Centre of Mass and Rotational Motion brings together several important ideas in Physics, from the motion of a system as a whole to the rotation of individual bodies. Focus on centre of mass, momentum conservation, relative velocity, torque, moment of inertia, angular momentum and rolling motion. Understanding how these concepts are connected can make problems involving collisions, rotating bodies and energy conservation easier to approach.

Use Physics Wallah notes for Centre of Mass and Rotational Motion to revise key concepts through examples and question practice. Important areas include centre-of-mass systems, conservation laws, moment-of-inertia theorems, pure rolling, and centre-of-mass frame applications. Practising these concepts can help you understand which principle to apply while solving related questions. 

Conservation of Linear Momentum in Collisions

Linear momentum is conserved when the net external force acting on a system is zero. The total momentum before the interaction remains equal to the total momentum after the interaction.

If

F_external = 0

then,

P_initial = P_final

For example, suppose two balls initially have momenta

p₁ = p î

and

p₂ = −p î

The total initial momentum is zero. Therefore, after the collision,

p₁′ + p₂′ = 0

The final momentum vectors must have a zero vector sum.

Component-Wise Testing

Momentum is a vector quantity, so its components must be considered separately along different directions.

• Components along the same direction can cancel when they have equal magnitudes and opposite signs.

• A non-zero component without an equal and opposite component cannot become zero.

• Two identical non-zero components in the same direction add rather than cancel.

For example,

a î + b î = 0

only when

b = −a

Similarly,

c k̂ − c k̂ = 0

When the final momentum vectors do not add up to zero, that configuration is not possible for a system whose initial total momentum is zero.

Memory Tip: In a collision where no external force acts, total momentum remains constant. If the initial total momentum is zero, the final vector sum of momenta must also be zero.

Relative Velocity in Counter-Rotating Discs

When two identical discs rotate in opposite directions with the same angular speed, points at the same distance from their centres have equal speeds.

The speed of a point at distance r from the centre is

v = rω

However, equal speeds do not necessarily mean equal velocities. Velocity is a vector quantity, so its direction must also be considered.

For two points P and Q, the relative velocity is

v_rel = v_P − v_Q

and its magnitude is

|v_P − v_Q|

Therefore, questions involving counter-rotating discs should be solved by considering both the magnitude and direction of the velocity vectors.

Important Time Conditions

Suppose the velocity vectors of points P and Q are being compared over one complete rotation.

• At t = 0, the velocity vectors are equal, so the relative velocity is zero.

• At t = T/2, the velocity vectors become equal again, so the relative velocity is zero.

• At t = T, the system returns to its initial state, so the relative velocity is again zero.

Therefore, the correct graph of relative velocity should:

• Start from zero.

• Become zero at T/2.

• Return to zero at T.

• Remain non-zero between these instants.

The distinction between relative speed and the magnitude of relative velocity is important because speed does not contain information about direction.

Centre of Mass and External Force

The motion of the centre of mass depends only on the net external force acting on the system.

The acceleration of the centre of mass is given by

a_CM = F_external / M

where M is the total mass of the system.

Therefore:

• Internal forces cannot change the velocity of the centre of mass of an isolated system.

• Mutual attraction between particles does not change the centre-of-mass velocity when no external force acts.

• If the system starts from rest and the net external force remains zero, its centre of mass remains at rest.

• If the centre of mass initially has a particular velocity and the net external force is zero, it continues moving with that velocity.

Example

Consider two blocks of masses 10 kg and 4 kg. If the 10 kg block moves with a velocity of 14 m/s and the 4 kg block is initially at rest, then

v_CM = (10 × 14 + 4 × 0)/(10 + 4)

Therefore,

v_CM = 10 m/s

A change in the relative velocity between the two blocks does not necessarily mean that the centre-of-mass velocity has changed.

Moment of Inertia of Composite Bodies

Moment of inertia depends on the distribution of mass about the axis of rotation. For composite bodies, symmetry and standard moment-of-inertia theorems can simplify the calculation.

For a planar lamina, the perpendicular-axis theorem states that

I_z = I_x + I_y

If symmetry gives

I_x = I_y

then,

I_z = 2I_x

Suppose a symmetric lamina has

I_z = 1.6 ma²

Then,

I_x = 0.8 ma²

If the required axis is parallel to the centre-of-mass axis and lies at a distance d = 2a, the parallel-axis theorem is used:

I = I_CM + Md²

Therefore,

I = 0.8 ma² + m(2a)²

I = 0.8 ma² + 4ma²

Hence,

I = 4.8 ma²

Important Moment of Inertia Theorems

The perpendicular-axis theorem is used for a planar lamina when the moment of inertia about an axis perpendicular to its plane is related to the moments about two mutually perpendicular axes in the plane.

The parallel-axis theorem is used when the required axis is parallel to an axis passing through the centre of mass.

The parallel-axis theorem is

I = I_CM + Md²

where d is the perpendicular distance between the two parallel axes.

These problems may involve complicated shapes, but identifying symmetry and applying the appropriate theorem can simplify the calculation.

Mass Conservation During Reshaping

When a body is melted and recast into another shape, its mass remains unchanged.

Therefore,

Initial mass = Final mass

The shape, volume and dimensions may change, but the same mass is present before and after reshaping.

For a solid sphere,

I_sphere = 2/5 MR²

For a disc about a tangential axis perpendicular to its plane, first use the moment of inertia about the central axis and then apply the parallel-axis theorem:

I_tangent = 1/2 Mr² + Mr²

Therefore,

I_tangent = 3/2 Mr²

If the two moments of inertia are equated,

2/5 MR² = 3/2 Mr²

After cancelling M,

r² = 4R²/15

Therefore,

r = 2R/√15

The important ideas in such problems are mass conservation, standard moment-of-inertia formulas and the parallel-axis theorem.

Rotational Motion of a Hinged Plate

Consider a horizontal plate hinged along one edge. The weight of the plate produces a torque about the hinge.

If elastic balls strike the plate and reverse their direction, the change in velocity of each ball is twice its initial speed.

For a ball of mass m moving with speed v, the change in velocity is

Δv = 2v

Therefore, the magnitude of change in momentum is

Δp = 2mv

If N balls strike the plate every second, the average force exerted on the plate is

F = 2Nmv

The plate remains horizontal when the net torque about the hinge is zero.

Thus,

Net torque about the hinge = 0

The torque produced by the weight of the plate is balanced by the torque produced by the reflected balls.

Important: When a ball reverses its direction, its change in velocity is 2v, not zero.

Conservation of Angular Momentum

Angular momentum is conserved when the net external torque acting on a system is zero.

τ_external = 0 ⇒ Angular momentum = constant

Consider a rotating platform initially at rest. If two balls are fired in opposite directions from positions on opposite sides of the centre, their angular momenta can act in the same rotational sense.

The platform then rotates in the opposite direction so that the total angular momentum of the system remains conserved.

If the balls each have mass m, speed v, and are at distance r from the centre, the conservation equation can be written as

0 = Iω + 2mvr

For a uniform disc,

I = 1/2 MR²

The negative sign in the angular velocity indicates that the platform rotates opposite to the angular momentum of the balls.

Memory Tip: If the initial angular momentum of an isolated system is zero, the angular momenta produced after an interaction must have a zero vector sum.

Pure Rolling and Velocity of Disc Points

Pure rolling occurs when a body rolls without slipping. In this case, the translational and rotational motions are related by

V_CM = Rω

For a rolling disc, different points have different instantaneous velocities.

The bottom point is instantaneously at rest because its translational velocity and rotational velocity are equal in magnitude and opposite in direction.

The centre of mass moves with velocity

V_CM = Rω

The top point has both translational and rotational velocities acting in the same direction. Therefore, its speed is

V_top = 2V_CM = 2Rω

For points A, B and C representing the bottom, centre and top respectively:

V_C − V_A = 2V_CM

V_C − V_B = V_CM

V_B − V_A = V_CM

Therefore,

V_C − V_B = V_B − V_A

Understanding the velocity of different points on a rolling body is important for questions involving pure rolling and relative motion.

Kinetic Energy of a Rolling Disc

The total kinetic energy of a rolling disc is the sum of its translational and rotational kinetic energies.

K = 1/2 mv² + 1/2 Iω²

For a disc,

I = 1/2 mr²

For pure rolling,

v = rω

Substituting these relations,

K = 1/2 mv² + 1/2 × 1/2 mr² × ω²

Since

rω = v

we get

K = 1/2 mv² + 1/4 mv²

Therefore,

K = 3/4 mv²

This expression can be used in energy-conservation problems involving a disc rolling without slipping.

Centre-of-Mass Frame

The centre-of-mass frame is useful when studying the relative motion of two particles.

For two particles, the kinetic energy in the centre-of-mass frame can be written as

K_CM = 1/2 μv_rel²

where the reduced mass is

μ = m₁m₂/(m₁ + m₂)

and v_rel is the relative velocity of the two particles.

In the centre-of-mass frame:

• The motion of the system can be described using relative velocity.

• The kinetic energy depends on the relative motion of the particles.

• If both particles have the same velocity in the laboratory frame, their relative velocity is zero.

• When v_rel = 0, the kinetic energy in the centre-of-mass frame is also zero.

The work-energy relation can be expressed as

Work done by internal forces + Work done by external forces = Final K_CM − Initial K_CM

Applications of the Centre-of-Mass Frame

The centre-of-mass frame can simplify problems involving:

• Maximum compression of a spring.

• Sliding of a block on a cart.

• Motion of a block on a smooth wedge.

• Relative motion between interacting bodies.

At the final condition in many such problems, the relative motion between the bodies stops. Therefore,

v_rel = 0

and the final kinetic energy in the centre-of-mass frame becomes zero.

Important Questions 

1. Two blocks of masses 10 kg and 4 kg are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulse giving a velocity of 14 m/s is given to the heavier block in the direction of the lighter block. What is the velocity of the centre of mass?

Answer: 10 m/s

Explanation:

The velocity of the centre of mass is:

V_CM = (m₁v₁ + m₂v₂) / (m₁ + m₂)

Here,

V_CM = [(10 × 14) + (4 × 0)] / (10 + 4)

V_CM = 140 / 14

V_CM = 10 m/s

Therefore,

V_CM = 10 m/s

2. Two identical particles A and B, initially at rest, move towards each other under a mutual force of attraction. At the instant when the speed of A is vv and the speed of B is 2v2v, what is the speed of the centre of mass of the system?

Answer: Zero

Explanation:
The particles are acted upon only by their mutual internal force. Since the system was initially at rest, its total momentum remains zero.

V_CM = 0

Therefore, V_CM = 0.

3. Two particles of mass 1 kg and 3 kg move towards each other under their mutual force of attraction. No other force acts on them. When the relative velocity of approach is 2 m/s, their centre of mass has a velocity of 5 m/s. When the relative velocity of approach becomes 3 m/s, the velocity of the centre of mass is 0.75 m/s. State whether the statement is true or false.

Options:

• A. True

• B. False

Answer: B. False

Explanation:
Since no external force acts on the system, the total momentum remains constant. Therefore, the velocity of the centre of mass cannot change from 5 m/s to 0.75 m/s.

The centre-of-mass velocity remains:

V_CM = 5 m/s

Therefore, V_CM = 5 m/s.

4. A solid sphere of radius R has moment of inertia II about its geometrical axis. It is melted and recast into a disc of radius r and thickness t. If its moment of inertia about the tangential axis, which is perpendicular to the plane of the disc, is also equal to I, what is the value of r?

Answer:

r = 2R / √15

Therefore,

r = 2R / √15

Explanation:

The mass and volume remain unchanged when the sphere is melted and recast into the disc. Using the moment of inertia expressions for the sphere and disc, along with conservation of volume, we get:

r = 2R / √15

5. A rectangular plate of mass M and dimensions a×ba\times b is held in a horizontal position by striking nn small balls uniformly, each of mass mm, per unit area per unit time. The balls strike the shaded half region of the plate and collide elastically with velocity v. Find v. Given M=3 kg, m=0.01 kg, n=100, a=1 m, b=2 m and g=10 m/s².

Answer: 10 m/s

Explanation:
For the plate to remain horizontal, the upward force produced by the elastic collisions must balance its weight.

Using the rate of change of momentum:

v = 10 m/s

Therefore, v = 10 m/s.

6. A ball is projected from the ground at an angle of 45°  with the horizontal. It reaches a maximum height of 120 m and returns to the ground. Upon hitting the ground for the first time, it loses half of its kinetic energy. Immediately after the bounce, the velocity of the ball makes an angle of 30° with the horizontal. What is the maximum height it reaches after the bounce?

Answer: 30 m

Explanation:
The initial vertical component of velocity is related to the maximum height by:

H = u_y² / (2g)

After the collision, the ball loses half of its kinetic energy. Considering the new velocity and its 30° angle with the horizontal, the maximum height after the bounce is:

H′ = 30 m

Therefore, H′ = 30 m.

7. A horizontal circular platform of radius 0.5 m and mass 0.45 kg is free to rotate about its axis. Two massless spring toy guns, each carrying a steel ball of mass 0.05 kg, are attached to the platform at a distance of 0.25 m from the centre on either side along its diameter. Each gun simultaneously fires its ball horizontally and perpendicular to the diameter in opposite directions. After leaving the platform, the balls have a horizontal speed of 9 m/s with respect to the ground. What is the rotational speed of the platform in rad/s after the balls leave the platform?

Answer: 4 rad/s

Explanation:
Initially, the total angular momentum of the system is zero. By conservation of angular momentum, the angular momentum of the balls is balanced by the angular momentum of the rotating platform.

Thus,

ω = 4 rad/s

Therefore, ω = 4 rad/s.

8. A body undergoes pure rolling. Which of the following statements are correct?

Options:

• V_C − V_A = 2(V_C − V_B)

• V_C − V_B = V_B − V_A

• |V_C − V_A| = 2|V_C − V_B|

• |V_C − V_A| = 4|V_B|

Answer: Options 1, 2 and 3 are correct.

Explanation:
For pure rolling, the velocity of the centre of mass is related to angular velocity by:

V_C = ωR

Using the velocity relations between the different points of the rolling body, statements 1, 2 and 3 satisfy the pure rolling condition.

9. A block of mass M has a circular cut with a frictionless surface and rests on a horizontal frictionless table. A point mass mm is released from the topmost point of the cut and slides down. When the mass loses contact with the table, its position is x and its velocity is v. Which of the following statements are correct?

Options:

Options:

1. The velocity of the point mass m is v=2gR1+mM

2. The x component of displacement of the centre of mass of the block M is -mRM+m

3. The position of the point mass is x=-2mRM+m

4. The velocity of the block M is V=-mM2gR

Answer: Options 1 and 2 are correct.

Explanation:

Since the table is frictionless, there is no external horizontal force. Therefore, horizontal momentum and the horizontal position of the centre of mass are conserved.

Using conservation of energy and horizontal momentum gives:

v = √[2gR / (1 + m/M)]

and

x_M = −mR / (M + m)

Hence, options 1 and 2 are correct.

10. A uniform bar of length 6a and mass 8m lies on a smooth horizontal table. Two point masses m and 2m, moving in the same horizontal plane with speeds 2v and v, respectively, strike the bar and stick to it. After the collision, ω\omega is the angular velocity about the centre of mass, E is the total energy, and VC is the velocity of the centre of mass. Which results are correct?

Answer:

ω = v / (5a)

V_C = 0

E = (3/5)mv²

Explanation:

The collision is analysed using conservation of linear momentum and conservation of angular momentum about the centre of mass.

Since the initial total linear momentum is zero:

V_C = 0

The resulting angular velocity is:

ω = v / (5a)

The total energy after the collision is:

E = (3/5)mv²

Key Takeaways for NEET 2026

• Centre-of-mass motion depends on the net external force acting on the system.

• Internal forces cannot change the centre-of-mass velocity of an isolated system.

• Linear momentum is conserved when the net external force is zero.

• Angular momentum is conserved when the net external torque is zero.

• Relative velocity must be treated as a vector because direction is important.

• Use the perpendicular-axis theorem for suitable planar lamina problems.

• Use the parallel-axis theorem when shifting between parallel axes.

• In pure rolling, V_CM = Rω.

• The bottom point of a rolling disc is instantaneously at rest, while the top point moves with speed 2V_CM.

• Total kinetic energy of a rolling disc includes both translational and rotational components.

• The centre-of-mass frame can simplify problems involving collisions, springs, wedges and relative motion.

Centre of Mass and Rotational Motion brings together important Physics concepts such as linear momentum, angular momentum, torque, moment of inertia, rolling motion, kinetic energy and relative motion. Understanding when to apply conservation laws and how to identify the appropriate axis, frame of reference or velocity relation can make numerical problems easier to solve. Regular revision and question practice through PW’s NEET Physics notes can help strengthen these concepts and improve your problem-solving approach for NEET 2026.

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