mu n \sin \theta\)

= \(n(\cos \theta+\mu \sin \theta)\)…(1)

and \(m \omega_1^2 r=n \sin \theta-f \cos \theta=n \sin \theta-\mu n \cos \theta\)

= \(n(\sin \theta-\mu \cos \theta)\)….(2)

From equations (1) and (2) we have, \(\frac{\omega_1^2 r}{g}=\frac{\sin \theta-\mu \cos \theta}{\cos \theta+\mu \sin \theta}\)

or, \(\left(2 \pi \nu_1\right)^2=\frac{g}{r}\left(\frac{\sin \theta-\mu \cos \theta}{\cos \theta+\mu \sin \theta}\right)\)

or, \(\nu_1=\frac{1}{2 \pi} \sqrt{\frac{g}{r}\left(\frac{\sin \theta-\mu \cos \theta}{\cos \theta+\mu \sin \theta}\right)}\)

Now, let v₂ be the maximum value of the frequency of rotation for which the cube remains static with respect to the funnel. Then the direction of the frictional forces will be downward along the walls of the funnel.

Balancing the forces on the cube, we get mg = \(n \cos \theta-f \sin \theta=n(\cos \theta-\mu \sin \theta)\)…(3)

and \(m \omega_2^2 r=n(\sin \theta+\mu \cos \theta)\)…(4)

From equations (3) and (4) we get, \(\frac{\omega_2^2 r}{g}=\frac{\sin \theta+\mu \cos \theta}{\cos \theta-\mu \sin \theta}\)

or, \(\nu_2=\frac{1}{2 \pi} \sqrt{\frac{g}{r}\left(\frac{\sin \theta+\mu \cos \theta}{\cos \theta-\mu \sin \theta}\right)}\)

Question 17. Why are the bends of a road or of a railway track banked?
Answer:

At the bend of horizontal roads or railway tracks, frictional force provides the necessary centripetal force for a car or a train to take a turn.

• If the radius of curvature at the bend is low and if the frictional force is not large enough, the speed of the car or train at the bend has to be lowered to avoid skidding.
• To avoid this difficulty, instead of depending on friction alone, the roads or railway tracks are banked to provide additional centripetal force. The necessary centripetal force is supplied by the horizontal component of the normal force of the plane of the road or railway tracks.
• Hence, the outer side of the road or railway tracks is slightly elevated with respect to the inner side.

Question 18. Why do we feel lighter and heavier at the highest and lowest points of a Ferris wheel? Suppose, the Ferris wheel is revolving with a constant angular velocity.
Answer:

At the highest point of the wheel, we have mg – n = \(\frac{m v^2}{r}\)

where m = mass of the passenger, v = linear velocity of the passenger at the rotating Ferris wheel, r = radius of curvature of the Ferris wheel and n is the normal reaction.

∴ n = \(m g-\frac{m v^2}{r}\)

So, we feel lighter at the highest point.

At the lowest point of the trajectory \(n-m g=\frac{m v^2}{r} \quad \text { or, } n=m g+\frac{m v^2}{r}\)

Hence, we feel heavier at the lowest point of the Ferris wheel.

Centrifugal Force A Pseudo Force

WBBSE Class 11 Pseudo Force Notes

Let us consider a merry-go-round, capable of revolving in a horizontal plane, to be at rest. A person is sitting in that merry-go-round and has a stone in his hand. When the merry-go-round begins to revolve with uniform angular velocity, the stone also begins revolving in a circular path along with the person.

The following two observers will describe the motion of the stone in two different ways.

Observer Standing On The Surface Of The Earth At Rest: Since the stone revolves in a circular path, according to this observer, a centripetal force is acting on the stone. Due to the application of centripetal force, a body can move in a circular path.

If the person sitting in the merry-go-round releases the stone from his hand at point A, then the observer will see the stone flying off along the tangent (AB) of the circular path. Due to the inertia of the stone, it seems to move along the path AB according to the observer.

Read and Learn More: Class 11 Physics Notes

Observer Sitting In The Merry-Go-Round: This observer is moving with the same angular velocity as that of the stone, and hence, he observes the stone to be always at the same place. This means that the stone seems to be at rest to this observer.

• If the observer releases the stone from his hand at point A, then after some time when the stone reaches the point B, the observer revolving along with the merry-go-round will reach point C. So, to him, the motion of the stone will be along the straight line CB, away from the centre.
• The stone seems to be moving radially outwards under the influence of some force. This is called the centrifugal force.
• If the person sitting in the whirling merry-go-round holds the stone tightly in his hands, then the stone cannot fly off.

Here, the person exerts a force on the stone and it seems to him that this real force (which is the real centripetal force to the observer standing on the surface of the earth at rest) and the centrifugal force keep the stone in equilibrium. Hence, these two forces are equal but opposite in direction.

• We know that Newton’s laws of motion are not valid in the non-inertial frames. If the frame translates with respect to an inertial frame with an acceleration \(\vec{a}^{\prime}\), an apparent force -m\(\vec{a}^{\prime}\) acts on a particle of mass m, whose motion is described using a non-inertial frame of reference or rotating frame of reference.
• This force is called pseudo force. Once this pseudo force is included, one can use Newton’s laws in their usual form. This is not a real force. This force only exists in the non-inertial frame of reference or rotating frame of reference.

If a frame of reference rotates at a constant angular velocity ω with respect to an inertial frame and we analyse the dynamics of a particle of mass m kept at a distance r from the axis of rotation, we have to assume that a force mω²r acts radially outward on the particle. Only then we can apply Newton’s laws of motion in the rotating frame.

This radially outward force is the centrifugal force which is an example of pseudo force. It should be mentioned that centrifugal force acts on a particle because we describe the particle from a rotating frame which is non-inertial and still we use Newton’s laws.

So, centrifugal force = ma’

[where a’ is the centripetal acceleration]

= \(\frac{m v^2}{r}=m \omega^2 r\)

This force can be defined in the following way:

Central Force Definition: When a body rotates with an angular velocity along a circular path and an observer rotates with the same angular velocity with the body, then to that observer, a force equal but opposite to the centripetal force appears to be acting on the body. This force is called the centrifugal force.

• When a bus, full of passengers, turns at a bend on the road, the passengers feel a push opposite in direction to that of the bend and they lean towards that side. A person standing on the road, i.e., in an inertial frame of reference, explains this occurrence as an effect of inertia.
• While taking a turn at the bend, the bus behaves as a rotating frame of reference and the passengers belonging to that frame of reference it appears to that a force pushes them away from the centre of the circular path. This force is nothing but centrifugal force.

Class 11 Physics Unit 3 Laws Of Motion Chapter 3 Circular Motion

A Few Examples Of Centrifugal Force

Key Concepts of Pseudo Force for Class 11

The existence of centrifugal force in a rotational motion is assumed in various aspects of our daily lives.

1. Centrifuge: It is usually a container capable of rotating about an axle with a high angular velocity. Particles suspended in a liquid can be separated with this instrument.
   • Generally, the density of a liquid and that of the particles suspended in that liquid are different. From the expression mω²r of centrifugal force, it can be said that the greater the mass (m) of a body, greater the centrifugal force acting on it if co and r are kept constant.
   • As a result, the heavier particles move farther from the axle than the lighter particles. This instrument is used for separating cream from milk, blood corpuscles from blood, etc.
   • When separating cream from milk, cream particles, being lighter than milk, gather round the axle rod and skimmed milk gets separated.
   • Antonin Prandtl invented die first dairy centrifuge.
2. A drying machine used for drying wet clothes is another kind of centrifuge. The wet clothes are placed in a container consisting of a large number of perforations. This container is rotated with a high angular velocity. Due to centrifugal force, the water droplets present in the wet clothes are driven off, and the clothes are gradually dried up.
3. Loss of weight of a body due to the earth’s diurnal motion: The earth rotates about its own axis in 24 hours. This is known as the diurnal motion or daily rotation of the earth. Due to this rotational motion of the earth about its own axis, every object on the earth’s surface rotates with the same angular velocity.
   • Let the latitude of a point A on the earth’s surface be 6 and a body of mass m be situated at that point. The weight of the body mg acts along the line AO towards the centre of the earth.
   • Again, due to diurnal motion of the earth, the body revolves around the axis of the earth in a circular path of radius r with angular velocity ω and feels an outward centrifugal force mω²r.
   • The component of this centrifugal force in the direction AC is mω²rcosθ. As this component acts in the opposite direction of the weight mg, there is an effective loss of weight.
   • It should be noted that if the earth was at rest, then no centrifugal force would act, and hence, there Would have been no apparent loss of weight.
     • So, the weight of the body situated at the point A, W = mg- mω²rcosθ = m(g-ω²Rcos²θ)
     • [R is the radius of the earth and r = Rcosθ]
     • At the equatorial region, θ = 0
     • ∴ W = m(g- ω²R)
     • So, the decrease in weight of a body is maximum at the equatorial region due to the diurnal motion of the earth.
     • At the poles, θ = 90°, so W = mg.
     • So, no loss in weight of the body occurs at the polar regions due to the diurnal motion of the earth.
4. Reason For Flattening Of The Earth At The Poles: The shape of the Earth is not a perfect sphere, but an oblate spheroid, i.e., flattened slightly at the poles and bulged at the equator. This is caused by the daily rotation of the earth about its axis and the generation of corresponding centrifugal force.
   • At the time of formation of the earth, its temperature was very high and it was mainly made up of fused and gaseous matter. Since the magnitude of the centrifugal force acting outwards is maximum at the equatorial region, the fused and gaseous particles in this region had the tendency to move away from the axis.
   • On the other hand, as the magnitude of the centrifugal force at the poles is zero, the particles at the poles did not tend to move away from the axis. From the beginning, the mutual force of cohesion between the material particles on the earth was large.
   • As a result, the material particles at the equatorial region bulged out due to centrifugal force, whereas the material particles at the polar regions were pulled inwards due to the force of cohesion.
   • Later, when the earth’s crust hardened, this specific irregularity in shape became a permanent feature. For this reason, the earth is slightly flattened at the poles and bulged at the equator.

Class 11 Physics Unit 3 Laws Of Motion Chapter 3 Circular Motion

Centrifugal Force Numerical Examples

Examples of Pseudo Forces in Everyday Life

Example 1. A tube of length L is filled completely with an incompressible liquid of mass M and its two open ends are closed. The tube is then rotated with an angular velocity ω with one end of it as the centre. What will be the force exerted by the liquid on the other end?
Solution:

Given

A tube of length L is filled completely with an incompressible liquid of mass M and its two open ends are closed. The tube is then rotated with an angular velocity ω with one end of it as the centre.

Mass of the liquid = M. The centre of mass for the whole liquid is situated at the mid-point of the tube. Hence, the radius of the circular path along which the centre of mass rotates is, r =\(\frac{L}{2}\)

So, the centrifugal force acting on the centre of mass = mω²r = Mω\(\frac{L}{2}\)

This is the centrifugal force acting on the liquid in the tube. This is the force that acts on the other end of the tube.

Example 2. A hemispherical bowl of radius 0.1 m is rotated about a vertical axis passing through the centre of the bowl with an angular velocity ω. A particle of mass m = 10^-2 kg placed inside the bowl also rotates with the bowl. If the height of the position of the particle from the bottom of the bowl is h, find the relation between h and ω.
Solution:

Given

A hemispherical bowl of radius 0.1 m is rotated about a vertical axis passing through the centre of the bowl with an angular velocity ω. A particle of mass m = 10^-2 kg placed inside the bowl also rotates with the bowl. If the height of the position of the particle from the bottom of the bowl is h

The particle is at a height h from the bottom of the bowl. With the rotation of the bowl, the particle also rotates about the vertical axis along a circular path of radius r.

In this situation, the weight mg of the particle of mass m, the centrifugal force mω²r and the normal force (R) on the particle by the surface of the bowl keep the particle in equilibrium.

So, \(R \sin \theta=m \omega^2 r\) and \(R \cos \theta=m g\)

∴ \(\tan \theta=\frac{\omega^2 r}{g}\)

If the radius of the bowl is a, then we get \(\tan \theta=\frac{r}{a-h}\)

∴ \(\frac{r}{a-h}=\frac{\omega^2 r}{g} \quad \text { or, } \quad a-h=\frac{g}{\omega^2}\)

or, \(h=a-\frac{g}{\omega^2}=0.1-\frac{9.8}{\omega^2}=0.1\left(1-\frac{98}{\omega^2}\right) \mathrm{m} .\)

Example 3. The radius of the earth is 6400 km. What will be the value of the centrifugal acceleration at the equatorial region due to the earth’s diurnal motion?
Solution:

Given

The radius of the earth is 6400 km.

If the radius of the earth is R and its angular velocity is ω, then at the equatorial region, centrifugal acceleration

= \(\omega^2 R=\left(\frac{2 \pi}{T}\right)^2 \times R=\left(\frac{2 \pi}{24 \times 60 \times 60}\right)^2 \times 6400 \times 10^3\)

= \(0.0338 \mathrm{~m} \cdot \mathrm{s}^{-2} .\)

Class 11 Physics Unit 3 Laws Of Motion Chapter 3 Circular Motion

Comparison Between Linear And Rotational Motion Numerical Examples

Understanding Pseudo Force: Class 11 Physics

In the discussion of rotational motion, some of the physical quantities are the rotational analogues of the corresponding physical quantities of linear motion. These quantities are given below:

Example 1. A round table Is rotated with an angular velocity of 10 rad · s^-1 about Its axis. Two blocks of mass m₁ = 10 kg and m₂ = 5 kg, connected to each other by a weightless inextensible string of length 0. 3 m, are placed along the diameter of the table. The coefficient of friction between the table and m₁ is 0. 5, while there is no friction between m₂ and the table. Mass m₁ is at a distance of 0.124 m from the centre of the table. The masses are at rest with respect to the table.

1. Calculate the frictional force on m₁.
2. What should be the minimum angular speed of the table so that these masses will slip from 1 the table?
3. How should these masses be kept so that the string remains taut but no frictional force; acts on m₁?

Solution:

Given

A round table Is rotated with an angular velocity of 10 rad · s^-1 about Its axis. Two blocks of mass m₁ = 10 kg and m₂ = 5 kg, connected to each other by a weightless inextensible string of length 0. 3 m, are placed along the diameter of the table. The coefficient of friction between the table and m₁ is 0. 5, while there is no friction between m₂ and the table. Mass m₁ is at a distance of 0.124 m from the centre of the table.

The top view of the table is shown.

According to the problem,

AB = 0.3 m;

OA = 0.124 m

So, OB = 0.3 – 0.124 = 0.176 m

1. Centrifugal force acting on mass m₁ along the direction OA, m₁ω²r = 10 x (10)² x 0.124 = 124 N

Again, centrifugal force acting on mass m₂ along the direction OB, m₂ω²r = 5 x (10)² x 0.176 = 88 N

So, the resultant force along the direction OA = 124 – 88 = 36 N

In spite of this resultant force, the two masses remain stationary with respect to the table.

Hence, the frictional force is equal and opposite to this force. Hence, the required frictional force = 36 N.

2. Reaction force of the table on mass m₁ = weight of the mass = m₁g = 10 x 9.8 = 98 N

As the coefficient of friction is 0.5, the limiting frictional force = 0.5 x 98 =49 N.

If the minimum angular velocity of the table is ω, the resultant of the two opposite centrifugal forces acting on the two masses

= m₁ω²r₁ – m₂ω²r₂ = ω²(10 x 0.124 – 5 x 0.176)

= ω²(1.24-0.88)= 0.36ω² N

According to the problem, the resultant of the two centrifugal forces = limiting frictional force

∴ 0.36 \(\omega^2=49 \text { or, } \omega^2=\frac{4900}{36}\)

or, \(\omega=\frac{70}{6}=11.67 \mathrm{rad} \cdot \mathrm{s}^{-1}\)

3. If there is no frictional force acting on the masses, the resultant of the two centrifugal forces should be zero. In that case, if mass m₁ is placed at a distance r from the centre, then the distance of m₂ becomes (0.3-r).

So, for any value of ω, \(m_1 \omega^2 r=m_2 \omega^2(0.3-r) \quad \text { or, } 0.3-r=\frac{m_1}{m_2} r\)

or, \(0.3-r=2 r \quad or, 3 r=0.3 \quad or, \quad r=0.1 \mathrm{~m}\)

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Example 2. What should be the maximum speed of a motor car of mass 2000 kg when it takes a circular turn of radius 100 m on a plane road? Coefficient of friction between the tyre and the road = 0.25. Given, g = 10 m· s^-2.
Solution:

Let the maximum speed of the motor car = v; mass of the car = m, radius of the circular path = r; coefficient of friction between the tyre and the road = μ.

In the case of maximum speed of the car, centripetal force = limiting frictional force mv2

or, \(\frac{m v^2}{r}=\mu m g\)

or, v \(=\sqrt{\mu r g}=\sqrt{0.25 \times 100 \times 10}=15.81 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

Example 3. The roadway bridge over a stream is in the form of an arc of a circle of radius r. A car is crossing the bridge with a speed v. Prove that the limiting speed of the car with which it can cross the bridge without leaving the ground at the highest point of the bridge is v ≤ √gr.
Solution:

Given

The roadway bridge over a stream is in the form of an arc of a circle of radius r. A car is crossing the bridge with a speed v.

Let the mass of the car be m and the speed of the car be v.

∴ Weight of the car = mg and centrifugal force = \(\frac{m v^2}{r}\)

At the highest point on the bridge, the car will not jump up from the road if the weight of the car ≥ centrifugal force.

∴ mg ≥ \(\frac{m v^2}{r}\) or, gr ≥ \(v^2\)

or, \(v^2\) ≤ gr  or, ν ≤ \(\sqrt{g r}\).

Example 4. What will be the angular velocity of the diurnal motion of the earth, so that the weight of a body at the equatorial region is 0.6 of the present weight? Radius of the earth = 6400 km.
Solution:

Due to the diurnal motion of the earth, a centrifugal force acts on a body on the surface of the earth and hence the body suffers an apparent loss of weight.

Apparent weight of the body at the equatorial region = mg- mω²R

(ω = angular velocity, R = radius of the earth)

According to the problem, 0.6 mg = mg- mω²R

or, \(0.4 g=\omega^2 R\)

or, \(\omega=\sqrt{\frac{0.4 \times g}{R}}=\sqrt{\frac{0.4 \times 9.8}{6400 \times 1000}}=7.8 \times 10^{-4} \mathrm{rad} \cdot \mathrm{s}^{-1}\)

Short Answer Questions on Pseudo Forces

Example 5. A cylindrical drum made of steel and of diameter 20 cm is rotating about its own vertical axis. A small body made of steel remains stuck inside the cylinder when the drum rotates at the rate of 200 rpm. When the velocity of rotation decreases, the body falls down. What is the coefficient of friction between the body and the surface of the drum?
Solution:

Given

A cylindrical drum made of steel and of diameter 20 cm is rotating about its own vertical axis. A small body made of steel remains stuck inside the cylinder when the drum rotates at the rate of 200 rpm. When the velocity of rotation decreases, the body falls down.

Let the point where the body remains stuck be P

For the equilibrium of the body, weight of the body = limiting frictional force or, mg = F = μR [R = normal force on the body by the wall of the drum]

Again, R supplies the necessary centripetal force to the body for its revolution.

Hence, R = \(\frac{m v^2}{r}=m \omega^2 r\)

∴ \(m g=\mu m \omega^2 r\)

or, \(\mu=\frac{g}{\omega^2 r}=\frac{980 \times 9}{(20 \pi)^2 \times 10}\) (because \(\omega=\frac{200 \times 2 \pi}{60}=\frac{20 \pi}{3}\))

= 0.2234

Example 6. Four balls of mass 5 kg each are placed on the top of a horizontal turntable and fastened together with four strings of length 1 m each to form a square of side 1 m. The axis of rotation passes through the centre of the square. Find the tension in the strings when the turntable is rotated at the rate of \(\frac{30}{\pi}\) rpm.
SolutiIton:

Given

Four balls of mass 5 kg each are placed on the top of a horizontal turntable and fastened together with four strings of length 1 m each to form a square of side 1 m. The axis of rotation passes through the centre of the square.

The table is shown from the top. When the turntable keeps on rotating, the centrifugal force experienced by each ball will be mω²r.

Here, \(m=5 \mathrm{~kg}, \omega=2 \pi \times \frac{30}{\pi \times 60}\) = \(1 \mathrm{rad} \cdot \mathrm{s}^{-1}\)

As the length of each side of the square is 1m,

r = \(\frac{1}{2} \times \text { length of the diagonal or, } r=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}} \mathrm{~m}\)

According to the figure, tension in each string, T = \(m \omega^2 r \cos 45^{\circ}=5 \times(1)^2 \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}=\frac{5}{2}=2.5 \mathrm{~N} \text {. }\)

Example 7. A hemispherical bowl of radius r is rotated with an angular velocity ω about a vertical axis passing through Its centre. An object rotates with the same angular velocity remaining attached to the inner surface of the bowl. The straight line Joining the body and the centre of the bowl makes an angle 45° with the vertical. The coefficient of friction between the body and the bowl is 0.2. Assuming the body is just about to move down along the curved surface of the bowl, prove that ω²r = 0.943 g.
Solution:

Given

A hemispherical bowl of radius r is rotated with an angular velocity ω about a vertical axis passing through Its centre. An object rotates with the same angular velocity remaining attached to the inner surface of the bowl. The straight line Joining the body and the centre of the bowl makes an angle 45° with the vertical. The coefficient of friction between the body and the bowl is 0.2. Assuming the body is just about to move down along the curved surface of the bowl,

The hemisphere is rotated about the vertical axis AO with an angular velocity ω. The forces acting on the body at point P are shown.

When the body is just about to move down, mg – ncosθ + μnsinθ

or, mg = n(cosθ + μsinθ) ….(1)

and mω²rsinθ = nsinθ – μncosθ

or, mω²r = n(1 -μcotθ)…(2)

From (1) and (2), we get \(\frac{\omega^2 r}{g}=\frac{1-\mu \cot \theta}{\cos \theta+\mu \sin \theta}\)

= \(\frac{1-0.2 \cot 45^{\circ}}{\cos 45^{\circ}+0.2 \sin 45^{\circ}}\left[because \mu=0.2 \text { and } \theta=45^{\circ}\right]\)

∴ \(\omega^2 r=0.943 \mathrm{~g}\)

Example 8. A car of mass m is moving with a velocity v over a bridge. What will be the values of the force Fat the highest point of a convex bridge and at the lowest point of a concave bridge?
Solution:

Given

A car of mass m is moving with a velocity v over a bridge.

1. A centripetal force is required to move the car over a convex bridge. The value of this centripetal force is mv²

Here, r is the radius of curvature of the convex bridge.

At the highest point of the bridge, the resultant of the weight of the car mg and F will supply the necessary centripetal force.

∴ \(m g-F=\frac{m v^2}{r}\)

or, F = \(m\left(g-\frac{v^2}{r}\right)\)

2. A centripetal force is required to move the car over a concave bridge also. At the lowest point of the concave bridge, the resultant of the weight of the car mg and F will supply the necessary centripetal force.

∴ \(F-m g=\frac{m v^2}{r} \quad \text { or, } F=m\left(g+\frac{v^2}{r}\right)\)

Example 9. The roadway bridge over a stream is in the form of an arc of a circle of radius 50 m. What is the maximum speed with which a car can cross the bridge without leaving the ground at the highest point?
Solution:

Given

The roadway bridge over a stream is in the form of an arc of a circle of radius 50 m.

Suppose the car is moving with a speed v on the bridge so that it can cross the bridge without leaving the ground at the highest point.

If the normal force at the highest point on the bridge is F, then \(m g-F=\frac{m v^2}{r}\)

For the maximum speed of the car, at the highest point on the bridge, F = 0

∴ \(\frac{m v^2}{r}=m g \quad \text { or, } \quad v^2=r g\)

or, \(v=\sqrt{r g}=\sqrt{50 \times 9.8}\)

= \(22 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

Example 10. A car passes over a convex bridge. The centre of gravity of the car follows an arc of a circle of radius 30 m. Assuming that the car has a mass of 1000 kg, find the force that the car will exert at the highest point on the bridge if the velocity of the car is 15 m · s^-1. At what speed will the car lose contact with the road?
Solution:

Given

A car passes over a convex bridge. The centre of gravity of the car follows an arc of a circle of radius 30 m. Assuming that the car has a mass of 1000 kg, find the force that the car will exert at the highest point on the bridge if the velocity of the car is 15 m · s^-1.

Mass of the car, m = 1000 kg,

velocity, v = 15 m · s^-1

and radius of the circular path, r = 30 m

The weight of the car = mg and the necessary centripetal force =c \(\frac{m v^2}{r}\)

Let at the highest point on the convex bridge, the effective normal force by the road be F.

Then F= \(m g-\frac{m v^2}{r}=m\left(g-\frac{v^2}{r}\right)=10^3\left(9.8-\frac{15^2}{30}\right)\)

= \(2.3 \times 10^3 \mathrm{~N}\)

(because g = \(9.8 \mathrm{~m} \cdot \mathrm{s}^{-2}, v=15 \mathrm{~m} \cdot \mathrm{s}^{-1} \text { and } r=30 \mathrm{~m}\))

When F = 0, the car loses contact with the road. In that situation, if the speed of the car is u, then

⇒ \(m g-0=\frac{m u^2}{r} \text { or, } u^2=g r\)

∴ u = \(\sqrt{9.8 \times 30}=17.1 \mathrm{~m} \cdot \mathrm{s}^{-1} .\)

Pseudo Force Formula and Applications

Example 11. A spotlight is rotating with a uniform angular velocity of 0.1 rad · s^-1 on a horizontal plane. The light spot moves on a wall at a distance of 3m. If the path of light inclines at an angle of 45° with the wall, then calculate the velocity of the light-spot.

Solution:

Given

A spotlight is rotating with a uniform angular velocity of 0.1 rad · s^-1 on a horizontal plane. The light spot moves on a wall at a distance of 3m. If the path of light inclines at an angle of 45° with the wall,

Angular velocity of the spotlight S, ω = 0.1 rad · s^-1

Suppose at some moment, the light spot is at point L on the wall

∴ SL = r = \(\frac{3}{\sin 45^{\circ}}=3 \sqrt{2} \mathrm{~m}\)

If the linear velocity of the light-spot is v [v = ωr], then the velocity of the light-spot along the wall

= component of v along the wall

= \(v \sin 45^{\circ}=\frac{\omega r}{\sqrt{2}}=\frac{0.1 \times 3 \sqrt{2}}{\sqrt{2}}=0.3 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

Example 12. A piece of stone of mass 1 kg is tied with a thread of length 10 m and is rotating along a horizontal circular path making an angle θ with the vertical. The thread can withstand a maximum tension of 12 N. What is the maximum velocity with which the stone can be rotated without snapping the thread?
Solution:

Given

A piece of stone of mass 1 kg is tied with a thread of length 10 m and is rotating along a horizontal circular path making an angle θ with the vertical. The thread can withstand a maximum tension of 12 N.

Let the length of the thread be l, mass and maximum linear velocity of the stone be m and v respectively and the maximum tension in the thread be T while rotating along a horizontal plane.

The required centripetal force for rotation, \(\frac{m v^2}{l \sin \theta}=T \sin \theta\)…(1)

and \(m g=T \cos \theta\)…(2)

From (1) and (2), we get, \(\sin ^2 \theta+\cos ^2 \theta=\frac{m v^2}{l T}+\left(\frac{m g}{T}\right)^2\)

or, \(\frac{m v^2}{l T}=\left(1-\frac{m^2 g^2}{T^2}\right) or, v=\left\{\frac{l T}{m}\left(1-\frac{m^2 g^2}{T^2}\right)\right\}^{1 / 2}\)

Given, \(l=10 \mathrm{~m}, T=12 \mathrm{~N}\) and \(m=1 \mathrm{~kg}\)

∴ v = \(\left\{\frac{10 \times 12}{1}\left(1-\frac{1^2 \times 9.8^2}{12^2}\right)\right\}^{1 / 2}\) (because \(g=9.8 \mathrm{~m} \cdot \mathrm{s}^{-2}\))

or, \(\nu=6.32 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

Centripetal Force

WBBSE Class 11 Centripetal Force Notes

We have seen that the value of normal or centripetal acceleration in uniform circular motion is \(\omega^2 r \text { or } \frac{v^2}{r}\). So, if the mass of a body is m, then effective force on the body

= mass x acceleration = \(m \omega^2 r=\frac{m v^2}{r}\)

• Since the acceleration is towards the centre, the effective force acting on the body will also be towards the centre. This force is known as centripetal force.
• According to Newton’s first law of motion, we know that if no external force is acting on a body, the body will remain at rest or will move with uniform linear velocity, i.e., the body will not move in a circular path.
• So, to move a body in a circular path some external force must be acted on it. During motion in a circular path, the acceleration of the body must be towards the centre.
• So, the force acting on the body must be towards the centre of the circular path, i.e., radially inwards. This external force is nothing but the centripetal force.

Read and Learn More: Class 11 Physics Notes

Centripetal Force Definition: Centripetal force is the force which moves the body in a circular path. It acts perpendicular to the direction of linear velocity and is directed radially towards the centre of that circular path.

If the mass of the body is m, the radius of the circular path is r, the linear velocity of the body is v and its angular velocity is ω, then centripetal force = \(m \omega^2 r=\frac{m v^2}{r}\).

Here, it is to be mentioned that the centripetal force is a no-work force. In the direction of centripetal force, no displacement of the body occurs, and hence, this force does not do any work.

Centripetal Force Some Practical Examples:

1. A stone is rotated along a circular path by a string. The string pulls the stone towards the centre. This pull or tension is the centripetal force.
2. While riding a bicycle along a circular path, the frictional force acting between the tyre of the cycle and the road supplies the centripetal force necessary to move the bicycle along that circular path.
3. The gravitational attraction of the sun on any planet provides the planet with the necessary centripetal force to revolve around the sun.
4. The electrostatic attraction between the positively charged nucleus inside an atom and the negatively charged electrons supplies the electrons with the necessary centripetal force to revolve around the nucleus.

 

Some Practical Examples Of Centripetal Force

Motion Of A Cyclist On A Horizontal Circular Track: When a cyclist goes around a horizontal circular track, a centripetal force is required. The force of friction between the tyres and the road is too small to provide the necessary centripetal force.

• As a result, a cyclist going round a curve leans inward, because then the horizontal component of the normal reaction provides the necessary centripetal force.
• Let the angle with the vertical at which the cyclist leans be θ. In this situation, the vertical component of the reaction R offered by the ground on the cyclist = OB = Rcosθ and the horizontal component of R = OA = Rsinθ.
• The vertical component balances the weight of the cycle along with the cyclist and the horizontal component supplies the necessary centripetal force required to take the turn on the circular path.

Key Concepts of Centripetal Force in Circular Motion

If the weight of the cycle along with the cyclist is mg, the radius of the circular path is r and the velocity of the cyclist is v, then Rcosθ = mg …(1)

R \(\sin \theta=\frac{m v^2}{r}\) (necessary centripetal force)…(2)

Dividing equation (2) by equation (1) we get, \(\tan \theta=\frac{v^2}{r g}\)….(3)

From equation (3), it is evident that the more the velocity (v) of the cyclist or less the radius (r) of the circular path, the more the value of θ will be, i.e., the cyclist has to incline more towards the bend.

Centripetal Force Formula and Examples

Since the normal force of the ground, Rcosθ = mg, the limiting frictional force between the tyre and the road is fμmg (μ = coefficient of friction). It is clear that this limiting frictional force is the maximum value of the centripetal force.

If the speed of the cyclist is v_m for a circular path of certain radius r, the centripetal force will be maximum when, \(\frac{m v_m^2}{r}=\mu m g \text { or, } v_m^2=\mu r g \text { or, } v_m=\sqrt{\mu r g}\)…(4)

which it can turn on a circular path of radius r. If the cyclist tries to attain a speed more than this maximum speed on that road, the frictional force cannot provide the necessary centripetal force, and hence, the cycle will skid. If the maximum value of the angle of inclination with the vertical is θ_m, then \(\tan \theta_m=\frac{v_m^2}{r g}=\frac{\mu r g}{r g}=\mu\)….(5)

It should be noted that none of the equations (3), (4) and (5) depend on the mass (m) of the cycle along with the cyclist.

Centripetal Force Numerical Examples

Example 1. A cyclist speeding at 18 km/h on a level rod takes a sharp circular turn of a radius 3 m without reducing the speed and without spending towards the centre of the circular path. The coefficient of static friction between the tyres and the road is 0.1. Will the cyclist slip while taking the turn?
Solution:

Given

A cyclist speeding at 18 km/h on a level rod takes a sharp circular turn of a radius 3 m without reducing the speed and without spending towards the centre of the circular path. The coefficient of static friction between the tyres and the road is 0.1.

The maximum safe speed is given by, \(v_{\max }=\sqrt{\mu \mathrm{rg}}=\sqrt{0.1 \times 3 \times 9.8}=1.715 \mathrm{~m} / \mathrm{s}\)

Actual speed of the cyclist, \(v=18 \mathrm{~km} / \mathrm{h}=\frac{18 \times 1000}{60 \times 60} \mathrm{~m} / \mathrm{s}=5 \mathrm{~m} / \mathrm{s}\)

Since the actual speed is greater than the maximum safe speed, the cyclist will slip while taking the turn.

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Example 2. A cyclist is moving in a circular path of radius 20 m with a velocity of 18 km ·h^-1. What is his inclination toward the vertical?
Solution:

Given

A cyclist is moving in a circular path of radius 20 m with a velocity of 18 km ·h^-1.

Let the angle of inclination of the cyclist with the vertical be θ.

The magnitude of the velocity, i.e., speed of the cyclist, \(\nu=\frac{18 \times 1000}{60 \times 60}=5 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

and radius of the circular path ( r) = 20 m.

We know that, \(\tan \theta=\frac{v^2}{r g}=\frac{5 \times 5}{20 \times 9.8}=0.1276\)

∴ \(\theta=\tan ^{-1}(0.1276)=7.27^{\circ}\)

Short Answer Questions on Centripetal Force

Example 3. A mass is suspended from the ceiling by a string revolving in a horizontal circle of radius 5 cm. The tangential speed of the mass is 0.7 m · s^-1. What is the angle between the string and the vertical? (Consider acceleration due to gravity as 9.8 m · s^-2)
Solution:

Given

A mass is suspended from the ceiling by a string revolving in a horizontal circle of radius 5 cm. The tangential speed of the mass is 0.7 m · s^-1.

Let the angle between the string and the vertical be θ.

From the \(\tan \theta=\frac{\frac{m v^2}{r}}{m g}=\frac{v^2}{r g}\)

Here, \(r=5 \mathrm{~cm}=0.05 \mathrm{~m}\),

v = \(0.7 \mathrm{~m} \cdot \mathrm{s}^{-1}, g=9.8 \mathrm{~m} \cdot \mathrm{s}^{-2}\)

So, \(\tan \theta=\frac{(0.7)^2}{0.05 \times 9.8}=1 \quad or, \theta=45^{\circ}\).

Example 4. A coin is placed on a horizontal turntable rotating at 33 1/3 rpm. The coin revolves with the table without slipping provided the coin is not more than 10 cm away from the axis. How far from the axis can the coin be placed so that it revolves with the table without slipping if the turntable rotates at 45 rpm? (g = 980 cm · s^-2)
Solution:

Given

A coin is placed on a horizontal turntable rotating at 33 1/3 rpm. The coin revolves with the table without slipping provided the coin is not more than 10 cm away from the axis.

The magnitude of the limiting frictional force between the coin and the table is the maximum value of centripetal force. So, if the coin of mass m is rotating along the turntable in a circular path of maximum radius r with angular velocity co, limiting frictional force,

f = mω²r

In the first case of the given problem, \(f=m \omega_1^2 r_1\)

and in the second case, \(f=m \omega_2^2 r_2\)

∴ \(m \omega_1^2 r_1=m \omega_2^2 r_2\)

or, \(r_2=\frac{\omega_1^2}{\omega_2^2} \times r_1=\left(\frac{100}{3 \times 45}\right)^2 \times 10=5.5 \mathrm{~cm}\).

Example 5. A small body is kept at a distance of 7 cm from the centre of a gramophone disc. The disc starts rotating with gradually increasing speed and the body is just on the verge of being thrown off when the disc rotates at 60 rpm. What will be the rate of rotation of the disc when the body, kept at a distance of 12 cm from the centre, is just being thrown off?
Solution:

Given

A small body is kept at a distance of 7 cm from the centre of a gramophone disc. The disc starts rotating with gradually increasing speed and the body is just on the verge of being thrown off when the disc rotates at 60 rpm.

The magnitude of the limiting frictional force between the body and the disc gives the maximum value of the centripetal force. So, if the body of mass m rotates along a circular path of maximum radius r with an angular velocity, then the limiting frictional force, f = mω²r.

In the given problem, f = \(m \omega_1^2 r_1\), in the first case

and \(f=m \omega_2^2 r_2\), in the second case

∴ \(m \omega_1^2 r_1=m \omega_2^2 r_2 \quad or, \omega_2^2=\omega_1^2 \cdot \frac{r_1}{r_2}\)

or, \(\omega_2=\omega_1 \sqrt{\frac{r_1}{r_2}}=60 \times \sqrt{\frac{7}{12}}=45.8 \mathrm{rpm}\).

Real-Life Applications of Centripetal Force Examples

Example 6. The driver of a truck travelling with a velocity v suddenly notices a wall in front of him at a distance d. Is it better for him to apply brakes or to make a circular turn without applying brakes in order to just i avoid crashing into the wall?
Solution:

Given

The driver of a truck travelling with a velocity v suddenly notices a wall in front of him at a distance d.

To stop the truck by applying brakes, the final velocity should be zero.

So, applying the equation v² = u² – 2 as we get, 0 = v² -2ad

or, a = \(\frac{v^2}{2 d}\) = retardation of the truck.

If the mass of the truck is m, then the force applied by the brakes,  F = ma = \(\frac{m v^2}{2 d}\)

• If the truck turns along a circular path to avoid collision, without applying brakes, then the radius of that circular path = d. For this circular motion, the necessary centripetal force is, \(F^{\prime}=\frac{m v^2}{d}\)
• Here, F< F’; hence, it is better to stop the truck by applying brakes so that crashing can be avoided by the application of comparatively less force.
• Here, it should be noted that due to the frictional force exerted by the road on the lyre of the wheels of the truck, it is possible to stop the truck.
• After the application of brakes, if the road does not exert any frictional force on the tyre then (1) due to the application of frictional force by the brakes on the wheels, they stop rotating and (2) the wheels go on skidding on the road.

(There is a chance of skidding of wheels on wet road when brakes are applied in a fast-moving vehicle). Hence, for any vehicle, if the frictional force necessary to stop it is less, then the chance of skidding is lowered.

Example 7. A body is kept at rest at a distance of 10 cm from the centre of a gramophone disc. If the coefficient of friction between the body and the disc is 0.3, then for what maximum rps of the disc, will the body not be thrown off the disc?
Solution:

Given

A body is kept at rest at a distance of 10 cm from the centre of a gramophone disc. If the coefficient of friction between the body and the disc is 0.3,

The magnitude of the limiting frictional force between the body and the gramophone disc is the maximum value of the centripetal force. If the mass of the body is m and it rotates with the disc along a circular path of radius r with angular velocity co such that the body will not be thrown off the disc, then

limiting frictional force, f = mω²r

or, μmg = mω²r [ω = 2πn, n = number of revolutions of the disc per second (rps); μ = coefficient of friction between the body and the disc]

or, \(\mu g=(2 \pi n)^2 r=4 \pi^2 n^2 r\) or, \(\quad n^2=\frac{\mu g}{4 \pi^2 r}\)

or, \(n=\sqrt{\frac{\mu g}{4 \pi^2 r}}=\sqrt{\frac{0.3 \times 980}{4 \times \pi^2 \times 10}}\)

= \(0.863 \mathrm{rps}\)

Motion Of A Train Or A Car On A Circular Track

Motion Along A Horizontal Path: Let O be the point about which a train or a car takes a turn in a circular path while moving on a horizontal path. 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 = OA = radius of the circular path.

Since the frictional force f between the wheel and the road supplies the necessary centripetal force, f = \(\frac{m v^2}{r}\)…(1)

Again the normal force R on the wheel by the ground is equal to the weight of the car, i.e., R = mg….(2)

Condition For No Skidding: If the frictional force f between the wheels and the road becomes equal to the limiting friction and if the coefficient of friction is f then f = μR

Using the equations (1) and (2) we get, \(\frac{m v^2}{r}\) = μmg or, v² = μrg

So, the maximum velocity of the car with which it can turn on a circular path of radius r will be \(v_m=\sqrt{\mu r g}\)…(3)

Banking In A Circular Path: A moving car, while taking a turn, requires a centripetal force. At the turning point if the road is banked, i.e., the road does not remain horizontal, then the horizontal component of the normal force offered by the road on the moving car supplies the necessary centripetal force.

• If this component is not sufficient to provide the necessary centripetal force, then the remaining part of it comes from the horizontal component of the frictional force between the road and the car.
• So, if the road is properly banked, then the horizontal component of the reaction force alone can provide the necessary centripetal force. In that case, frictional force does not come into play.
• To avoid accidents of cars near a bend, the centripetal force is supplied by raising the outer edge of the road above its inner edge and this method of construction of a road is known as banking.
• The angle made by the road with the horizontal is known as the angle of banking.

The angle of banking of the road ∠BAC = θ, speed of the car = v, mass of the car = m and the radius of the bend of the road = r (i.e., the radius of the circular path of the car).

In the above-mentioned case, there are two forces acting on the car:

1. Weight of the car mg acting vertically downwards and
2. Reaction force R offered by the road on the car, normal to the surface of the road.

The horizontal component Rsinθ of R provides the necessary centripetal force and the vertical component Rcos8 balances the weight of the car.

So, \(R \sin \theta=\frac{m v^2}{r}, R \cos \theta=m g\)

∴ \(\tan \theta=\frac{v^2}{r g}\)…(4)

Applications of Centripetal Force in Daily Life

From the above relation, the angle of banking (θ) required for a car of certain speed (v) moving on a particular road can be ascertained. For a particular speed, the value of θ is fixed, i.e., a certain road is properly banked only for a definite value of the speed.

Moreover, according to the figure, tan0 = \(\frac{h}{x}\)

(h = the height of the outer edge of the road with respect to its inner edge; it should be noted that the inner edge of the road lies towards the centre of the circular path)

∴ \(\frac{v^2}{r g}=\frac{h}{x}\) (x = horizontal distance between the outer and the inner edges of the road)

or, \(v=\sqrt{\frac{h r g}{x}}\)…..(5)

• This relation expresses the maximum speed of the car with which it moves along a banked circular path. If the speed is more than this, then there will be a chance of the car skidding outwards.
• Similarly, railway tracks are also banked near a bend. Here, the outer rail near the bend is placed at a slightly higher level than the inner rail.
• At the site of every bend, the maximum permissible speed is displayed on a board. To avoid an accident, the driver of the train should be acquainted with the maximum permissible speed at that place.

Condition For No Skidding: If the speed of the car exceeds the maximum permissible speed, then the car tends to skid towards point B and the frictional force f acting along the direction BA tries to oppose this skidding.

Let the maximum permissible speed of the car for limiting friction is v_m. Then limiting frictional force is f = μR (μ = coefficient of friction between the wheel and the road).

According to the given figure, the condition for no skidding is

R \(\sin \theta+f \cos \theta=\frac{m v_m^2}{r}\) or, \(R \sin \theta+\mu R \cos \theta=\frac{m v_m^2}{r}\)

or, \(R(\sin \theta+\mu \cos \theta)\) = \(\frac{m v_m^2}{r}\)….(6)

Understanding Centripetal Force in Physics

Again, \(R \cos \theta=m g+f \sin \theta\) or, \(R \cos \theta-\mu R \sin \theta=m g\)

or, \(R(\cos \theta-\mu \sin \theta)=m g\)….(7)

(6)+(7) gives, \(\frac{\nu_m^2}{r g}=\frac{\sin \theta+\mu \cos \theta}{\cos \theta-\mu \sin \theta}=\frac{\tan \theta+\mu}{1-\mu \tan \theta}\)

or, \(v_m=\left[r g\left(\frac{\mu+\tan \theta}{1-\mu \tan \theta}\right)\right]^{1 / 2}\)….(8)

So, when the speed of the car exceeds v_m, it skids towards the point B. Naturally, for θ = 0, i.e., on an unbanked road, \(v_m=\sqrt{\mu r g}\) which is nothing but equation (3).

In equation (8) the quantity (1 – μtanθ) is less than 1, and hence, the value of v_m becomes greater due to the presence of banking. Hence, a car can take a sharp turn on a banked road with speed than on an unbanked horizontal road.

Circular Motion

Circular Track Numerical examples

Example 1. The radius of curvature of a railway track at a bend is 500 m. The distance between the two tracks is 1 m and the outer line is 4 cm higher than the inner. At what maximum speed can a train bend around the curve without exerting lateral pressure on the outer line?
Solution:

Given

The radius of curvature of a railway track at a bend is 500 m. The distance between the two tracks is 1 m and the outer line is 4 cm higher than the inner.

If the speed of the train is v, the radius of curvature of the tracks is r and the angle of banking is θ, then \(\tan \theta=\frac{v^2}{r g}\)

If the distance between the two lines is x and the difference in their heights is h, then \(\tan \theta=\frac{h}{x}\)

∴ \(\frac{v^2}{r g}=\frac{h}{x} \text { or, } v^2=\frac{h r g}{x} \quad \text { or, } \quad v=\sqrt{\frac{h r g}{x}}\)

∴ v = \(\sqrt{\frac{0.04 \times 500 \times 9.8}{1}}=14 \mathrm{~m} \cdot \mathrm{s}^{-1}=50.4 \mathrm{~km} \cdot \mathrm{h}^{-1}\)

Uniform Circular Motion Centripetal Acceleration

We know that the axis of rotation of a rotating particle passes through the centre of the circular path and also lies normally on the plane of rotation. In the case of uniform circular motion, the particle travels equal distances in equal intervals of time, i.e., the speed of the particle remains constant.

Uniform Circular Motions Centripetal Acceleration Definition: if the angular velocity of a particle rotating in a circular path remains constant, then its motion is called a uniform circular motion.

• The direction of motion of a particle under uniform circular motion changes at every moment, and hence, the direction of the linear velocity v of the particle changes continuously. However due to the constant speed of the particle, the magnitude of linear velocity always remains constant. Since linear velocity is a vector quantity, it has both magnitude and direction.
• So, in this case, it cannot be said that the linear velocity of the particle is constant. Hence, uniform circular motion is an example of uniform speed but not of uniform velocity.
• The direction of the linear velocity of the particle at every point on the circular path is along the tangent drawn at that point. For this reason, during the revolution of a stone tied with a thread, when the thread snaps, the stone flies off along the tangent.
• When a bicycle moves along a muddy road speedily, the mud particles seem to fly off from the wheel of the bicycle tangentially. Sparks of fire seem to fly off tangentially when knives, scissors, etc., are sharpened on a rotating sharpening machine.
• We have learnt that the rate of change of velocity of a body is called its acceleration. In a uniform circular motion, the linear velocity of a body varies in direction and hence, the body possesses an acceleration.

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With the help of the following calculation, we can determine the magnitude and direction of the acceleration of the body.

Understanding Centripetal Force in Circular Motion

Calculation Of Centripetal Acceleration: Suppose a particle of mass m is rotating along a circular path of radius r with uniform speed v. In a very small time interval t, the particle moves from A to B and as a result, the angular displacement θ(=∠AOB) is very small. So, the angular velocity of the particle is, \(\omega=\frac{\theta}{t}\).

• At point A, the direction of linear velocity v is along the tangent AP. The component of this velocity along the radius AO is zero because AO and AP are mutually perpendicular.
• At point B, the velocity v of the particle is along the tangent BQ. This velocity is resolved into two mutually perpendicular components. The component of this velocity parallel to AP and in the direction BR is vcosθ and the component parallel to AO and in the direction BS is vsinθ.
• If θ is very small, then sinθ ≈ θ and cosθ ≈ 1. Again, if ∠AOB is very small, then the straight lines BR and BS will just coincide with the straight lines AP and AO, respectively.

So, the initial velocity of the particle in the direction AP = v and its final velocity = vcosθ = v.

Hence, change in velocity = v- v = 0

or, acceleration = \(\frac{\text { change in velocity }}{\text { time }}=0\)

So, in the direction AP, i.e., along the tangent of the circle, the particle has no acceleration.

Again, the initial velocity of the particle along AO = 0

Its final velocity = vsinθ = vθ

∴ Change in velocity = vθ – 0 = vθ

or, acceleration = \(\frac{\text { change in velocity }}{\text { time }}=\frac{v \theta}{t}=v \omega\)

= \(\omega r \cdot \omega=\omega^2 r=\left(\frac{v}{r}\right)^2 r=\frac{v^2}{r}\)

So, along the direction AO (radially towards the centre of the circle), the particle has an acceleration whose value is (ω²r or\(\frac{v^2}{r}\). This acceleration is called the radial or, normal or centripetal acceleration.

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WBBSE Class 11 Uniform Circular Motion Notes

Calculation Of Centripetal Acceleration Definition: When a particle moves along a circular path with constant speed, it possesses an acceleration towards the centre of the circle and this acceleration is called its radial normal or centripetal acceleration.

It is to be mentioned here that the velocity of the particle is always along the tangent of the circle but its acceleration is always towards the centre. So, in the case of uniform circular motion, velocity and acceleration are always perpendicular to each other.

 

Angular Velocity

WBBSE Class 11 Angular Velocity Notes

Angular Velocity Definition: The rate of change of the angular position of a particle with time is called the angular velocity of that particle.

The angle subtended by a particle, revolving in a circular path, at the centre of rotation in unit time is called the angular velocity of the particle.

So, angular velocity = \(\frac{\text { change of angular position }}{\text { time }}\)

= \(\frac{\text { angular displacement }}{\text { time }}\)

• Angular velocity is expressed by the symbol ω (Greek letter ‘omega’).
• While revolving in a circular path, if a particle subtends equal angles at the centre in equal intervals of time, the particle is said to be in uniform circular motion.
• During a uniform circular motion, if a particle subtends an angle θ at the centre in time t, then the angular velocity of the particle will be = \(\omega=\frac{\theta}{t}\)
• If the circular motion is not uniform, then the total angle subtended (θ) by the particle divided by the total time (t) is called the average angular velocity of the particle, i.e., the average angular velocity, \(\omega=\frac{\theta}{t}\)

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Instantaneous Angular Velocity: The instantaneous angular velocity of a particle at a given point is the limiting value of the rate of the angular displacement, from that point with respect to a time interval when the time interval tends to zero.

In a time interval Δt, if the angular displacement of a particle is Δθ, then the instantaneous angular velocity,

= \(\lim _{\Delta t \rightarrow 0} \frac{\Delta \theta}{\Delta t}=\frac{d \theta}{d t}\)

Since angular displacement is an axial vector, angular velocity is also an axial vector. The direction of angular velocity is the same as that of angular displacement, i.e., along the axis of rotation.

Understanding Angular Velocity in Circular Motion

Unit And Dimension Of Angular Velocity: Usually, the unit of angular displacement is radian; hence the unit of angular velocity

= \(\frac{\text { unit of angular displacement }}{\text { unit of time }}\)

= \(\text { radian } / \text { second }=\mathrm{rad} \cdot \mathrm{s}^{-1}\)

Since angular displacement is a dimensionless quantity, the dimension of angular velocity

= \(\frac{\text { dimension of angular displacement }}{\text { dimension of time }}=\frac{1}{\mathrm{~T}}=\mathrm{T}^{-1}\)

Time-Period: The time taken by a particle to make one complete revolution along a circular path is called the time period of that particle in rotational motion.

In one complete revolution of the particle along the circular path, the angular displacement of the particle, θ = 2π rad. So, if the time period is T, then the angular velocity of the particle is, \(\omega=\frac{2 \pi}{T} \quad \text { or, } \quad T=\frac{2 \pi}{\omega}\)

The unit of time period is second (s) and its dimension is T.

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Angular Velocity Formula and Examples

Frequency: The number of revolutions completed by a particle in unit time is called the frequency (n) of that particle. It represents the rotational speed of the particle.

For 1 complete revolution, angular displacement = 2π

So, for n complete revolutions, angular displacement =2πn

Since the time taken for these n revolutions is 1 second, the angular velocity,

⇒ \(\omega=\frac{2 \pi n}{1}=2 \pi n \quad \text { or, } \quad n=\frac{\omega}{2 \pi}\)

Hence, comparing with the time period we see that, n = 1/T

∴ The dimension of frequency = T^-1

The unit of frequency is s^-1, commonly known as Hertz (Hz). It is also called revolution per second or rps. For practical purposes, a widely used unit is revolution per minute or rpm. For very slow revolutions, revolution per hour or rph is also used.

It is clear that 1 rps = 60 rpm = 3600 rph

The frequencies of the second, minute and hour hands of a clock are 1 rpm, 1 rph and 1/12rph respectively.

Relation Between rpm And rad · s^-1:

1 rpm = \(\frac{1 \text { complete revolution }}{1 \text { minute }}=\frac{2 \pi}{60} \mathrm{rad} \cdot \mathrm{s}^{-1}=\frac{\pi}{30} \mathrm{rad} \cdot \mathrm{s}^{-1}\)

Key Concepts of Angular Velocity in Physics

Relation Between Instantaneous Linear Velocity And Angular Velocity: Let us consider that a particle moving with angular velocity ω along a circular path of radius r, covers a distance s along the circular path in time t. In that time, the angular displacement of the particle is θ.

According to the definition, \(\omega=\frac{\theta}{t}\).

Now, if the linear speed of the particle is v, then,

v = \(\frac{s}{t}=\frac{r \theta}{t} \quad \text { or, } \quad v=\omega r\)

i.e., speed = angular velocity x radius of the circular path

• In this case, the magnitude of instantaneous linear velocity is equal to the magnitude of instantaneous speed; hence, it can be inferred that at any moment, magnitude of instantaneous linear velocity = magnitude of instantaneous angular velocity x radius of the circular path
• If an extended body undergoes perfect revolution (i.e., different points on the body revolve in different circular paths around a definite axis), then the speed of different particles situated at different distances from the axis of rotation will be different, though the magnitude of the angular velocity of every particle remains the same.

Greater the distance of a particle from the axis, greater is its linear velocity.

Proof Of v = ωr With The Help Of Calculus:

We know that, s = rθ

Differentiating with respect to time we get, \(\frac{d s}{d t}=r \frac{d \theta}{d t}\) (because r is constant)

or, v = rω (instantaneous linear velocity, v = \(\frac{d s}{d t}\); instantaneous angular velocity, = \(\frac{d \theta}{d t}\)

Vector Notation: The vector notation of the relation between linear velocity and angular velocity is \(\vec{v}=\vec{\omega} \times \vec{r}\)

The vectors \(\vec{v}\), \(\vec{w}\) and \(\vec{r}\) are shown.

Angular Velocity Numerical Examples

Short Answer Questions on Angular Velocity

Example 1. Determine the angular velocities of the second, minute and hour hands of a clock.
Solution:

The angular velocities of the second, minute and hour hands of a clock are as follows

Since the second hand of a clock completes a circular revolution in 60 seconds, the angular velocity of the second hand is, \(\omega=\frac{2 \pi}{60}=\frac{\pi}{30}=0.105 \mathrm{rad} \cdot \mathrm{s}^{-1}\)

Since the minute hand of a clock completes a circular revolution in 60 minutes, the angular velocity of the minute hand is, \(\omega=\frac{2 \pi}{60 \times 60}=\frac{\pi}{1800}=1.75 \times 10^{-3} \mathrm{rad} \cdot \mathrm{s}^{-1} .\)

Since the hour hand of a clock completes a circular revolution in 12 hours, the angular velocity of the \(\omega=\frac{2 \pi}{60 \times 60}=\frac{\pi}{1800}=1.75 \times 10^{-3} \mathrm{rad} \cdot \mathrm{s}^{-1} .\)

Example 2. A car is moving along a circular path of radius 20 m with a speed of 40 km · h^-1. Find its angular velocity.
Solution:

Given

A car is moving along a circular path of radius 20 m with a speed of 40 km · h^-1.

Linear velocity, \(v=\frac{40 \times 1000}{60 \times 60}=\frac{100}{9} \mathrm{~m} \cdot \mathrm{s}^{-1}\)

∴ Angular velocity, \(\omega=\frac{v}{r}=\frac{100}{9 \times 20}=0.56 \mathrm{rad} \cdot \mathrm{s}^{-1}\).

Real-Life Applications of Angular Velocity

Example 3. Calculate the angular velocity of the earth about its own axis.
Solution:

The angular velocity of the earth about its own axis

Earth completes one rotation about its axis in 24 h i.e., it covers an angle 2π in 24 x 60 x 60 s.

∴ Angular velocity of the earth = \(\frac{2 \pi}{24 \times 60 \times 60}\)

= \(\frac{\pi}{43200} \mathrm{rad} / \mathrm{s}\)

Example 4. Find the time interval between two successive overlaps of the hour hand and the minute hand of a clock.
Solution:

The time interval between two successive overlaps of the hour hand and the minute hand of a clock

Let the time interval after which the two hands meet each other be t hours.

In the case of the hour hand, the angular velocity \(\omega=\frac{2 \pi}{12}=\frac{\pi}{6} \mathrm{rad} \cdot \mathrm{h}^{-1}\)

Hence, in t hours, the angular displacement of the hour hand, \(\theta=\omega t=\frac{\pi}{6} t \mathrm{rad}\)

In the case of the minute hand, the angular velocity \(\omega^{\prime}=\frac{2 \pi}{1}=2 \pi \mathrm{rad} \cdot \mathrm{h}^{-1} .\)

Hence, the angular displacement of the minute hand, \(\theta^{\prime}=\omega^{\prime} t=2 \pi t \mathrm{rad} .\)

After meeting each other, when the hands coincide again after t hours, the minute hand completes one revolution more than the hour hand, i.e., the angular displacement of the minute hand is 2π rad more than that of the hour hand.

So, \(\theta^{\prime}-\theta=2 \pi\) or, \(2 \pi t-\frac{\pi t}{6}=2 \pi\)

or, \(t \times \frac{11}{6}=2 or, \quad t=\frac{12}{11} \mathrm{~h}=1 \mathrm{~h} 5 \min 27 \mathrm{~s}(approx.).\)

Angular Displacement

WBBSE Class 11 Angular Displacement Notes

Angular Displacement Definition: The angle subtended at the centre by the initial and final positions of a particle revolving in a circular path is called the angular displacement of the particle.

Let us consider that the initial position of a particle is point A whose angular coordinate is θ_1, and its final position is B whose angular coordinate is θ₂. So, for the movement of the particle from A to B, i.e., for its path of motion AB, the angular displacement is

θ = ∠AOB = θ₂ -θ₁ = change in angular position

So, the value of ∠AOB indicates the value of the angular displacement.

Measurement Of Angles: The most commonly used unit for the measurement of angles is degree. Moreover, an angle can also be measured by the ratio of the arc-length subtending that angle at the centre of the circle to the radius of the circle.

For example, \(\angle A O B=\frac{\text { arc length } A B}{\text { radius } O A}\)

Actually, the relation, \(\theta=\frac{\text { arc }}{\text { radius }}\), provides the defination of an angle d. Arc and radius—both have the dimension of length. So, their ratio, an angle is a dimensionless quantity.

The unit of the angle, defined as \(\frac{\text { arc }}{\text { radius }}\), is radian (abbreviated as rad). Hence, the angle formed, by an arc of a circle equal in length to the radius of the circle, at its centre is one radian.

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1 revolution =360° = 2π radian

∴ 1 rad = \(\frac{180^{\circ}}{\pi}=57.296^{\circ}\)

As radian is the ratio of two lengths, it is dimensionless. It is only a number. For this reason, the use of radians while measuring angles by any method is advantageous.

The angular displacement is a dimensionless physical quantity when measured in radians.

Polar Vector Axial Vector: We know that linear displacement, velocity and acceleration are vector quantities.

Angular Displacement Formula and Derivation

Similarly, angular quantities like angular displacement, angular velocity and angular acceleration are also vectors.

• In order to express them completely, we need to mention their definite directions along with their magnitudes. By convention, the directions of these vector quantities are taken along the axis of rotation.
• Vectors like linear displacement, linear velocity, linear acceleration, linear momentum, and force have real directions. These are known as polar vectors. The initial point of any polar vector is known as the pole of the vector.

On the other hand, the direction of the vectors associated with rotational motion (like angular displacement, angular velocity, angular acceleration, etc.) is imagined to be along a real axis, which is nothing but the axis of rotation. These vectors are called axial vectors.

• Along the same circular path, the motion of a particle may be clockwise or anticlockwise. These two kinds of motion are opposite to each other. Hence, the two opposite directions of the axis of rotation are considered as the directions of the axial vectors in these two cases.
• The convention Is If the direction of rotation is along the direction of rotation of a right-handed screw then the direction of advancement of the screw-head indicates the direction of the axial vector.
• For example, while opening the cap of a bottle placed on a table, if the rotation of the cap is anticlockwise (as seen from the above), then the advancement of the cap will be in the upward direction.

This direction is chosen as the direction of the axial vectors like angular displacement. This is shown. The opposite is shown. The angular displacement Δθ along the circular path and the corresponding vector \(\Delta \vec{\theta}\) along the axis of rotation are shown.

Key Concepts of Angular Displacement

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Relation Between Linear Displacement And Angular Displacement: Let us assume that the length of arc AB is s and the radius of the circular path is r. Then, the angular displacement of the particle can be expressed as

θ = \(\frac{\text { arc length } A B}{\text { radius } O A}=\frac{s}{r}\)

i.e., s = rθ

or, distance travelled = radius x angular displacement

This distance s cannot be termed as the linear displacement of the particle, because it is a scalar quantity but displacement is a vector quantity. In circular motion, the direction of displacement changes continuously and hence, the magnitude of the distance travelled is not equal to the magnitude of displacement.

Understanding Angular Displacement in Circular Motion

In the case of circular motion, the use of angular displacement is more advantageous than linear displacement for the following reasons:

1. Linear displacement changes direction continuously but the direction of angular displacement remains the same.
2. Different particles of an extended body may travel different distances, but the angular displacement of every particle is the same. When an electric fan rotates, a particle at a greater distance from the centre of the fan covers more distance than a particle nearer to the centre, although both of them subtend equal angles at the centre of rotation, i.e., both have the same angular displacement.

 Friction Short Answer Type Questions

WBBSE Class 11 Short Answer Questions on Friction

Question 1. A block of mass M rests on an inclined plane. If the coefficient of friction between the block and the plane is μ, then the block will slide down the plane under its own weight when the angle of inclination is

1. θ >tan^-1 (μ)
2. θ >tan^-1 (1/μ)
3. θ <tan^-1 (μ)
4. θ <tan^-1 (1/μ)

Answer: The option 1 is correct.

Question 2. A block of mass 5 kg rests on a table. 8 N horizontal force is applied to push the block. Find out the force of friction between the block and the table. Given, μ_s = 0.3 and μ_k = 0.2, between the block and the table and g = 10m · s^-2
Answer:

Given

A block of mass 5 kg rests on a table. 8 N horizontal force is applied to push the block.

Given, μ_s = 0.3 and μ_k = 0.2, between the block and the table and g = 10m · s^-2

Normal reaction = 5×10 = 50N

So, limiting friction = μ_s x normal reaction = 0.3 x 50 = 15

Hence, on the application of 8 N horizontal force, the block remains at rest

So, frictional force = applied force = 8 N

Question 3. A block is placed on a horizontal surface. The block is pushed by applying a force F which acts at an angle of θ with the vertical. How much force will be required to move the block if the frictional coefficient μ? Discuss the fact if tanθ < μ.
Answer:

Given

A block is placed on a horizontal surface. The block is pushed by applying a force F which acts at an angle of θ with the vertical.

Normal reaction force, N = (mg+ Fcosθ)

∴ Minimum required force (horizontal) to move the block, p_min = μ(mg + Fcosθ)

Now, \(\frac{\text { required force }\left(P_{\min }\right)}{\text { applied horizontal force }(P)}=\frac{\mu(m g+F \cos \theta)}{F \sin \theta}\)

Key Concepts of Friction: Short Answer Format

If \(\tan \theta<\mu\) then \(\frac{P_{\min }}{P}>\frac{m g \tan \theta+\tan \theta \cdot F \cos \theta}{F \sin \theta}>\frac{m g}{F \cos \theta}+1>1\) [because cosθ >0]

i.e., P<P_min

∴ The block remains static.

Question 4.

1. Establish the relation between the angle of friction and the angle of repose.
2. Explain with reason, whether the coefficient of friction between two surfaces can be zero.

Answer:

1. \(\vec{F}\) = limiting friction, \(\vec{N}\) = normal reaction, \(\vec{R}\) = resultant of \(\vec{F}\) and \(\vec{N}\).

The angle λ, between \(\vec{R}\) and \(\vec{N}\) is called the angle of friction.

∴ tanλ = \(\frac{F}{N}\) = μ = coefficient of friction

When a body just starts to move downward due to its own weight along an inclined plane, the angle of inclination of the plane at that moment is called the angle of repose.

Now, coefficient of friction, \(\mu=\frac{F}{N}=\frac{W \sin \theta}{W \cos \theta}=\tan \theta\)

∴ \(\mu=\tan \lambda=\tan \theta \quad \text { or, } \theta=\lambda\)

i.e., the angle of friction and the angle of repose are equal.

2. Coefficient of friction, μ = \(\frac{F}{N}\); for p to be zero F has to be zero. In practice, no matter how smooth two surfaces may be, there will always be some friction between the two surfaces whenever one is made to move over the other. Hence, μ can never be zero.

Applications of Friction in Daily Life: Short Answers

Question 5. An object of weight W rests on an inclined plane. The coefficient of friction between the object and the plane is μ. For what value of the angle of inclination θ, will the object move downward with uniform speed under its own weight?
Answer:

Given

An object of weight W rests on an inclined plane. The coefficient of friction between the object and the plane is μ.

The forces acting on the body along with their components are shown.

When the body is about to slide down along the incline then, \(W \sin \theta=\mu R=\mu W \cos \theta\)

or, \(\tan \theta=\mu \quad \text { or, } \theta=\tan ^{-1} \mu\)

Question 6. To determine the coefficient of friction between a rough surface and a block, the surface is kept inclined at 45° and the block is released from rest. The block takes a time t in moving a distance d. The rough surface is then replaced by a smooth surface and the same experiment is repeated. The block now takes a time t/2 in moving down the same distance d. The coefficient of friction is

1. 3/4
2. 5/4
3. 1/2
4. 1/2

Answer:

Given

The surface is kept inclined at 45° and the block is released from rest. The block takes a time t in moving a distance d. The rough surface is then replaced by a smooth surface and the same experiment is repeated. The block now takes a time t/2 in moving down the same distance d.

For a smooth surface, the block takes a time t/2 in moving a distance d with an acceleration gsinθ.

∴ d = \(\frac{1}{2} g \sin \theta \cdot \frac{t^2}{4}=\frac{1}{8} g \sin \theta \cdot t^2\)…(1)

For rough surfaces, the block now takes a time t in moving down the same distance with an acceleration (gsinθ-f) or (gsinθ – μgcosθ).

∴ d = \(\frac{1}{2}(g \sin \theta-\mu g \cos \theta) \cdot t^2\)…(2)

Comparing equations (1) and (2), \(\frac{1}{4} \sin \theta=\sin \theta-\mu \cos \theta\)

or, \(\mu=\tan \theta-\frac{1}{4} \tan \theta=\frac{3}{4} \tan \theta=\frac{3}{4} \tan 45^{\circ}=\frac{3}{4}\)

The option 1 is correct.

Factors Affecting Friction: Short Answer Questions

Question 7. Block B lying on a table weighs W. The coefficient of static friction between the block and the table is μ. Assume that the cord between B and the knot is horizontal. The maximum weight of block A for which the system will be stationary is

1. \(\frac{W \tan \theta}{\mu}\)
2. \(\mu W \tan \theta\)
3. \(\mu W \sqrt{1+\tan ^2 \theta}\)
4. \(\mu W \sin \theta\)

Answer:

Given

Block B lying on a table weighs W. The coefficient of static friction between the block and the table is μ. Assume that the cord between B and the knot is horizontal.

The normal reaction of the table on block B = W

So the effective frictional force acting on the left = μW

Again when the maximum weight of A is W₀, it acts downwards.

Therefore, when in equilibrium, the cord between the knot and the wall will be along the resultant of μW and W₀.

In that case, \(\tan \theta=\frac{W_0}{\mu W} \quad \text { or, } W_0=\mu W \tan \theta\)

The option 2 is correct.

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Question 8. A block of mass m₂ is placed on a horizontal table and another block of mass m₁ is placed on top of it. An increasing horizontal force F = αt is exerted on the upper block but the lower block never moves as a result. If the coefficient of friction between the blocks is μ₁ and that between the lower block and the table is μ₂, then what is the maximum possible value of fi μ₁/μ₂?

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

Answer:

Given

A block of mass m₂ is placed on a horizontal table and another block of mass m₁ is placed on top of it. An increasing horizontal force F = αt is exerted on the upper block but the lower block never moves as a result. If the coefficient of friction between the blocks is μ₁ and that between the lower block and the table is μ₂

∴ \(\mu_2\left(m_1+m_2\right) g\) ≥ \(\mu_1 m_1 g\)

or, \(\frac{m_1+m_2}{m_1}\) ≥ \(\frac{\mu_1}{\mu_2}\)

or, \(\frac{\mu_1}{\mu_2}\) ≤ \(1+\frac{m_2}{m_1} or, \left(\frac{\mu_1}{\mu_2}\right)_{\text {max }}=1+\frac{m_2}{m_1}\)

The option 2 is correct.

Types of Friction Explained: Short Answers

Question 9. Given two blocks A and B of weight 20 N and 100 N, respectively. These are being pressed against a wall by a force F as shown. If the coefficient of friction between the blocks is 0.1 and between block B and the wall is 0.15, the frictional force applied by the wall on block B is 

1. 100 N
2. 80 N
3. 120 N
4. 150 N

Answer:

Given

Given two blocks A and B of weight 20 N and 100 N, respectively. These are being pressed against a wall by a force F as shown. If the coefficient of friction between the blocks is 0.1 and between block B and the wall is 0.15,

On the plane of contact of A and B, the upward frictional force acting on A = μF = W_A = 20 N.

So the downward frictional force on B on that plane = 20 N

This 20 N force and the weight of B is balanced by μ₂F (the upward frictional force applied by the wall on B)

∴ μ₂F = 20 + W_A = 20 + 100 = 120 N

The option 3 is correct.

Question 10. Two masses m₁ = 5 kg and m₂ = 10 kg, connected by an inextensible string over a frictionless pulley, are moving as shown. The coefficient of friction of the horizontal surface is 0.15. The minimum weight m that should be put on top of m₂ to stop the motion is

1. 43.3 kg
2. 10.3 kg
3. 18.3 kg
4. 27.3 kg

Answer:

Given

Two masses m₁ = 5 kg and m₂ = 10 kg, connected by an inextensible string over a frictionless pulley, are moving as shown. The coefficient of friction of the horizontal surface is 0.15.

Let the mass m₂ come to rest if a mass m is placed on it. For the mass m₂ to come to rest, an acceleration (due to friction) should act in the opposite direction of its motion i.e., when f≤ μN

Also f = T = 5 g [when m and m₂ are at rest]

∴ m₁g ≤ μ(m +m₂)g or, 5 ≤ 0.15(m +10)

or, 23.33 <≤ m

The option 4 is correct.

Question 11. A system consists of three masses m₁, m₂, and m₃ connected by a string passing over a pulley P. The mass m₁ hangs freely and m₂ and m₃ are on a rough horizontal table (the coefficient of friction = μ). The pulley is frictionless and of negligible mass. The downward acceleration of mass ml is (Assume m₁ = m₂ = m₃ = m)

1. \(\frac{g(1-g \mu)}{9}\)
2. \(\frac{1 g \mu}{3}\)
3. \(\frac{g(1-2 \mu)}{3}\)
4. \(\frac{g(1-2 \mu)}{2}\)

Answer:

Given

A system consists of three masses m₁, m₂, and m₃ connected by a string passing over a pulley P. The mass m₁ hangs freely and m₂ and m₃ are on a rough horizontal table (the coefficient of friction = μ). The pulley is frictionless and of negligible mass.

Acceleration = \(\frac{\text { net force in the direction of motion }}{\text { total mass of system }}\)

= \(\frac{m_1 g-\mu\left(m_2+m_3\right)} g= {m_1+m_2+m_3}=\frac{g(1-2 \mu)}{3}\)

The option 3 is correct.

Mathematical Concepts in Friction: Short Answer Questions

Question 12. A block A of mass m₁ rests on a horizontal table. A light string connected to it passes over a frictionless pulley at the edge of the table and from its other end, another block B of mass m₂ is suspended. The coefficient of kinetic friction between the block and the table is μ_k. When the block A is sliding on the table, the tension in the string is

1. \(\frac{\left(m_2+\mu_k m_1\right) g}{\left(m_1+m_2\right)}\)
2. \(\frac{\left(m_2-\mu_k m_1\right) g}{\left(m_1+m_2\right)}\)
3. \(\frac{m_1 m_2\left(1+\mu_k\right) g}{\left(m_1+m_2\right)}\)
4. \(\frac{m_1 m_2\left(1-\mu_k\right) g}{\left(m_1+m_2\right)}\)

Answer:

Given

A block A of mass m₁ rests on a horizontal table. A light string connected to it passes over a frictionless pulley at the edge of the table and from its other end, another block B of mass m₂ is suspended. The coefficient of kinetic friction between the block and the table is μ_k

The frictional force acting on block A = μ_km₁g

Hence, the resultant force on the blocks = m₂g-μ_km₁g

Their acceleration, a = \(\frac{m_2 g-\mu_k m_1 g}{m_1+m_2}\)

If the tension in the string is T, then for block B, \(m_2 g-T=m_2 a\)

∴ T = \(m_2(g-a)=m_2 g\left(1-\frac{m_2-\mu_k m_1}{m_1+m_2}\right)\)

= \(m_2 g \frac{m_1+\mu_k m_1}{m_1+m_2}\)

= \(\frac{m_1 m_2\left(1+\mu_k\right) g}{m_1+m_2}\)

The option 3 is correct.

Question 13. Which one of the following statements is incorrect?

1. Frictional force opposes the relative motion
2. The limiting value of static friction is directly proportional to the normal reaction
3. Rolling friction is smaller than sliding friction
4. The coefficient of sliding friction has dimensions of length

Answer: 4. The coefficient of friction is a dimensionless quantity.

The option 4 is correct.

Question 14. The inclination θ of a rough plane is increased gradually. The ply on the plane just comes into motion when inclination θ becomes 30°. Find the coefficient of friction. If the inclination is further increased to 45° then find the acceleration of the body along the plane, (g = 10 m · s^-2)
Answer:

Given

The inclination θ of a rough plane is increased gradually. The ply on the plane just comes into motion when inclination θ becomes 30°.

Here, angle of repose =30°

∴ Coefficient of friction, μ = tan30° = 1/√3

If the angle of inclination is 45°, the acceleration of the body,

a = \(g\left(\sin 45^{\circ}-\mu^{\prime} \cos 45^{\circ}\right)=10 \times \frac{1}{\sqrt{2}}\left(1-\mu^{\prime}\right)\)

= \(\frac{10}{\sqrt{2}}\left(1-\frac{1}{\sqrt{3}}\right) \approx 2.99 \mathrm{~m} \cdot \mathrm{s}^{-2} \quad\left[because \mu^{\prime} \approx \mu\right]\)

Question 15. Give the magnitude and direction of the net force acting on a car moving with a constant velocity of 30 km · h^-1 on a rough road.
Answer:

The velocity does not change, so, there is no acceleration.

The relation F = ma shows that, as a = 0, the net force F = 0.

Here, the force applied by the engine is balanced exactly by the friction of the road.

Short Answer Questions on Laws of Friction

Question 16. Two bodies A, B of masses 5 kg and 10 kg in contact with each other rest on a table against a rigid wall. The coefficient of friction between the bodies and the table is 0.15. A force of 200N is applied horizontally to A.

1. What is the reaction of the partition?
2. The action-reaction forces between A, and B?
3. What happens when the wall is removed?

Does the answer 2. Change when bodies are in motion? [Ignore the difference between μ_s and μ_k]
Answer:

We take the rounded-off value, g = 10 m · s^-2

Normal reaction on A, N_A = its weight (mg)

= 5 x 10 = 50 N

Normal reaction on B, N_B = 10 x 10 = 100 N

So the limiting friction on A, f_A = μN_A = 0.15 x 50 = 7.5 N

and limiting friction on B, f_B = μN_B = 0.15 x 100 = 15 N

(As the applied force, 200 N, is higher, the force of friction attains its limiting value)

1. We take the combination of A and B as a single body. It is at rest due to the presence of the wall. If R is the reaction of the wall, the force equation in the horizontal direction is, 200-f_A-f_B-R=0

or, R = 200- 7.5- 15 = 177.5 N

2. Let F_AB = action force of A on B; F_BA = reaction force of B on A; F_AB and F_BA are equal and opposite. From the equilibrium of A, we get 200-f_A-F_BA = 0

or, F_BA = 200-7.5 = 192.5 N

So, F_AB = -192.5 N

3. Now, if the wall is removed, the reaction force R is absent; due to the applied force, the combination of A and B would move forward with an acceleration a (say). Then the force equation in the horizontal direction would be, 200 -f_A-f_B = (m_A+m_B)a

or, a = \(\frac{200-7.5-15}{5+10}=\frac{177.5}{15}=11.83 \mathrm{~m} \cdot \mathrm{s}^{-2}\)

Now, the force equation for the body A is, 200-f_A-F’_BA = m_Aa

[F’_BA = new value of force by B on A]

or, F’_BA = 200-7.5-5 x 11.83 – 133.3 N

Similarly, F’_AB = -133.3 N

So due to the accelerated motion, the action-reaction pair changes to a much reduced value.

Question 17. Define:

1. Stopping distance of a vehicle,
2. Reaction time during a free fall.

Answer:

1. Let at a certain moment, the brakes are applied on a vehicle running on a road. If the vehicle then travels a distance x until it comes to rest, then x is called the stopping distance of the vehicle.
2. Many natural events demand immediate action from the observers. An observer observes an event, registers it in his brain, and then takes the required action. The time spent between the first observation and the action taken is called his/her reaction time.
   • For example, suppose an observer sees an object falling freely from the ceiling of a room towards a glass plate kept on a table. He/she would try either to intercept the falling object or to remove the plate from the table. In this context, if the reaction time is greater than the time of fall, then the glass would break.

Question 18. Give two methods of reducing friction. Show that kinetic friction is less than static friction.
Answer:

1. Application of lubricating oils, and
2. Use of ball bearings and roller bearings, between the surfaces in contact, are two of many methods that can reduce friction.

• When a force, applied to move a body over a surface, just exceeds the limiting friction, the body starts to move. Now, even if the applied force is not increased any more, the body is observed to accelerate.
• This means that the resultant force is positive, i.e., the kinetic friction during the motion is less than the applied force. This indicates that the kinetic friction falls below the limiting friction (the limiting value of static friction) as soon as the body starts to move.

Question 19. What is the acceleration of the block and trolley system shown given below, if the coefficient of friction between the trolley and the surface is 0.04? What is the tension in the string? (Take g = 10 m · s^-2). Neglect the mass of the string.

Answer:

The 3 kg block and the 20 kg trolley both have the same magnitude of acceleration.

Hence, for 3 kg block, 30 – T = 3a…..(1)

where T is tension in the string and a is acceleration.

For the trolley, T  – f = 20a

Now for friction, f = μR = 0.04 x 20 x 10 = 8N

where, f = force of friction

μ = coefficient of friction between trolley and the surface

and R = normal force on the trolley

Hence, T – 8 = 20a…..(2)

Solving equations (1) and (2), we get, a = \(\frac{22}{23}\) m s = 0.96 m s and T = 27.2 N

Question 20. A block of mass 3 kg slides down an incline of angle 30° with acceleration g/4. Complete the free-body diagram and find the coefficient of kinetic friction.

Answer:

Given

A block of mass 3 kg slides down an incline of angle 30° with acceleration g/4.

The forces on the block of mass m are as shown.

Taking components along the inclined plane, \(m g \sin 30^{\circ}-f=\frac{m g}{4}\)[because a=g]

or, f = \(\frac{m g}{4}\)

There is no acceleration perpendicular to inclin plane, so R = \(m g \cos 30^{\circ}=\frac{\sqrt{3}}{2} m g\)

Then coefficient of friction, \(\mu=\frac{f}{R}=\frac{m g}{4\left(m g \frac{\sqrt{3}}{2}\right)}=\frac{1}{2 \sqrt{3}}\)

Class 11 Physics Friction Multiple Choice Questions And Answers

WBBSE Class 11 Friction MCQs with Answers

Question 1. During the acceleration of a bicycle, the direction of the force of static friction, exerted by the ground on the wheels is

1. Against the direction of motion of the front wheel, and along the direction of motion of the rear wheel
2. Along the motion of the front wheel, and against the motion of the rear wheel
3. Against the motion of both the wheels
4. Along the motion of both the wheels

Answer: 1. Against the direction of motion of the front wheel, and along the direction of motion of the rear wheel

Question 2. A block of mass 0.1 kg is kept pressed against a wall, by applying a force of 5 N horizontally on the block. The coefficient of friction between the block and the wall is 0.5. Friction acting on the block is

1. 2.5 N
2. 0.98 N
3. 4.9 N
4. 0.49 N

Answer: 2. 0.98 N

Question 3. A piece of stone of mass 1 kg slides over ice at 2 m · s^-1 and comes to rest in 10 s. In this case, the frictional force is

1. 0.2N
2. 20N
3. 10N
4. 1N

Answer: 1. 0.2N

Read And Learn More WBCHSE Class 11 Physics MCQs

Question 4. A uniform chain of length l is lying on a rough table and 1/nth of its length is hanging from the table’s edge. If the chain is about to slide off the table, the coefficient of friction between the chain and the table is

1. \(\frac{1}{n}\)
2. \(\frac{1}{n-1}\)
3. \(\frac{1}{n+1}\)
4. \(\frac{n-1}{n+1}\)

Answer: 2. \(\frac{1}{n-1}\)

Question 5. If the coefficient of friction is 1/√3, the height up to which a particle can rise and stay inside a hollow sphere of radius r, is (the inner surface of the sphere is rough)

1. 0.5 r
2. 0.75 r
3. 0.95 r
4. 0.134 r

Answer: 4. 0.134 r

Conceptual MCQs on Friction for Class 11

Question 6. A block of mass 5 kg is placed on a rough horizontal surface. The coefficients of static and sliding friction between the body and the surface are 0.7 and 0.5 respectively. A horizontal force is applied on the block so that the block just starts moving. If the applied force continues to act even after the block is set in motion, the acceleration of the block will be [g = 10 m · s^-2]

1. 1 m s^-2
2. 2 m s^-2
3. 3 m s^-2
4. 4 m s ^-2

Answer: 2. 2 m s^-2

Question 7. Graphs below show the variation of the frictional force, against the force on a block, applied parallel to the surface of contact of the block with a rough horizontal plane. Out of 1, 2, 3, 4 which one is the correct graph?

Answer: 2

Question 8. A tram is moving with an acceleration of 49 cm · s^-2. If 50% of the engine power is used up to overcome friction, and the remaining 50% is spent for increasing velocity, the coefficient of friction between the wheel and the track should be

1. 0.03
2. 0.02
3. 0.05
4. 0.10

Answer: 3. 0.05

Question 9. There is no slipping between the two blocks shown. What is the force of friction between the blocks?

1. Zero
2. 9N
3. 12 N
4. 6N

Answer: 4. 6N

Practice Questions on Static and Kinetic Friction

Question 10. A block of weight 5 N is pushed against a vertical wall by a force of 12 N. The coefficient of friction between the wall and the block is 0.6. The magnitude of the force exerted by the wall on the block is

1. 12 N
2. 5N
3. 7.2 N
4. 13 N

Answer: 3. 7.2 N

Question 11. In order to stop a car in the shortest distance on a horizontal road, one should

1. Apply the brakes very hard so that the wheels stop rotating
2. Apply the brakes hard enough to just prevent slipping
3. Pump the brakes (press and release)
4. Shut the engine off and do not apply brakes

Answer: 2. Apply the brakes hard enough to just prevent slipping

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Question 12. A boy of mass M is applying a horizontal force to slide a box of mass M’ on a rough horizontal surface. The coefficient of friction between the shoes of the boy and the floor is μ and that between the box and the floor is μ’. In which of the following cases it is certainly not possible to slide the box?

1. μ<μ’, M< M’
2. μ> μ’, M<M’
3. μ < μ’, M>M’
4. μ>μ’, M>M’

Answer: 1. μ < μ’, M< M

Question 13. Angle of an inclined plane with the horizontal is θ. The first half of the plane is smooth and the other half is rough. When a body is released from the top of the plane, it slides down and comes to rest at the bottom. The coefficient of friction between the body and the inclined plane is

1. μ = 2 tanθ
2. μ = tanθ
3. \(\mu=\frac{2}{\tan \theta}\)
4. \(\mu=\frac{1}{\tan \theta}\)

Answer: 1. μ = 2 tanθ

Question 14. Two blocks of masses m₁ = 5 kg and m₂ = 6 kg, are connected by a weightless frictionless string passing over a pulley and are kept as shown. While m2 is hanging vertically, is resting on an incline.

Key MCQs on Laws of Friction

The angle of inclination θ = 30°. In this case, the magnitude and direction of the frictional force on mass m₁, are [g = 10 m ·s^-2]

1. 35 N, upward along the inclined plane
2. 35 N, downward along the inclined plane
3. 85 N, upward along the inclined plane
4. 85 N, downward along the inclined plane

Answer: 2. 35 N, downward along the inclined plane

Question 15. A body slides down an inclined plane of inclination θ. While sliding downwards, the coefficient of friction is directly proportional to the displacement. The body slides down the plane with a

1. Constant acceleration g sinθ
2. Constant acceleration (g sinθ – μg cosθ)
3. Constant retardation (μg cosθ – g sinθ)
4. Variable acceleration

Answer: 4. Variable acceleration

Question 16. For an object sliding on a plane, the force of friction is less if the plane is inclined, instead of being horizontal, because,

1. The coefficient of friction decreases
2. Normal force decreases
3. Effective mass decreases
4. For an angle of inclination θ, friction is inversely proportional to tan θ

Answer: 2. Normal force decreases

Question 17. A body of mass m is pushed up with a velocity u along a plane of inclination θ. If the coefficient of friction between the body and the inclined plane is μ, displacement of the body before coming to rest is then the maximum inclination of the plane with the horizontal is

1. \(\frac{u^2 \mu}{2 g \sin \theta}\)
2. \(\frac{u^2 \mu}{2 g \cos \theta}\)
3. \(\frac{u^2}{4 g \sin \theta}\)
4. \(\frac{u^2}{4 g \cos \theta}\)

Answer: 3. \(\frac{u^2}{4 g \sin \theta}\)

Question 18. A mass placed on an inclined plane is just in equilibrium. If μ is the coefficient of friction of the surface, then the maximum inclination of the plane with the horizontal is

1. tan^-1μ
2. tan^-1(μ/2)
3. sin^-1μ
4. cos^-1μ

Answer: 1. tan^-1μ

Sample Questions on Coefficient of Friction

Question 19. A 13 m ladder is placed against a smooth vertical wall with its lower end 5 m from the wall. What should be the minimum coefficient of friction between the ladder and the floor so that it remains in equilibrium?

1. 0.36
2. 0.72
3. 0.21
4. 0.52

Answer: 3. 0.21

Question 20. A box of mass 8 kg is placed on a rough inclined plane of inclination θ. Its downward motion can be prevented by applying an upward pull F and it can be made to slide upwards by applying a force 2 F. The coefficient of friction between the box and the inclined plane is

1. 1/3 tanθ
2. 3tanθ
3. 1/2tanθ
4. 2tanθ

Answer: 1. 1/3 tanθ

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

Question 21. Three blocks are arranged as shown. The whole system is placed on a smooth horizontal plane. The coefficient of friction between the plane and the 5 kg block is 0.1, between the 5kg block and 3kg block is a smooth horizontal surface of 0.1, and between the 3 kg and 2 kg block is 0.2. The total frictional force on the 3 kg block is

1. Towards right
2. Towards left
3. Zero
4. Non zero

Answer:

2. Towards the left

4. Non-zero

Question 22. If the force of friction is equal to the force applied, then friction may be

1. Static
2. Kinetic
3. Limiting
4. No conclusions can be drawn

Answer:

1. Static

3. Limiting

WBBSE Class 11 Practice Tests on Friction

Question 23. If the object is at rest, then friction may be

1. Static
2. Kinetic
3. Limiting
4. No conclusion can be drawn

Answer:

1. Static

3. Limiting

Question 24. Two particles A and B, each of mass m are kept stationary by applying a horizontal force F = mg on particle B as shown. Then

1. tanβ = 2 tanα
2. 2T₁ = 5T₂
3. √2T₁ = √5T₂
4. None of these

Answer:

1. tan/3 = 2 tan

3. √2T₁ = √5T₂

Interactive MCQs on Friction for Students

Question 25. Mark the correct statements about the friction between two bodies.

1. Static friction is always greater than kinetic friction
2. The coefficient of static friction is always greater than the coefficient of kinetic friction
3. Limiting static friction is always greater than kinetic friction.
4. Limiting friction is never less than static friction

Answer:

2. The coefficient of static friction is always greater than the coefficient of kinetic friction

3. Limiting static friction is always greater than the kinetic friction.

4. Limiting friction is never less than static friction

 Friction Long Answer Type Questions

WBBSE Class 11 Long Answer Questions on Friction

Question 1. Why should one take short footsteps while walking on ice (or oily surface)?
Answer:

To walk, one exerts a force F obliquely on the ground. The ground also exerts an equal and opposite reaction force R on the man. It is the horizontal component H, of R, which makes walking possible. Frictional force f supplies this horizontal component.

As an icy surface is very smooth, the frictional force is very small. Hence, any attempt to take longer steps can make H greater than the limiting friction. As a result, the person may slip forward.

Question 2. In rainy season sand is sometimes thrown on railway tracks. Why?
Answer:

In rainy season sand is sometimes thrown on railway tracks.

The force impressed on the wheel is generally greater than the force of rolling friction; this produces the rolling of the wheel on the railway track. On the other hand, the kinetic friction is fairly higher than the impressed force, and there is no sliding of the wheels.

But in the rainy season, on the wet railway track, there is a danger that the kinetic friction may drop below the impressed force, resulting in a sliding of the wheels. To avoid this, sand is sometimes thrown on the railway track to increase the kinetic friction.

Read and Learn More Class 11 Physics Long Answer Questions

Question 3. Show that the coefficient of friction is a dimensionless quantity.
Answer:

From the definition, the coefficient of friction,

= \(\frac{\text { limiting friction }(f)}{\text { normal force }(R)}\)

Both f and R are forces and have the same dimension MLT^-2.

∴ Dimension of μ = \(\frac{\mathrm{MLT}^{-2}}{\mathrm{MLT}^{-2}}=1\)

Hence, μ is a dimensionless quantity.

Detailed Explanation of Friction in Physics

Question 4. A chair is kept on the floor. When does friction act between them? Where does this force act? Is the magnitude of this force a constant?
Answer:

When the chair is at rest on the floor, frictional force does not act. Friction comes into play when one tries to drag the chair over the floor and it continues to act when the chair is actually in motion.

• The frictional force acts parallel to the surface of contact between the chair and the floor and opposite to the direction of motion or relative motion.
• The magnitude of the force of friction is not constant. As the force on the chair is gradually increased, the force of friction also increases and reaches a limit, called the force of limiting friction.
• On further increase of the applied force, the chair will start moving. Now, the magnitude of the frictional force decreases a little and thereafter remains constant.

Question 5. Can the value of the coefficient of friction be greater than 1?
Answer:

The value of the coefficient of friction (μ) is usually less than 1. But in some special cases, its value can be equal to or even greater than 1. The value of μ between two metal surfaces, cleaned scientifically and kept in a vacuum, may rise up to 10 approximately. Under these conditions, μ ≈ 1.6 between two copper plates.

Question 6. Explain why it is difficult to write on a paper surface that is too smooth or too rough.
Answer:

A force of friction acts at the point of contact of the pen and the paper, against the motion of the pen. For a very smooth paper, the frictional force is low, and the pen slips on the paper’s surface. For a very rough paper, frictional force is quite high, and this makes it difficult to move the pen over the paper. Thus, these surfaces are not suitable for writing.

Question 7. A body of mass m is kept over an object of mass M. The object is at rest on smooth ground. The coefficient of static friction between the two bodies is μ. What is the minimum force that needs to be applied to the object, so that the body will be able to slip over it?
Answer:

Given

A body of mass m is kept over an object of mass M. The object is at rest on smooth ground. The coefficient of static friction between the two bodies is μ.

Let the force applied on the object horizontally, so that the body is about to slide be F.

Friction and Its Importance in Daily Life

Hence, the common acceleration of the system, a = \(\frac{F}{M+m}\)

so the force on the body = ma = \(\frac{Fm}{M+m}\)

When the body is about to slip, this force is balanced by the force of limiting friction acting at the plane of contact.

∴ Frictional force, f = μmg = \(\frac{Fm}{M+m}\)

or, F = μ(m+m)g.

Applications of Friction: Long Answer Questions

Question 8. The effectiveness of the brake of a car does not depend on the area of contact of the brakeshoe with the rim of the wheel—explain.
Answer:

The effectiveness of the brake of a car does not depend on the area of contact of the brakeshoe with the rim of the wheel

A car brake is applied to generate a frictional force against the rolling of the wheel, and thereby to decrease the velocity of the car.

Frictional force does not depend on the area of contact but depends on the nature of the surfaces in contact and the normal reaction. The area of contact of the brake has no effect on its functioning.

Question 9. While polishing a substance, if the polishing cloth is pressed hard, a considerable amount of heat is developed. Why?
Answer:

An increase in pressure increases the normal force. Hence, the force of friction also increases. So, a greater amount of work has to be done against friction while polishing, which in turn, generates a greater amount of heat.

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Factors Affecting Friction: Long Answer Format

Question 10. Will there be any change in the coefficient of friction between the surfaces of two objects, if they are taken to the moon?
Answer:

The coefficient of friction depends on the materials and the smoothness of the contact planes of the two objects. As these factors remain unchanged even when the objects are taken to the moon, the coefficient of friction will also remain the same.

Question 11. When the wheels of a car are bogged down in mud, why cannot the car move forward?
Answer:

The car engine rotates the wheels; and due to friction between the road and the wheels, the car moves forward. When the friction cannot provide a sufficient reaction force, the engine cannot make the car move forward by rotating the wheels; they continue to rotate in the same place without any translation.

Question 12. After a rainfall, one should not drive very fast on a wet asphalt road. Explain.
Answer:

The coefficient of static friction between an asphalt road and the tires plays an important role in controlling the motion of a car. If the value of the coefficient is small, it becomes difficult to accelerate or steer the car.

In some cases, the wheels may start to skid, and the driver may lose control. Such a situation arises while driving on a wet road at high speed, as, the coefficient of static friction between the tyres and the road falls appreciably when this road gets wet.

Mathematical Analysis of Friction Forces

Question 13. A body of mass m is placed on a platform of mass M (m«M), moving with a velocity v. If the coefficient of friction between the platform and the mass is μ, then for how long will the body continue to slide on the platform, and what distance will it cover during that time?
Answer:

Given

A body of mass m is placed on a platform of mass M (m«M), moving with a velocity v. If the coefficient of friction between the platform

When the platform moves, the body on it tends to slide backward. So the frictional force on the body acts in the direction of motion of the platform. It is given by, f= μmg.

∴ Its acceleration, a = \(\frac{f}{m}\) = μg

When the velocity of the body becomes v, it no longer slides over the platform.

We know, \(v=u+a t\)or, \(t=\frac{v-u}{a}\)

Here, u=0, a = \(\mu g\)

t = \(\frac{\nu}{\mu g}\)

Again, \(v^2=u^2+2 a s\)

or, \(s=\frac{v^2-u^2}{2 a}=\frac{v^2}{2 \mu g}\)

∴ The body will slide for a time of \(\frac{v}{\mu g},\), and in this interval of time, it will cover a distance of \(\frac{v^2}{\mu g},\).

Long Answer Questions on Laws of Friction

Question 14. How does the accelerator increase the speed of a car?
Answer:

The accelerator increase the speed of a car as follows

The accelerator, on being pressed, increases the angular velocity of the wheel, and thereby the rolling friction increases. At the point of contact with the ground, the wheel rotates backward so the rolling friction acts in the forward direction. The car accelerates due to the increase of this rolling friction.

Disadvantages Of Friction And Their Remedies

WBBSE Class 11 Advantages of Friction Notes

Disadvantages Of Friction: Friction is disadvantageous in many fields. For example:

1. When a machine is in use, work has to be done against friction due to the relative motion among its different parts. Hence, a part of the applied energy that is spent to overcome friction changes into heat energy, and the efficiency of the machine reduces.
2. Different parts of a running machine usually corrode due to friction. The machine may become useless due to heavy wear and tear.

Remedy To Minimise The Disadvantages Caused By Friction:

Read and Learn More: Class 11 Physics Notes

1. If the contact surfaces are very smooth, then friction is less. The surfaces can be made slippery by lubrication. This deposits a thin layer of oil between the two surfaces.
   • Different mineral oils, vaseline, graphite, wax, and fat are a few commonly used lubricants.
2. The coefficient of friction between two steel surfaces is experimentally found to be greater than that between steel and, for example, an alloy of lead and antimony.
   • So, these types of alloys are used in machines made of steel. These are called antifriction alloys. The process of decreasing friction by using antifriction alloys was invented by the scientist Babbit, and so the process is known as babbling.
3. Rolling friction is less than kinetic friction. Hence, ball bearings or roller bearings are used in machines, wherever possible.
4. Streamlining is another method of reducing frictional drag when an object moves through a fluid and the fluid closest to the object opposes that motion.
   • Vehicles are driven through fluid with high speed example, aeroplanes, jets, and spacecrafts are given special shapes or streamlined to reduce fluid friction. Birds, fishes also have streamlined bodies to reduce frictional drag as they move through air or water.
5. A large frictional force is encountered by a spacecraft during its high-speed journey through the atmosphere, which produces excessive heat on its outer surface. To protect its body against this enormous heat, a special type of thermal barrier is used to cover the body.

WBCHSE Class 11 Physics Notes For Advantages Of Friction

To run different machines efficiently, the frictional force between the movable parts should be minimized. However, it is wrong to think that machines would have been more efficient if friction was totally absent.

As an illustration, we may cite the following examples:

1. Various parts of the machine are able to rotate, and hold together with nuts and bolts because of friction.
2. The tires of a vehicle are made rough to increase friction between the road and the tires and this prevents skidding.
3. Chains are attached to the tires of automobiles to increase friction while driving through snow or ice.

A few other examples from day-to-day life, where friction is involved, are given below:

1. Human beings and animals can walk on the ground. A force is exerted obliquely. While walking, the reaction force generated due to friction makes it possible for a man to move forward without slipping.
2. Trees can hold the ground tightly with their roots.
3. Otherwise, plants could have been uprooted easily.
4. Striking ignites a matchstick.
5. Objects can be held with the fingers and palms of our hands.
6. Ladders can be supported on vertical walls.
7. Sand is thrown on tracks covered with snow to increase friction so that driving or walking on snow becomes safer. Similarly, on a rainy day sand is thrown on the slippery ground to increase the friction between our feet and the ground and thus the chances of slipping is reduced.

Hence, in spite of the inconveniences caused by friction, it plays an important role in nature and in our day-to-day lives.

WBCHSE Class 11 Physics – Disadvantages and Advantages Of Friction Numerical Examples

Disadvantages of Friction in Physics

Example 1. A block of mass 5 kg is kept on a horizontal table. It is connected to a weight of mass 2 kg by a weightless string passing over a smooth pulley. The part of the string on the table is horizontal, and the weight is; hanging freely from the pulley. If the coefficient of kinetic friction between the table and the block is 0. 2, find the acceleration of the block. What will be the tension in the string?
Solution:

Given

A block of mass 5 kg is kept on a horizontal table. It is connected to a weight of mass 2 kg by a weightless string passing over a smooth pulley. The part of the string on the table is horizontal, and the weight is; hanging freely from the pulley. If the coefficient of kinetic friction between the table and the block is 0. 2

Weight of the block, Mg = 5 x 9.8 N

∴ The normal force of the table on the block, R = 5×9.8 N

Kinetic friction against the motion of the block, f’ = μ’R = 0.2 x 5 x 9.8 = 9.8 N

Let the tension in the string be T and the acceleration of the block be a.

Hence, for the motion of the block, T-f’ = Ma

or, T-9.8 = 5a….(1)

The weight of mass m (say) moves downwards with the same acceleration.

∴ mg- T = ma

or, 2 x 9.8- T = 2a [m = 2]….(2)

Adding (1) and (2) we get, 2 x 9.8-9.8 = 7 a

or, a = \(\frac{9.8}{7}\) = 1.4 m·s^-2

Substituting this value in (1), we get, T – 9.8 = 5 x 1.4 or, T= 16.8 N.

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Example 2. A block of mass m is sliding down a stationary inclined plane. The base of the inclined plane has a length of l, and the coefficient of kinetic friction between the block and the inclined surface is 0.14. What should be the inclination of the plane so that the block can slide down to the ground in minimum time?
Solution:

Given

A block of mass m is sliding down a stationary inclined plane. The base of the inclined plane has a length of l, and the coefficient of kinetic friction between the block and the inclined surface is 0.14.

Acceleration of the block along the plane, a = g(sinθ – μ’cosθ)

Let the time required to slide down from A to B be t.

∴ AB = \(\frac{1}{2} a t^2=\frac{l}{\cos \theta}\)

or, \(\frac{1}{2} t^2 g\left(\sin \theta-\mu^{\prime} \cos \theta\right)=\frac{l}{\cos \theta}\)

or, \(t^2=\frac{2 l}{g} \frac{1}{\left(\sin \theta \cos \theta-\mu^{\prime} \cos ^2 \theta\right)}\)

= \(\frac{4 l}{g}\left[\frac{1}{\sin 2 \theta-\mu^{\prime}(1+\cos 2 \theta)}\right]\)

When t is minimum, \(t^2\) is also minimum.

∴ \(\sin 2 \theta-\mu^{\prime}(1+\cos 2 \theta)\)=x (say) is maximum.

∴ \(\frac{d x}{d \theta}=0 \quad or, 2 \cos 2 \theta-\mu^{\prime}(0-2 \sin 2 \theta)=0\),

or, \(\cos 2 \theta+\mu^{\prime} \sin 2 \theta=0\)

or, \(\tan 2 \theta=-\frac{1}{\mu^{\prime}}=-\frac{1}{0.14}=-\tan 82^{\circ}\)

or, \(\tan 2 \theta=\tan \left(180^{\circ}-82^{\circ}\right)=\tan 98^{\circ}\)

∴ \(2 \theta=98^{\circ} and \theta=49^{\circ}\).

Key Benefits of Friction in Everyday Life

Example 3 A coin slides down an inclined plane of inclination ø at a constant speed. Prove that if the coin is pushed up with a velocity u on that plane, it can rise up to \(\frac{u^2}{4 g \sin \phi}\), and from there it will not slide down again.
Solution:

Given

A coin slides down an inclined plane of inclination ø at a constant speed.

Since, the coin slides down with a uniform speed, it has no acceleration along the plane. So, fμ’R = mg sinø

The downward force acting on the coin as it is pushed up, F = mg sinø + μ’R = mg sinø + mg sinø = 2mg sinø

Retardation, a = \(\frac{2 m g \sin \phi}{m}=2 g \sin \phi .\)

If the coin moves up to s, then, \(u^2=2 a s \text { or, } s=\frac{u^2}{2(2 g \sin \phi)}=\frac{u^2}{4 g \sin \phi} \text {. }\)

As the coin stops and attempts to come down, limiting friction acts on it. Downward force mg sinø along the plane is equal to μ’R which is always less than μR, as μ > μ’ (since pL is the coefficient of static friction).

Hence, the coin cannot slide down again.

Example 4. The velocity of a 2.5 kg block sliding down an inclined plane (μ  = 0.2) is found to be 1.5 m • s^-1. One second later, it has a velocity of 5 m • s^-1. What is the angle of the plane with respect to the horizontal?
Solution:

Given

The velocity of a 2.5 kg block sliding down an inclined plane (μ  = 0.2) is found to be 1.5 m • s^-1. One second later, it has a velocity of 5 m • s^-1.

Downward resultant force on the block along the inclined plane

= mgsinθ – μN = mgsinθ – μmgcosθ = mg(sinθ – μcosθ)

Downward acceleration, a = g(sinθ- μcosθ) =9.8(sinθ-0.2cosθ)m · s^-1

From the relation, v = u+at, a = \(\frac{v-u}{t}=\frac{5-1.5}{1}=3.5 \mathrm{~m} \cdot \mathrm{s}^{-2}\)

∴ \(9.8(\sin \theta-0.2 \cos \theta)=3.5\)

or, \(\sin \theta-0.2 \cos \theta=\frac{3.5}{9.8}=\frac{5}{14}\)

Solving this equation we get, θ = 32°

It is the angle of the plane with respect to the horizontal.

Example 5. A, B, and C are the three blocks of masses 3 kg, 4 kg, and 8kg respectively. The blocks are placed over one another. The coefficient of friction between each pair of surfaces in contact is 0.25. A is connected to the wall by a massless rigid rod, and B and C are connected by an inextensible thread passing over a rigidly fixed smooth pulley. Find the force F required to pull C at a constant speed.

Solution:

Given

A, B, and C are the three blocks of masses 3 kg, 4 kg, and 8kg respectively. The blocks are placed over one another. The coefficient of friction between each pair of surfaces in contact is 0.25. A is connected to the wall by a massless rigid rod, and B and C are connected by an inextensible thread passing over a rigidly fixed smooth pulley.

Let the weights of the blocks A, B, and C be denoted by W_A, W_B, and W_C respectively. When block C moves to the left with uniform velocity, B moves to the right and A remains stationary. Hence, the tension on the thread connected to B is equal to the sum of the forces of friction acting on the upper and lower surfaces of B.

Negative Impacts of Friction: Class 11 Notes

∴ From the free body diagram of B, tension, T =f_s + f_k1 =μ X W_A + μ(W_A+ W_B)

= 0.25(3 + 3 + 4) x g = 2.5 xg

For C to move at a constant speed, T and frictions on the upper and lower surfaces of C, together, should be equal to F.

∴ From the free-body diagram of C,

F = T+f_k1+f_k2

= T+μ(W_A+W_B) + μ(W_A+W_B+W_C)

= 2.5 x g+ 0.25(3 + 4) x g+ 0.25(3 + 4 + 8) x g = 8×9.8 = 78.4N.

Example 6. A block of mass M is at rest on a table. The coefficient of friction between the block and the table is μ. What can be the maximum weight of B so that the system remains in equilibrium?

Solution:

Given

A block of mass M is at rest on a table. The coefficient of friction between the block and the table is μ.

Let the tension in the thread in part OC be T. Resolving the tension T into components, we get,

1. Tcosθ along AO and
2. Fsinθ at right angles to AO along BO.

Let the maximum permitted weight of B = W

In equilibrium, T sinθ = W, T cosθ = μMg

⇔ \(\frac{T \sin \theta}{T \cos \theta}=\frac{W}{\mu M g} \text { or, } W=\mu M g \tan \theta \text {. }\)

Applications of Friction: Pros and Cons

Example 7. Calculate the minimum force required to drag a body of mass m, resting on a horizontal surface. The coefficient of friction between the body and the surface is μ.
Solution:

Let the applied force be F, acting at an angle θ with the horizontal.

f_s = force of limiting friction;

mg = the weight of the body and R = normal force.

For equilibrium, R+ F sinθ = mg or, R = mg-F sinθ

and F cosθ = f_s = μR = μ(mg- F sinθ)

or, F cos# + ftF sin# = μmg

∴ F = \(\frac{\mu m g}{\cos \theta+\mu \sin \theta}\)

The value of F will be minimum when (cosθ + μsinθ) is maximum.

∴ \(\frac{d}{d \theta}(\cos \theta+\mu \sin \theta)=0\)

or, \(-\sin \theta+\mu \cos \theta=0 \text { or, } \tan \theta=\mu\)

∴ \(\sin \theta=\frac{\mu}{\sqrt{1+\mu^2}}, \text { and } \cos \theta=\frac{1}{\sqrt{1+\mu^2}}\)

∴ \(F_{\min }=\frac{\mu m g}{\frac{1}{\sqrt{1+\mu^2}}+\frac{\mu^2}{\sqrt{1+\mu^2}}}=\frac{\mu m g}{\sqrt{1+\mu^2}} \text { and } \theta=\tan ^{-1} \mu\)

Short Answer Questions on Friction’s Effects

Example 8. A block of mass 4 kg is kept on a smooth horizontal table surface. Another body of mass 1 kg is placed over one end of the block. The length of the block is 150 cm. The coefficient of friction between the block and the body is 0.1. If a force of 106 dyn is applied on the block, when does the body fall off the block?

Solution:

Given

A block of mass 4 kg is kept on a smooth horizontal table surface. Another body of mass 1 kg is placed over one end of the block. The length of the block is 150 cm. The coefficient of friction between the block and the body is 0.1. If a force of 106 dyn is applied on the block

Let the force applied on the block, so that the smaller body begins to slide be F = 10⁶ dyn = 10 N. Let the mass of the block be M and that of the smaller body be m.

Hence, the friction f, acting in the direction of F and opposing the slipping of the smaller body, is equal to μmg.

Let the acceleration of masses M and m be a₁ and a₂(both taken in the direction of F) respectively, with respect to the table.

Hence, for the 4 kg block, F-f = Ma₁…(1)

and for the 1 kg body, f = ma₂…..(2)

From equations (1) and (2), we get the acceleration of the body with respect to the block,

∴ \(a_2-a_1 =\frac{f}{m}-\frac{F-f}{M}\)

= \(\frac{\mu m g}{m}-\frac{F-\mu m g}{M}=\frac{\mu g(m+M)-F}{M}\)

Substituting the values, \(a_2-a_1=\frac{0.1 \times 9.8(1+4)-10}{4}\)

= \(-\frac{5.1}{4}=-1.275 \mathrm{~m} \cdot \mathrm{s}^{-2}\)

As, \(s=\frac{1}{2} a t^2\),

t = \(\sqrt{\frac{2 s}{a}}=\sqrt{\frac{2 \times 1.5}{1.275}}\)

[s=150 cm =1.5 m and a = 1.275 m · s^-2]

=1.53 s

If calculations show that a₂ – a₁ is positive, i.e., a₂ >a₁, it means that F is less than the limiting friction. In that case, the force of friction actually adjusts itself and remains less than the limiting value, such that a₂ = a₁. Then, both the blocks move together without any sliding between them.

Example 9. At what maximum height, with respect to the lowest point of a hollow sphere of radius r, can a particle stay at rest inside it? Given the coefficient of friction between the sphere and the particle is μ.
Solution:

Let m = mass of the particle, h = AC = maximum height of the point B where the particle can stay at rest

Normal force of the sphere on the particle,

R = mg cosθ = component of the particle’s weight in the radial direction.

mgsinθ = tangential component of this particle’s weight = force responsible for this particle’s motion.

For the highest equilibrium point B, this force is equal and opposite to the force of limiting friction, F.

∴ F = mg sinθ

Again F = μR = μmgcosθ

∴ mg sinθ = μmg cosθ

or, tanθ = μ

Now, from the triangle OBC, we have OC = OB cosθ = r cosθ

∴ h = AC = OA- OC = r- r cosθ

= r(1 – cosθ)

As \(\tan \theta=\mu\), we have \(\cos \theta=\frac{1}{\sqrt{1+\mu^2}}\)

∴ h = \(r\left(1-\frac{1}{\sqrt{1+\mu^2}}\right)\)