CBSE Class 11 Physics Motion of System of Particles and Rigid Body Assignment

1 MARK QUESTIONS

Question 1. What is a rigid body?
Answer: A rigid body is a body with a perfectly definite and unchanging shape. The distance between any two pairs of particles belonging to the body remains invariant under the action of external forces.
In simple words: A rigid body does not deform or change its shape when forces are applied to it.

Exam Tip: Remember that real bodies are nearly rigid, but in theoretical physics, perfect rigidity is an idealization.

Question 2. State the principle of moments of rotational equilibrium.
Answer: According to the principle of moments, for a body in rotational equilibrium, the algebraic sum of all the torques acting on the body about any point is zero. Alternatively, the sum of clockwise torques is equal to the sum of counter-clockwise torques about any axis of rotation.
In simple words: Total turning force in one direction must balance the total turning force in the opposite direction.

Exam Tip: This principle is the underlying basis for the working of levers and seesaws.

Question 3. Is centre of mass of a body necessarily lie inside the body? Give any example
Answer: No, the centre of mass of a body does not necessarily lie inside the body.
Example: A circular ring or a hollow sphere, where the centre of mass lies at its geometrical centre, which is in empty space outside the solid material.
In simple words: The centre of mass can be located in empty space outside the physical boundaries of an object.

Exam Tip: Always associate symmetrical hollow objects like rings and cups with this condition.

Question 4. Can the couple acting on a rigid body produce translatory motion?
Answer: No. A couple consists of two equal and opposite forces acting along different lines of action. Since the vector sum of these forces is zero (\(\sum \vec{F} = 0\)), the net force is zero, meaning it cannot produce any translatory motion. It produces purely rotatory motion.
In simple words: A couple only causes rotation because the forces cancel each other out in linear motion.

Exam Tip: Net force zero means no linear acceleration; separate lines of action mean non-zero net torque.

Question 5. Which component of linear momentum does not contribute to angular momentum?
Answer: The radial component of linear momentum (directed along the position vector from the axis of rotation) does not contribute to angular momentum, because the cross product of collinear vectors is zero.
In simple words: Motion directly toward or away from the axis creates no turning effect.

Exam Tip: Only the tangential component of velocity or momentum contributes to angular momentum (\(L = r p \sin\theta\)).

Question 6. A system is in stable equilibrium? What can we say about its potential energy?
Answer: When a system is in stable equilibrium, its potential energy is at a minimum.
In simple words: Stable equilibrium corresponds to the lowest possible potential energy state.

Exam Tip: If displaced slightly, a system in stable equilibrium tends to return to its original position because energy would increase otherwise.

Question 7. Is radius of gyration a constant quantity?
Answer: No, the radius of gyration is not a constant quantity. It changes with the position and orientation of the axis of rotation relative to the body.
In simple words: Radius of gyration depends entirely on where you choose to spin the object.

Exam Tip: The mass of the body remains constant, but distribution of mass relative to the axis changes \(K\).

Question 8. Two solid spheres of the same mass are made of metals of different densities. Which of them has a large moment of inertia about the diameter?
Answer: Moment of inertia of a solid sphere about its diameter is given by \(I = \frac{2}{5}MR^2\).
Since mass \(M\) is the same, \(I\) depends directly on \(R^2\).
Density \(\rho = \frac{M}{\frac{4}{3}\pi R^3}\), so \(R \propto \rho^{-1/3}\).
Therefore, the sphere of smaller density has a larger radius \(R\), and hence a larger moment of inertia.
In simple words: The less dense sphere is larger in size for the same mass, placing mass further from the center and increasing moment of inertia.

Exam Tip: Relate density, volume, and radius carefully before comparing moments of inertia.

Question 9. The moment of inertia of two rotating bodies A and B are \(I_A\) and \(I_B\) (\(I_A > I_B\)) and their angular momenta are equal. Which one has a greater kinetic energy?
Answer: Rotational kinetic energy is given by \(K = \frac{L^2}{2I}\).
Since the angular momenta \(L\) are equal, \(K\) is inversely proportional to \(I\) (\(K \propto \frac{1}{I}\)).
Given \(I_A > I_B\), it follows that \(K_B > K_A\). Thus, body B has greater kinetic energy.
In simple words: The body with the smaller moment of inertia has higher kinetic energy for the same angular momentum.

Exam Tip: Always use the relation \(K = \frac{L^2}{2I}\) when angular momentum is kept constant.

Question 10. A particle moves on a circular path with decreasing speed. What happens to its angular momentum?
Answer: As \(\vec{L} = \vec{r} \times m\vec{v}\), the magnitude of angular momentum \(L = mvr\) decreases because the speed \(v\) is decreasing, but its direction (perpendicular to the plane of the circular path) remains constant.
In simple words: The magnitude gets smaller as speed drops, but the direction stays fixed.

Exam Tip: Speed changes linear velocity magnitude, directly affecting angular momentum magnitude.

Question 11. What is the value of instantaneous speed of the point of contact during pure rolling?
Answer: Zero.
In simple words: The lowest point of a rolling wheel momentarily rests against the ground without slipping.

Exam Tip: Pure rolling implies zero relative motion (slipping) at the point of contact.

Question 12. Which physical quantity is conserved when a planet revolves around the sun?
Answer: Angular momentum of the planet about the sun is conserved because the gravitational force is a central force and exerts zero torque about the sun.
In simple words: Angular momentum remains constant throughout planetary motion.

Exam Tip: Central forces produce zero torque since the line of action passes through the center of force.

Question 13. What is the value of torque on the planet due to the gravitational force of sun?
Answer: Zero, because the gravitational force is directed along the line joining the planet and the sun (central force), making the perpendicular distance \(r_\perp = 0\).
In simple words: The gravitational pull points straight toward the sun, causing no turning effect.

Exam Tip: Torque \(\vec{\tau} = \vec{r} \times \vec{F}\); when \(\vec{r}\) and \(\vec{F}\) are collinear, torque is zero.

Question 14. If no external torque acts on a body, will its angular velocity be constant?
Answer: Not necessarily. From conservation of angular momentum, \(I\omega = \text{constant}\), so \(\omega \propto \frac{1}{I}\). If the moment of inertia \(I\) changes (e.g., by changing shape or mass distribution), the angular velocity \(\omega\) will change.
In simple words: Angular velocity remains constant only if the moment of inertia also stays constant.

Exam Tip: Think of a ballet dancer pulling in their arms to spin faster while angular momentum stays conserved.

Question 15. Why there are two propellers in a helicopter?
Answer: Two propellers are used due to the law of conservation of angular momentum. If there were only a single main rotor, the body of the helicopter would spin in the opposite direction due to reaction torque. A second rotor (tail rotor or counter-rotating blade) balances this torque.
In simple words: The second propeller prevents the helicopter body from spinning uncontrollably in reverse.

Exam Tip: Relate this directly to Newton's third law and conservation of angular momentum.

Question 16. A child sits stationary at one end of a long trolley moving uniformly with speed \(V\) on a smooth horizontal floor. If the child gets up and runs about on the trolley in any manner, then what is the effect of the speed of the centre of mass of the (trolley + child) system?
Answer: There is no change in the speed of the centre of mass of the system because no external force acts on the (trolley + child) system; running around involves only internal forces.
In simple words: Internal movements cannot change the velocity of the centre of mass of an isolated system.

Exam Tip: Only external forces can change the velocity of the centre of mass.

2 MARKS QUESTIONS

Question 1. Show that in the absence of any external force, the velocity of the centre of mass remains constant.
Answer: The position vector of the centre of mass of a system of particles is given by:
\(\vec{R} = \frac{\sum m_i \vec{r}_i}{M}\)
Differentiating with respect to time \(t\), we get the velocity of the centre of mass:
\(\vec{V}_{\text{cm}} = \frac{d\vec{R}}{dt} = \frac{\sum m_i \vec{v}_i}{M} = \frac{\sum \vec{p}_i}{M} = \frac{\vec{P}}{M}\)
Differentiating again with respect to time to find acceleration:
\(\vec{A}_{\text{cm}} = \frac{d\vec{V}_{\text{cm}}}{dt} = \frac{1}{M}\frac{d\vec{P}}{dt} = \frac{\vec{F}_{\text{ext}}}{M}\)
In the absence of any external force, \(\vec{F}_{\text{ext}} = 0\), which implies:
\(\vec{A}_{\text{cm}} = 0 \implies \vec{V}_{\text{cm}} = \text{constant}\).
In simple words: If no outside push or pull acts on the system, the total momentum stays steady, so the centre of mass moves at a constant velocity.

Exam Tip: Clearly show the steps: position to velocity, velocity derivative to acceleration, and apply Newton's second law for systems.

Question 2. State the factors on which the position of centre of mass of a rigid body depends.
Answer: The position of the centre of mass of a rigid body depends on:
(i) The shape and geometric configuration of the body.
(ii) The mass distribution (how mass is distributed throughout the volume of the body).
In simple words: It depends on the overall shape of the object and where its mass is concentrated.

Exam Tip: Keep the points concise: shape/geometry and mass distribution.

Question 3. What is the turning effect of force called for? On what factors does it depend?
Answer: The turning effect of a force is called **torque** or **moment of force**.
It depends on:
(i) The magnitude of the applied force (\(F\)).
(ii) The perpendicular distance of the line of action of the force from the axis of rotation (\(r_\perp\)).
In simple words: Torque is the turning power of a force, which increases if you push harder or push further from the pivot.

Exam Tip: State the formula \(\tau = r F \sin\theta\) to quickly highlight dependence on force, distance, and angle.

Question 4. State the factors on which the moment of inertia of a body depends.
Answer: The moment of inertia of a body depends on:
(i) Mass of the body.
(ii) Size and shape of the body.
(iii) Distribution of mass about the axis of rotation.
(iv) Position and orientation of the axis of rotation.
In simple words: It depends on how much mass the object has and how far that mass is spread out from the spinning axis.

Exam Tip: Remember that moment of inertia is not an inherent constant; it changes when the axis of rotation changes.

Question 5. On what factors does radius of gyration of body depend?
Answer: The radius of gyration of a body depends on:
(i) Distribution of mass of the body about the axis of rotation.
(ii) Position and orientation of the axis of rotation.
In simple words: It tells us how the mass is spread relative to the rotation axis.

Exam Tip: Note that radius of gyration does *not* depend on the total mass of the body itself, only on its distribution.

Question 6. Why do we prefer to use a wrench of longer arm?
Answer: Torque is given by \(\tau = r F\). A longer wrench provides a larger perpendicular distance (\(r\)), which means a smaller force (\(F\)) is required to produce the same torque needed to loosen tight nuts or bolts.
In simple words: A longer handle gives more leverage, making it much easier to turn stubborn screws and bolts.

Exam Tip: Relate this directly to the principle of torque and perpendicular distance.

Question 7. Can a body be in equilibrium while in motion? If yes, give an example.
1Answer: Yes, a body can be in equilibrium while in motion, provided it is in dynamic equilibrium (both net force and net torque are zero).
Example: A rain drop falling with terminal velocity (constant velocity, zero acceleration) or a car moving at a constant speed along a straight, level highway.
In simple words: Yes, if an object moves at a steady speed in a straight line with no acceleration, it is in equilibrium.

Exam Tip: Equilibrium means zero acceleration (\(a = 0\) and \(\alpha = 0\)), not necessarily zero velocity.

Question 8. There is a stick half of which is wooden and half is of steel. (i) it is pivoted at the wooden end and a force is applied at the steel end at right angle to its length (ii) it is pivoted at the steel end and the same force is applied at the wooden end. In which case is the angular acceleration more and why?
Answer: Angular acceleration is given by \(\alpha = \frac{\tau}{I}\).
Case (i): Heavy steel end is further from the pivot, so moment of inertia \(I\) is large, leading to smaller \(\alpha\).
Case (ii): Heavy steel end is near the pivot, so moment of inertia \(I\) is smaller, leading to greater \(\alpha\).
Therefore, angular acceleration is more in **case (ii)**.
In simple words: Case (ii) has less mass concentrated far from the pivot, making it easier to rotate.

Exam Tip: Mass placed closer to the pivot decreases moment of inertia and increases angular acceleration for the same torque.

Question 9. If earth contracts to half its radius what would be the length of the day at equator?
Answer: Initial moment of inertia of Earth (assuming solid sphere): \(I_1 = \frac{2}{5}MR^2\).
When radius contracts to half (\(R_2 = \frac{R}{2}\)), the new moment of inertia is:
\(I_2 = \frac{2}{5}M\left(\frac{R}{2}\right)^2 = \frac{1}{4}\left(\frac{2}{5}MR^2\right) = \frac{I_1}{4}\)
By conservation of angular momentum (\(L = I_1\omega_1 = I_2\omega_2\)):
\(I_1 \left(\frac{2\pi}{T_1}\right) = \frac{I_1}{4} \left(\frac{2\pi}{T_2}\right)\)
\(\frac{1}{T_1} = \frac{1}{4T_2} \implies T_2 = \frac{T_1}{4}\)
Given original day length \(T_1 = 24\) hours:
\(T_2 = \frac{24}{4} = 6 \text{ hours}\).
In simple words: Shrinking the Earth makes it spin much faster, shortening the day to 6 hours.

Exam Tip: Use conservation of angular momentum and substitute \(\omega = \frac{2\pi}{T}\).

Question 10. An internal force can not change the state of motion of centre of mass of a body. How does the internal force of the brakes bring a vehicle to rest?
Answer: Strictly speaking, the braking force between the brake shoes and the wheels is internal to the vehicle system, but it generates friction between the tires and the road. This frictional force exerted by the road on the tires is an **external force** that brings the vehicle to rest.
In simple words: Brakes stop internal parts, but it is the external friction from the road that actually stops the car.

Exam Tip: Always distinguish between internal forces and external friction from the ground.

Question 11. When does a rigid body said to be in equilibrium? State the necessary condition for a body to be in equilibrium.
Answer: A rigid body is said to be in mechanical equilibrium if neither its linear momentum nor its angular momentum changes with time.
**Necessary conditions:**
1. Translational equilibrium: The vector sum of all external forces acting on the body must be zero:
\(\sum \vec{F}_{\text{ext}} = 0\)
2. Rotational equilibrium: The vector sum of all external torques acting on the body about any point must be zero:
\(\sum \vec{\tau}_{\text{ext}} = 0\)
In simple words: A body is in equilibrium when all forces cancel out (no sliding) and all torques cancel out (no spinning).

Exam Tip: Both net force and net torque must simultaneously equal zero for complete equilibrium.

Question 12. How will you distinguish between a hard boiled egg and a raw egg?
Answer: Spin both eggs on a smooth table top with the same force. A hard-boiled egg (solid throughout) will spin steadily and come to a stop smoothly. A raw egg (liquid inside) will experience internal fluid friction, moving liquid shifting its centre of mass, making it wobble and stop spinning much sooner.
In simple words: Spin them: the boiled egg spins smoothly, while the raw egg wobbles because of the sloshing liquid inside.

Exam Tip: Mention internal friction and shifting centre of mass for full credit.

Question 13. What are binary stars? Discuss their motion in respect of their centre of mass.
Answer: Binary stars are a system of two stars orbiting around their common centre of mass under their mutual gravitational attraction. Since no external force acts on the binary system, their centre of mass moves with a constant velocity (or remains at rest if the total momentum is zero), while both stars revolve in orbits around this fixed centre of mass.
In simple words: Two stars spinning around a shared balance point in space, while their combined centre of mass moves steadily.

Exam Tip: The more massive star stays closer to the common centre of mass than the lighter star.

Question 14. In which condition a body in gravitational field is in stable equilibrium?
Answer: A body in a gravitational field is in stable equilibrium when the vertical line passing through its centre of gravity falls within the base of support of the body.
In simple words: As long as the centre of gravity's downward line lands inside the object's base, it remains stable.

Exam Tip: This condition ensures that any tilting produces a restoring torque that brings the body back.

Question 15. Give the physical significance of moment of inertia.
Answer: Moment of inertia plays the exact same role in rotational motion that mass plays in translatory motion (inertia). It is a measure of the rotational inertia of a body, quantifying how much torque is required to produce a given angular acceleration about a specific axis.
In simple words: Moment of inertia is the rotational equivalent of mass; it measures how stubborn an object is to starting or stopping rotation.

Exam Tip: Think of mass as linear inertia and moment of inertia as rotational inertia.

3 MARKS QUESTIONS

Question 1. Derive the three equation of rotational motion:
(i) \(\omega = \omega_0 + \alpha t\)
(ii) \(\theta = \omega_0 t + \frac{1}{2}\alpha t^2\)
(iii) \(\omega^2 = \omega_0^2 + 2\alpha\theta\)
under constant angular acceleration. Here symbols have usual meaning.
Answer:
**(i) First Equation:**
Angular acceleration is defined as the rate of change of angular velocity:
\(\alpha = \frac{d\omega}{dt}\)
\(d\omega = \alpha \, dt\)
Integrating both sides from time \(t = 0\) (angular velocity \(\omega_0\)) to time \(t\) (angular velocity \(\omega\)):
\(\int_{\omega_0}^{\omega} d\omega = \alpha \int_0^t dt\)
\(\omega - \omega_0 = \alpha t \implies \omega = \omega_0 + \alpha t\)

**(ii) Second Equation:**
Angular velocity is the rate of change of angular displacement:
\(\omega = \frac{d\theta}{dt} \implies d\theta = \omega \, dt\)
Substituting \(\omega = \omega_0 + \alpha t\):
\(d\theta = (\omega_0 + \alpha t) dt\)
Integrating from \(t = 0\) to \(t\) and \(\theta = 0\) to \(\theta\):
\(\int_0^\theta d\theta = \int_0^t (\omega_0 + \alpha t) dt\)
\(\theta = \omega_0 t + \frac{1}{2}\alpha t^2\)

**(iii) Third Equation:**
Using the chain rule for angular acceleration:
\(\alpha = \frac{d\omega}{dt} = \frac{d\omega}{d\theta} \cdot \frac{d\theta}{dt} = \omega \frac{d\omega}{d\theta}\)
\(\alpha \, d\theta = \omega \, d\omega\)
Integrating both sides from \(\theta = 0\) to \(\theta\) and \(\omega_0\) to \(\omega\):
\(\int_0^\theta \alpha \, d\theta = \int_{\omega_0}^\omega \omega \, d\omega\)
\(\alpha \theta = \frac{\omega^2 - \omega_0^2}{2}\)
\(\omega^2 = \omega_0^2 + 2\alpha\theta\).
In simple words: These kinematic equations for spinning objects mirror the standard linear motion formulas by replacing linear variables with angular ones.

Exam Tip: Practice calculus derivations as they are frequently tested in board exams.

Question 2. Obtain an expression for the work done by a torque. Hence write the expression for power.
Answer:
Consider a rigid body rotating about a fixed axis under the action of a torque \(\tau\), turning through a small angular displacement \(d\theta\).
Small work done \(dW\) by the torque is given by:
\(dW = \tau \, d\theta\)
Total work done \(W\) when the body rotates from angular position \(\theta_1\) to \(\theta_2\) is:
\(W = \int_{\theta_1}^{\theta_2} \tau \, d\theta\)
If the torque is constant:
\(W = \tau (\theta_2 - \theta_1) = \tau \theta\)
**Power:**
Power \(P\) is defined as the rate of doing work:
\(P = \frac{dW}{dt} = \frac{\tau \, d\theta}{dt}\)
Since \(\frac{d\theta}{dt} = \omega\) (angular velocity):
\(P = \tau \omega\).
In simple words: Work in rotation is torque times angle turned, and power is torque times angular speed.

Exam Tip: Analogies help: \(W = Fs\) becomes \(W = \tau\theta\), and \(P = Fv\) becomes \(P = \tau\omega\).

Question 3. Prove that the rate of change of angular momentum of a system of particles about a reference point is equal to the net external torque acting on the system.
Answer:
The angular momentum of a system of particles is given by:
\(\vec{L} = \sum \vec{r}_i \times \vec{p}_i\)
Differentiating with respect to time \(t\):
\(\frac{d\vec{L}}{dt} = \frac{d}{dt} \sum (\vec{r}_i \times \vec{p}_i) = \sum \left( \frac{d\vec{r}_i}{dt} \times \vec{p}_i + \vec{r}_i \times \frac{d\vec{p}_i}{dt} \right)\)
Since \(\frac{d\vec{r}_i}{dt} = \vec{v}_i\) and \(\vec{p}_i = m_i \vec{v}_i\):
The first term is \(\sum (\vec{v}_i \times m_i \vec{v}_i) = 0\) (cross product of parallel vectors).
By Newton's second law, \(\frac{d\vec{p}_i}{dt} = \vec{F}_i\) (total force on particle \(i\)), which includes both internal forces \(\vec{F}_{ij}\) and external forces \(\vec{F}_i^{\text{ext}}\).
Sum of torques due to internal forces is zero by Newton's third law (equal and opposite forces acting along the same line).
Therefore, the remaining term is:
\(\frac{d\vec{L}}{dt} = \sum (\vec{r}_i \times \vec{F}_i^{\text{ext}}) = \vec{\tau}_{\text{ext}}\).
Hence proved.torque is equal to the rate of change of angular momentum.
In simple words: Just like force equals the rate of change of linear momentum, net external torque equals the rate of change of angular momentum.

Exam Tip: This is the rotational analogue of Newton's second law (\(\vec{F} = \frac{d\vec{p}}{dt}\)).

Question 4. Derive a relation between angular momentum, moment of inertia and angular velocity of a rigid body.
Answer:
Consider a rigid body rotating about a fixed axis with a uniform angular velocity \(\omega\). The body consists of particles of masses \(m_1, m_2, \dots, m_n\) at perpendicular distances \(r_1, r_2, \dots, r_n\) from the axis of rotation.
The linear velocity of the \(i\)-th particle is:
\(v_i = r_i \omega\)
The magnitude of angular momentum of the \(i\)-th particle about the axis of rotation is:
\(L_i = m_i v_i r_i = m_i (r_i \omega) r_i = m_i r_i^2 \omega\)
The total angular momentum \(L\) of the rigid body is the sum of angular momenta of all individual particles:
\(L = \sum L_i = \sum (m_i r_i^2 \omega) = \left( \sum m_i r_i^2 \right) \omega\)
Since the moment of inertia is \(I = \sum m_i r_i^2\), we get:
\(L = I\omega\).
In simple words: Angular momentum equals moment of inertia multiplied by angular speed.

Exam Tip: Compare this with linear momentum formula \(p = mv\).

Question 5. Show that moment of a couple does not depend on the point about which moment is calculated.
Answer:
Consider a couple consisting of two equal and opposite forces \(\vec{F}\) and \(-\vec{F}\) acting at points with position vectors \(\vec{r}_1\) and \(\vec{r}_2\) relative to an origin \(O\).
The moment of the couple (total torque \(\tau\)) about origin \(O\) is:
\(\vec{\tau} = (\vec{r}_1 \times \vec{F}) + (\vec{r}_2 \times (-\vec{F})) = (\vec{r}_1 - \vec{r}_2) \times \vec{F}\)
Let \(\vec{r} = \vec{r}_1 - \vec{r}_2\) be the vector joining the point of application of the two forces. Then:
\(\vec{\tau} = \vec{r} \times \vec{F}\)
This expression is independent of the choice of origin \(O\) because the position vectors \(\vec{r}_1\) and \(\vec{r}_2\) cancel out in the difference \(\vec{r}_1 - \vec{r}_2\). Thus, the moment of a couple is the same about any point in space.
In simple words: The turning power of a couple is purely determined by the forces and the distance between them, regardless of where you measure it from.

Exam Tip: Emphasize that origin independence arises because the net force of a couple is zero.

Question 6. A disc rotating about its axis with angular speed \(\omega_0\) is placed lightly (without any linear push) on a perfectly frictionless table. The radius of the disc is \(R\). What are the linear velocities of the points A, B and C on the disc shown in figure? Will the disc roll?
Answer:
For point A (at distance \(R\) from centre):
\(V_A = R\omega_0\) in the forward direction.
For point B (at distance \(R\) from centre, opposite side):
\(V_B = R\omega_0\) in the backward direction.
For point C (at distance \(R/2\) from centre):
\(V_C = \frac{R}{2}\omega_0\) in the forward direction.
**Will the disc roll?**
No, the disc will not roll because rolling requires friction at the point of contact to prevent slipping. Since the table is perfectly frictionless, the disc will only spin in place while sliding smoothly without rolling.
In simple words: The disc spins and slides on the table without rolling because there is no friction to grip the surface.

Exam Tip: Friction is essential for rolling motion; without it, pure rotation occurs alongside sliding.

Question 7. A uniform circular disc of radius \(R\) is rolling on a horizontal surface. Determine the tangential velocity (i) at the upper most point (ii) at the centre of mass and (iii) at the point of contact.
Answer:
In pure rolling, the velocity of the centre of mass is \(v_{\text{cm}} = R\omega\).
(i) At the uppermost point: The velocity is the sum of translational and rotational velocities:
\(v_{\text{top}} = v_{\text{cm}} + R\omega = v_{\text{cm}} + v_{\text{cm}} = 2v_{\text{cm}}\) (in forward direction).
(ii) At the centre of mass: The velocity is simply the velocity of translation:
\(v_{\text{cm}} = R\omega\).
(iii) At the point of contact: The translational and rotational velocities are equal and opposite, so the net velocity is:
\(v_{\text{contact}} = v_{\text{cm}} - R\omega = 0\).
In simple words: Top point moves at twice the speed, the centre moves at normal speed, and the bottom contact point has zero speed.

Exam Tip: Remember: top = \(2v\), centre = \(v\), bottom = \(0\) in pure rolling.

Question 8. Derive an expression for the total work done on a rigid body executing both translational and rotational motions.
Answer:
Consider a rigid body of mass \(M\) and moment of inertia \(I\) moving with velocity \(v\) and angular velocity \(\omega\).
The total kinetic energy of the body executing combined translatory and rotatory motion is:
\(E = K_{\text{trans}} + K_{\text{rot}} = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2\)
By the work-energy theorem, the total work done \(W\) on the body is equal to the change in its total kinetic energy:
\(W = \Delta E = E_{\text{final}} - E_{\text{initial}}\)
\(W = \left( \frac{1}{2}Mv_f^2 + \frac{1}{2}I\omega_f^2 \right) - \left( \frac{1}{2}Mv_i^2 + \frac{1}{2}I\omega_i^2 \right)\).
In simple words: Total work done equals the change in both linear and rotational kinetic energies combined.

Exam Tip: Link work-energy theorem directly to combined kinetic energy terms.

Question 9. Prove that the acceleration of a solid cylinder rolling without slipping down an inclined plane is \(\frac{2}{3} g \sin\theta\).
Answer:
For a body rolling down an inclined plane of inclination \(\theta\):
Acceleration \(a = \frac{g \sin\theta}{1 + \frac{I}{MR^2}}\)
For a solid cylinder, the moment of inertia about its geometric axis is:
\(I = \frac{1}{2}MR^2\)
Substituting \(I\) into the acceleration formula:
\(a = \frac{g \sin\theta}{1 + \frac{\frac{1}{2}MR^2}{MR^2}} = \frac{g \sin\theta}{1 + \frac{1}{2}} = \frac{g \sin\theta}{\frac{3}{2}} = \frac{2}{3} g \sin\theta\).
Hence proved.
In simple words: Substitute the solid cylinder's moment of inertia into the standard rolling acceleration formula to get \(\frac{2}{3}g\sin\theta\).

Exam Tip: Memorize the general acceleration formula for rolling bodies: \(a = \frac{g\sin\theta}{1 + K^2/R^2}\).

Question 10. Show that the angular momentum of a particle is the product of its linear momentum and moment arm. Also show that the angular momentum is produced only by the angular component of linear momentum.
Answer:
**(i) Product of linear momentum and moment arm:**
Angular momentum of a particle is given by:
\(L = |\vec{r} \times \vec{p}| = r p \sin\theta\)
Where \(r \sin\theta\) is the perpendicular distance (moment arm) from the axis of rotation to the line of action of linear momentum \(p\). Thus, \(L = (\text{linear momentum}) \times (\text{moment arm})\).
**(ii) Contribution of angular component:**
Resolving linear momentum \(p\) into two rectangular components: radial component \(p_r = p \cos\theta\) along the position vector, and transverse (tangential) component \(p_\theta = p \sin\theta\) perpendicular to the position vector.
Angular momentum magnitude is:
\(L = r (p \sin\theta) = r p_\theta\)
Since the radial component \(p_r\) has a zero cross product with the position vector (\(\vec{r} \times \vec{p}_r = 0\)), it contributes nothing to angular momentum. Thus, angular momentum is produced exclusively by the tangential (angular) component of linear momentum.
In simple words: Angular momentum is linear momentum times its lever arm, and only the sideways-moving momentum component creates it.

Exam Tip: Break momentum into radial and tangential components when explaining this concept.

NUMERICALS

Question 1. Three masses \(3\text{ kg}\), \(4\text{ kg}\) and \(5\text{ kg}\) are located at the corners of an equilateral triangle of side \(1\text{ m}\). Locate the centre of mass of the system.
Answer:
Let the \(3\text{ kg}\) mass be at the origin \((0,0)\).
Let the \(4\text{ kg}\) mass be along the x-axis at \((1, 0)\).
Let the \(5\text{ kg}\) mass be at the third vertex of the equilateral triangle, with coordinates \(\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right) = (0.5, 0.866)\).
Coordinates of centre of mass \((X_{\text{cm}}, Y_{\text{cm}})\):
\(X_{\text{cm}} = \frac{m_1x_1 + m_2x_2 + m_3x_3}{m_1 + m_2 + m_3}\)
\(X_{\text{cm}} = \frac{3(0) + 4(1) + 5(0.5)}{3 + 4 + 5} = \frac{0 + 4 + 2.5}{12} = \frac{6.5}{12} = 0.54\text{ m}\)
\(Y_{\text{cm}} = \frac{m_1y_1 + m_2y_2 + m_3y_3}{m_1 + m_2 + m_3}\)
\(Y_{\text{cm}} = \frac{3(0) + 4(0) + 5(0.866)}{12} = \frac{4.33}{12} = 0.36\text{ m}\)
Thus, \((X_{\text{cm}}, Y_{\text{cm}}) = (0.54\text{ m}, 0.36\text{ m})\).
In simple words: Set one mass at the origin, find coordinates for the others, and use the weighted average formula for \(X\) and \(Y\).

Exam Tip: Placing one mass at the origin simplifies the coordinate calculations significantly.

Question 2. Two particles mass \(100\text{ g}\) and \(300\text{ g}\) at a given time have velocities \(10\hat{i} - 7\hat{j} - 3\hat{k}\) and \(7\hat{i} - 9\hat{j} + 6\hat{k}\text{ ms}^{-1}\) respectively. Determine velocity of COM.
Answer:
Given:
\(m_1 = 100\text{ g} = 0.1\text{ kg}\)
\(\vec{v}_1 = 10\hat{i} - 7\hat{j} - 3\hat{k}\)
\(m_2 = 300\text{ g} = 0.3\text{ kg}\)
\(\vec{v}_2 = 7\hat{i} - 9\hat{j} + 6\hat{k}\)
Total mass \(M = m_1 + m_2 = 0.1 + 0.3 = 0.4\text{ kg}\)
Velocity of centre of mass \(\vec{V}_{\text{cm}} = \frac{m_1\vec{v}_1 + m_2\vec{v}_2}{m_1 + m_2}\):
\(m_1\vec{v}_1 = 0.1(10\hat{i} - 7\hat{j} - 3\hat{k}) = 1.0\hat{i} - 0.7\hat{j} - 0.3\hat{k}\)
\(m_2\vec{v}_2 = 0.3(7\hat{i} - 9\hat{j} + 6\hat{k}) = 2.1\hat{i} - 2.7\hat{j} + 1.8\hat{k}\)
Sum of momenta = \((1.0 + 2.1)\hat{i} + (-0.7 - 2.7)\hat{j} + (-0.3 + 1.8)\hat{k} = 3.1\hat{i} - 3.4\hat{j} + 1.5\hat{k}\)
\(\vec{V}_{\text{cm}} = \frac{3.1\hat{i} - 3.4\hat{j} + 1.5\hat{k}}{0.4} = \frac{31\hat{i} - 34\hat{j} + 15\hat{k}}{4}\text{ ms}^{-1}\).
In simple words: Multiply each mass by its velocity vector, add them up, and divide by the total mass.

Exam Tip: Convert grams to kilograms before calculating to maintain correct SI unit consistency.

Question 3. From a uniform disc of radius R, a circular disc of radius R/2 is cut out. The centre of the hole is at R/2 from the centre of original disc. Locate the centre of gravity of the resultant flat body.
Answer:
Let the original complete disc have mass \(M\) and radius \(R\), centred at the origin \((0,0)\).
Area of original disc \(A = \pi R^2\). Mass per unit area \(\sigma = \frac{M}{\pi R^2}\).
Radius of cut-out portion is \(R' = \frac{R}{2}\), and its area is \(A' = \pi \left(\frac{R}{2}\right)^2 = \frac{\pi R^2}{4} = \frac{A}{4}\).
Mass of cut-out portion \(M' = \frac{M}{4}\).
Centre of mass of the cut-out portion is at \(x = \frac{R}{2}\).
Using the centre of mass formula for a composite body (original mass minus removed mass):
\(X_{\text{cm}} = \frac{M(0) - M'\left(\frac{R}{2}\right)}{M - M'} = \frac{0 - \left(\frac{M}{4}\right)\left(\frac{R}{2}\right)}{M - \frac{M}{4}} = \frac{-\frac{MR}{8}}{\frac{3M}{4}} = -\frac{R}{6}\).
Thus, the COM of the resulting portion lies at \(\frac{R}{6}\) from the centre of the original disc in a direction opposite to the centre of the cut-out portion.
In simple words: Treat the removed circle as negative mass and apply the COM formula to find the shift.

Exam Tip: Negative mass technique simplifies problems involving cut-out cavities.

Question 4. The angular speed of a motor wheel is increased from \(1200\text{ rpm}\) to \(3120\text{ rpm}\) in \(16\text{ seconds}\). (i) What is its angular acceleration (assume the acceleration to be uniform) (ii) How many revolutions does the wheel make during this time?
Answer:
Initial angular speed \(\omega_1 = 1200\text{ rpm} = \frac{1200 \times 2\pi}{60} = 40\pi\text{ rad s}^{-1}\).
Final angular speed \(\omega_2 = 3120\text{ rpm} = \frac{3120 \times 2\pi}{60} = 104\pi\text{ rad s}^{-1}\).
Time \(t = 16\text{ s}\).
**(i) Angular acceleration \(\alpha\):**
\(\alpha = \frac{\omega_2 - \omega_1}{t} = \frac{104\pi - 40\pi}{16} = \frac{64\pi}{16} = 4\pi\text{ rad s}^{-2} \approx 12.57\text{ rad s}^{-2}\).
**(ii) Number of revolutions \(n\):**
Total angular displacement \(\theta = \omega_1 t + \frac{1}{2}\alpha t^2\)
\(\theta = (40\pi)(16) + \frac{1}{2}(4\pi)(16)^2 = 640\pi + 2\pi(256) = 640\pi + 512\pi = 1152\pi\text{ radians}\)
Number of revolutions \(n = \frac{\theta}{2\pi} = \frac{1152\pi}{2\pi} = 576\text{ revolutions}\).
In simple words: Convert rpm to rad/s, use \(\alpha = \Delta\omega / t\) for acceleration, and find total revolutions from angular displacement.

Exam Tip: Always convert rpm to rad/s by multiplying by \(\frac{2\pi}{60}\) before doing calculations.

Question 5. A metre stick is balanced on a knife edge at its centre. When two coins, each of mass \(5\text{ g}\) are put one on top of the other at the \(12.0\text{ cm}\) mark, the stick is found to be balanced at \(45.0\text{ cm}\), what is the mass of the metre stick?
Answer:
Let mass of the metre stick be \(M\), acting at its centre of gravity \(50.0\text{ cm}\).
The pivot (knife edge) is now at \(45.0\text{ cm}\).
Torque due to the two coins acting at \(12.0\text{ cm}\):
Distance of coins from pivot = \(45.0 - 12.0 = 33.0\text{ cm}\).
Mass of two coins \(m = 2 \times 5\text{ g} = 10\text{ g}\).
Torque of coins = \(m \times g \times 33.0 = 10 \times g \times 33\).
Torque due to the weight of the metre stick acting at its centre (\(50.0\text{ cm}\)):
Distance of centre of mass from pivot = \(50.0 - 45.0 = 5.0\text{ cm}\).
Torque of metre stick = \(M \times g \times 5.0\).
By the principle of moments (clockwise torque = counter-clockwise torque):
\(M \times g \times 5.0 = 10 \times g \times 33.0\)
\(5M = 330 \implies M = \frac{330}{5} = 66\text{ g}\).
In simple words: Equate the turning force of the coins on one side to the weight of the metre stick on the other side.

Exam Tip: Measure all lever arms relative to the new pivot position (\(45.0\text{ cm}\)).

Question 6. A \(3\text{ m}\) long ladder weighting \(20\text{ kg}\) leans on a frictionless wall. Its feet rest on the floor \(1\text{ m}\) from the wall as shown in figure. Find the reaction forces \(F_1\) and \(F_2\) of the wall and the floor.
Answer:
Given: Length of ladder \(L = 3\text{ m}\), weight of ladder \(W = mg = 20 \times 9.8 = 196\text{ N}\) (or taking \(g \approx 9.8\text{ or }10\)).
Distance of foot from wall \(x = 1\text{ m}\). Height of wall contact \(y = \sqrt{3^2 - 1^2} = \sqrt{8} = 2\sqrt{2}\text{ m}\).
Let \(F_1\) be the normal reaction force from the wall and \(F_2\) (or normal \(N\) and friction \(f\)) be the floor reaction.
For translational equilibrium:
Horizontal forces: \(F_1 = f\) (friction at floor)
Vertical forces: \(N = W = 196\text{ N}\)
Taking torque about the bottom end (A):
\(F_1 \times (2\sqrt{2}) - W \times (0.5) = 0\) (since COM is at midpoint of ladder, distance from wall is \(0.5\text{ m}\)).
\(F_1 \times 2.828 = 196 \times 0.5 = 98 \implies F_1 = \frac{98}{2.828} \approx 34.6\text{ N}\).
Floor reaction force \(F_2 = \sqrt{N^2 + f^2} = \sqrt{196^2 + 34.6^2} = \sqrt{38416 + 1197.16} = \sqrt{39613.16} \approx 199.0\text{ N}\).
Angle \(\tan\alpha = \frac{N}{f} = \frac{196}{34.6} = 5.6568 \implies \alpha \approx 80^\circ\).
In simple words: Balance horizontal forces, vertical forces, and take torques about the base to find all reaction forces.

Exam Tip: Taking torques about the foot of the ladder eliminates two unknown floor forces instantly.

Question 7. Calculate the ratio of radii of gyration of a circular ring and a disc of the same radius with respect to the axis passing through their centres and perpendicular to their planes.
Answer:
Moment of inertia of a ring about the perpendicular axis through its centre is \(I_{\text{ring}} = MR^2 = M K_{\text{ring}}^2 \implies K_{\text{ring}} = R\).
Moment of inertia of a disc about the perpendicular axis through its centre is \(I_{\text{disc}} = \frac{1}{2}MR^2 = M K_{\text{disc}}^2 \implies K_{\text{disc}} = \frac{R}{\sqrt{2}}\).
Ratio of radii of gyration:
\(\frac{K_{\text{ring}}}{K_{\text{disc}}} = \frac{R}{\frac{R}{\sqrt{2}}} = \frac{\sqrt{2}}{1}\).
In simple words: Divide the ring's radius of gyration by the disc's to get \(\sqrt{2} : 1\).

Exam Tip: Use \(I = MK^2\) to easily extract the radius of gyration \(K\) for any body.

Question 8. An automobile moves on a road with a speed of \(54\text{ kmh}^{-1}\). The radius of its wheels is \(0.35\text{ m}\). What is the average negative torque transmitted by its brakes to a wheel if the vehicle is brought to rest in \(15\text{s}\)? The moment of inertia of the wheel about the axis of rotation is \(3\text{ kg m}^2\).
Answer:
Initial speed \(v = 54\text{ kmh}^{-1} = 54 \times \frac{5}{18} = 15\text{ ms}^{-1}\).
Initial angular velocity of wheel \(\omega_0 = \frac{v}{R} = \frac{15}{0.35} = \frac{300}{7}\text{ rad s}^{-1}\).
Final angular velocity \(\omega = 0\) (since it comes to rest in \(t = 15\text{ s}\)).
Angular acceleration \(\alpha = \frac{\omega - \omega_0}{t} = \frac{0 - \frac{300}{7}}{15} = -\frac{20}{7}\text{ rad s}^{-2} \approx -2.857\text{ rad s}^{-2}\).
Torque \(\tau = I\alpha = 3 \times \left(-\frac{20}{7}\right) = -\frac{60}{7} \approx -8.57\text{ kg m}^2\text{s}^{-2}\).
In simple words: Find initial angular speed from linear speed, calculate angular deceleration, and multiply by moment of inertia to get the braking torque.

Exam Tip: Negative sign indicates retarding (braking) torque opposing rotation.

Question 9. A rod of length L and mass M is hinged at point O. A small bullet of mass m hits the rod, as shown in figure. The bullet get embedded in the rod. Find the angular velocity of the system just after the impact.
Answer:
By the principle of conservation of angular momentum (since no external torque acts on the (rod + bullet) system during collision):
Initial angular momentum of the bullet about the hinge O before impact:
\(L_{\text{initial}} = M V L\) (or \(m v L\), using standard notation where bullet mass is \(m\) and velocity is \(v\); following standard textbook notation with bullet mass \(m\) and speed \(v\)):
\(L_{\text{initial}} = m v L\)
Moment of inertia of the rod about hinge at end: \(I_{\text{rod}} = \frac{1}{3}ML^2\).
Moment of inertia of the embedded bullet at the end of the rod: \(I_{\text{bullet}} = mL^2\).
Total moment of inertia of the system after impact:
\(I_{\text{total}} = \frac{1}{3}ML^2 + mL^2 = \left(\frac{M + 3m}{3}\right)L^2\)
Equating initial and final angular momentum (\(L_{\text{initial}} = I_{\text{total}}\omega\)):
\(m v L = \left(\frac{M + 3m}{3}\right)L^2 \omega\)
Solving for angular velocity \(\omega\):
\(\omega = \frac{3mv}{(M + 3m)L}\).
In simple words: Use conservation of angular momentum: bullet's initial angular momentum equals the combined moment of inertia times final angular velocity.

Exam Tip: Perfectly inelastic collision of a bullet with a hinged rod conserves angular momentum about the hinge.

Question 10. A solid disc and a ring, both of radius \(10\text{ cm}\) are placed on a horizontal table simultaneously, with initial angular speed equal to \(10\pi\text{ rad s}^{-1}\). Which of the two will start to roll earlier? The coefficient of kinetic friction is \(\mu_k = 0.2\).
Answer:
Frictional force provides the retarding torque and accelerating linear force. Acceleration \(a = \mu_k g\) is the same for both. Retarding angular acceleration \(\alpha = \frac{\tau}{I} = \frac{\mu_k mg R}{I}\).
For disc: \(I_{\text{disc}} = \frac{1}{2}mR^2 \implies \alpha_{\text{disc}} = \frac{2\mu_k g}{R}\).
For ring: \(I_{\text{ring}} = mR^2 \implies \alpha_{\text{ring}} = \frac{\mu_k g}{R}\).
Time taken to achieve pure rolling condition (\(v = r\omega\)) is given by \(t = \frac{r\omega_0}{\mu_k g \left(1 + \frac{mR^2}{I}\right)}\) or comparing rates of deceleration.
Since the disc has a smaller moment of inertia, its angular speed decreases faster to satisfy the pure rolling condition \(v = r\omega\) earlier than the ring. Thus, **the disc begins to roll earlier than the ring**.
In simple words: The disc reaches pure rolling condition faster because its lower moment of inertia allows it to adjust its rotation quicker under friction.

Exam Tip: Objects with smaller mass distribution relative to the axis (like discs and spheres) start pure rolling before rings and hollow cylinders.

5 MARK QUESTIONS

Question 1. Obtain the expression for the linear acceleration of a cylinder rolling down an inclined plane and hence find the condition for the cylinder to roll down without slipping.
Answer:
Consider a cylinder of mass \(M\) and radius \(R\) rolling down an inclined plane of inclination \(\theta\) with acceleration \(a\).
Forces acting on the cylinder:
1. Component of gravitational force down the plane: \(mg \sin\theta\)
2. Normal reaction: \(N = mg \cos\theta\)
3. Frictional force \(f\) up the plane.
Equation of translational motion:
\(mg \sin\theta - f = Ma\) -- (1)
Equation of rotational motion about the centre of mass:
\(\tau = fR = I\alpha\) -- (2)
For pure rolling without slipping, \(a = R\alpha\), so \(\alpha = \frac{a}{R}\). Substituting into (2):
\(fR = I\left(\frac{a}{R}\right) \implies f = \frac{I}{R^2}a\) -- (3)
Substitute (3) into (1):
\(mg \sin\theta - \frac{I}{R^2}a = Ma\)
\(mg \sin\theta = \left(M + \frac{I}{R^2}\right)a\)
\(a = \frac{mg \sin\theta}{M + \frac{I}{R^2}} = \frac{g \sin\theta}{1 + \frac{I}{MR^2}}\)
**Condition for rolling without slipping:**
For the cylinder to roll without slipping, the required frictional force \(f\) must be less than or equal to the maximum static friction (\(f \le \mu_s N\)):
\(f = \frac{Ma}{1 + \frac{MR^2}{I}} \le \mu_s mg \cos\theta\).
In simple words: Combine Newton's second law for linear and rotational motion, substitute \(a = R\alpha\), and solve for acceleration and friction condition.

Exam Tip: This is a standard 5-mark derivation; write every step clearly separating translation and rotation equations.

Question 2. Prove the result that the velocity V of translation of a rolling body (like a ring, disc, cylinder or sphere) at the bottom of an inclined plane of a height h is given by:
\(v^2 = \frac{2gh}{1 + \frac{K^2}{R^2}}\)
where \(K\) = Radius of gyration of body about its symmetry axis, and \(R\) is radius of body. The body starts from rest at the top of the plane.
Answer:
By the law of conservation of mechanical energy, the potential energy at the top of the inclined plane is converted into total kinetic energy (translation + rotation) at the bottom.
Potential Energy at height \(h\):
\(\text{PE} = Mgh\)
Total Kinetic Energy at the bottom:
\(\text{KE}_{\text{total}} = K_{\text{trans}} + K_{\text{rot}} = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2\)
Since \(I = MK^2\) and \(\omega = \frac{v}{R}\):
\(\text{KE}_{\text{total}} = \frac{1}{2}Mv^2 + \frac{1}{2}(MK^2)\left(\frac{v}{R}\right)^2 = \frac{1}{2}Mv^2 \left(1 + \frac{K^2}{R^2}\right)\)
Equating Potential Energy to Total Kinetic Energy:
\(Mgh = \frac{1}{2}Mv^2 \left(1 + \frac{K^2}{R^2}\right)\)
Canceling mass \(M\) from both sides:
\(gh = \frac{1}{2}v^2 \left(1 + \frac{K^2}{R^2}\right)\)
\(v^2 = \frac{2gh}{1 + \frac{K^2}{R^2}}\).
Hence proved.
In simple words: Equate gravitational potential energy at the top to the sum of linear and rotational kinetic energies at the bottom.

Exam Tip: Energy conservation is much faster than force equations for finding velocity at the bottom of inclines.

Question 3. A light string is wound round a cylinder and carries a mass tied to it at the free end. When the mass is released, calculate:
(a) the linear acceleration of the descending mass
(b) angular acceleration of the cylinder
(c) Tension in the string.
Answer:
Let mass of the cylinder be \(M\), radius \(R\), and suspended mass be \(m\).
Let \(T\) be the tension in the string and \(a\) be the downward acceleration of mass \(m\).
**(a) Linear acceleration of descending mass:**
Equation of motion for the descending mass \(m\):
\(mg - T = ma\) -- (1)
Equation of rotational motion for the cylinder of moment of inertia \(I = \frac{1}{2}MR^2\):
Torque \(\tau = TR = I\alpha\) -- (2)
Since string does not slip, \(a = R\alpha \implies \alpha = \frac{a}{R}\). Substituting into (2):
\(TR = \left(\frac{1}{2}MR^2\right)\left(\frac{a}{R}\right) \implies T = \frac{1}{2}Ma\) -- (3)
Substitute (3) into (1):
\(mg - \frac{1}{2}Ma = ma \implies mg = \left(m + \frac{1}{2}M\right)a\)
\(a = \frac{mg}{m + \frac{1}{2}M} = \frac{2mg}{2m + M}\).
**(b) Angular acceleration of the cylinder:**
\(\alpha = \frac{a}{R} = \frac{2mg}{(2m + M)R}\).
**(c) Tension in the string:**
Substitute \(a\) into equation (3):
\(T = \frac{1}{2}M \left(\frac{2mg}{2m + M}\right) = \frac{Mmg}{2m + M}\).
In simple words: Set up Newton's second law for the falling weight and torque equation for the rotating cylinder, then solve simultaneously.

Exam Tip: Be careful with signs: weight pulls down, tension pulls up for the mass; tension provides the torque for the cylinder.

Question 4. State the theorem of (i) perpendicular axis (ii) parallel axis. Find the moment of inertia of a rod of mass M and length L about and axis perpendicular to it through one end. Given the moment of inertia about an axis perpendicular to rod and through COM is \(\frac{1}{12}ML^2\).
Answer:
**(i) Theorem of Perpendicular Axes:**
It states that the moment of inertia of a plane (two-dimensional) body about an axis perpendicular to its plane is equal to the sum of the moments of inertia of the lamina about any two mutually perpendicular axes in its plane and intersecting each other at the point where the perpendicular axis passes through the plane.
Formula: \(I_z = I_x + I_y\).
**(ii) Theorem of Parallel Axes:**
It states that the moment of inertia of a rigid body about any axis is equal to the moment of inertia of the body about a parallel axis passing through its centre of mass plus the product of the mass of the body and the square of the perpendicular distance between the two parallel axes.
Formula: \(I = I_c + Mh^2\).
**Moment of inertia of rod about one end:**
Given moment of inertia about centre of mass \(I_c = \frac{1}{12}ML^2\).
The distance between the axis passing through the centre of mass and the axis passing through one end is \(h = \frac{L}{2}\).
Applying the parallel axis theorem:
\(I_{\text{end}} = I_c + Mh^2 = \frac{1}{12}ML^2 + M\left(\frac{L}{2}\right)^2\)
\(I_{\text{end}} = \frac{1}{12}ML^2 + \frac{1}{4}ML^2 = \left(\frac{1}{12} + \frac{3}{12}\right)ML^2 = \frac{4}{12}ML^2 = \frac{1}{3}ML^2\).
In simple words: State both theorems clearly, then use the parallel axis theorem adding \(Mh^2\) to the centre-of-mass moment of inertia to find it at the end.

Exam Tip: State theorems word-for-word along with their mathematical expressions before applying them to derivations.