WBCHSE Class 11 Physics Notes For Centre Of Mass

Centre Of Mass Or C M

WBBSE Class 11 Centre of Mass Notes

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

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

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

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

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

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

Key Concepts of Centre of Mass in Physics

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

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

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

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

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

Centre of Mass Formula and Derivation

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Similarly, z_cm = 0

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

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Some Important Information About The Centre Of Mass:

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

Momentum Conservation Motion Of Centre Of Mass:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Understanding Centre of Mass in Systems of Particles

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

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

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

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

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

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

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

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

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

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

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

Applications of Centre of Mass in Real Life

Illustrative Examples:

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

Statics – Centre Of Mass Numerical Examples

Short Answer Questions on Centre of Mass

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

Given

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

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

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

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

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

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

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

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

Given

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

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

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

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

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

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

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

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

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

Hence, the centre of mass remains stationary.

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

Given

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

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

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

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

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

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

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

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

∴ a = L + R + R = L + 2R

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

Real-Life Examples Illustrating Centre of Mass

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

Given

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

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

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

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

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

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

The centre of mass of a triangular lamina

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

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

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

Given

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

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

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

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

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

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

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

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

Given

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

Let the mass of the H-atom be m.

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

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

So, the coordinate of the centre of mass,

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

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

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

Given

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

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

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

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

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

Given

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

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

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

Let the stones collide over time t.

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

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

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

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

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

∴ Displacement of the centre of mass

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

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