📌 Snapshot
- Rotation made simple: centre of mass, torque, moment of inertia and angular momentum with solved examples.
- Why internal forces never move the centre of mass, and how a rigid body is treated as a system of particles.
- The rotational twin of every linear law: $\tau = I\alpha$, $L = I\omega$ and $K = 1/2\,I\omega^2$, side by side in one table.
- Equilibrium, couples and levers, plus the skater effect that CUET loves to test in assertion-reason form.
📖 Detailed Notes
Written and reviewed by Akanksha Tiwari, Physics & Chemistry faculty at UniDrill (M.Sc. Physics, Banaras Hindu University · 8 years teaching Class XI–XII science). Based on NCERT Physics Class XI, Chapter 6.
Why does a spinning skater speed up when she pulls her arms in?
She spins slowly with arms stretched out. She draws them in, and without anyone touching her she whirls two or three times faster. No force pushed her, and yet her spin changed. The answer is a conservation law, the rotational cousin of conservation of momentum. This chapter builds that law step by step: first how to handle a whole body as a swarm of particles, then the rotational versions of force, mass, momentum and energy.
A rigid body as a system of particles
Real objects are not points. A rigid body is an idealised body in which the distance between every pair of particles stays fixed, whatever forces act. Its motion can be of three kinds. In pure translation every particle has the same velocity at every instant. In rotation about a fixed axis every particle moves in a circle in a plane perpendicular to the axis, with its centre on the axis. A body that is fixed at one point or along a line can only rotate. A free body can do both at once, like a rolling wheel, a spinning cricket ball in flight or a tumbling stone.
Centre of mass: the one point that behaves like the whole body
For two particles on a line, the centre of mass (CM) lies between them, closer to the heavier one:
$$X = \dfrac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$$
For equal masses it sits midway. For n particles, and in vector form:
$$R = \dfrac{\sum m_i r_i}{M}, \quad \text{where } M = \sum m_i$$
For a continuous body the sum becomes an integral, $R = (1/M)\int r\,dm$. If you place the origin at the CM, then $\int r\,dm = 0$: the mass-weighted displacements cancel. By symmetry, the CM of a uniform rod, ring, disc or sphere is at its geometric centre. Three equal masses at the corners of a triangle have their CM at the centroid. With unequal masses it moves toward the heavier corner, so it no longer coincides with the centroid.
Beyond the textbook. For the exam: the CM of a ring is its centre, and a CM is a point defined by the mass distribution. In reality: the CM need not lie inside the body at all. A ring, a bent boomerang and a high jumper arching over the bar all have their CM in empty space. In the Fosbury flop the jumper's CM can pass below the bar while every part of the body clears it. Why it matters: the CM is a calculated point, not a piece of matter.
Pause and predict: where is the centre of mass of a uniform ring, and is there any mass there?
Check your answer: it is at the centre of the ring, and there is no mass there. The CM is the balance point of the mass distribution. Every particle on one side is balanced by one on the opposite side.
Composite bodies. If a body is made of simple shapes with known CMs, treat each piece as a point mass at its own CM, then use the weighted-average formula.
Worked example: an L-shaped plate. An L-shaped uniform plate is made of three identical squares of side 1 m, arranged as: one at the corner, one to its right and one above it. Find the CM, with the origin at the outer corner.
| Step | What we do | Result |
|---|---|---|
| 1 | Replace each square by a point mass m at its centre | (0.5, 0.5), (1.5, 0.5), (0.5, 1.5) |
| 2 | $X = (m \times 0.5 + m \times 1.5 + m \times 0.5)/3m$ | $X = 2.5/3 = 5/6$ m |
| 3 | $Y = (m \times 0.5 + m \times 0.5 + m \times 1.5)/3m$ | $Y = 2.5/3 = 5/6$ m |
The CM is at $(5/6 \text{ m}, 5/6 \text{ m})$, about (0.83 m, 0.83 m). Common wrong step. Taking the centre of the whole 2 m × 2 m bounding square, (1 m, 1 m). The missing fourth square is exactly what pulls the true CM toward the corner.
Motion of the CM: why internal forces don't matter
Differentiate the CM definition twice and use Newton's second law for each particle. The forces between particles come in equal and opposite pairs (Newton's third law), so they cancel in the sum. Only external forces remain:
$MA = F(\text{external})$, and $P = MV$, so $dP/dt = F(\text{external})$
The CM moves as if all the mass were concentrated there and all external forces acted on it, no matter how complicated the internal motion is. If $F(\text{external}) = 0$, the total momentum P is constant: this is conservation of linear momentum for a system.
A shell flying along a parabola explodes in mid-air. The fragments scatter in all directions, but their CM keeps following the original parabola, because the explosion forces are internal and gravity is the only external force. A radioactive nucleus that decays while moving, or two stars orbiting each other with no outside force, show the same rule: the CM of the system moves uniformly.
Worked example: a man walking on a boat. A 60 kg man stands at one end of a 120 kg boat floating on still water. He walks 3 m along the boat to the other end. Ignoring water resistance, how far does the boat move, and how far does the man move relative to the water?
| Step | What we do | Result |
|---|---|---|
| 1 | No external horizontal force, and the system starts at rest, so the CM stays fixed | $\sum m \times (\text{displacement}) = 0$ |
| 2 | Let the boat move y backward. The man moves (3 − y) forward relative to the water | 60(3 − y) = 120 y |
| 3 | Solve | 180 = 180 y, so y = 1 m |
| 4 | Man's displacement relative to water | 3 − 1 = 2 m |
Common wrong step. Assuming the boat does not move at all, or that the man moves 3 m relative to the water. The man's push on the boat is an internal force, so it shifts the boat back while he moves forward. The lighter man moves twice as far as the heavier boat.
The vector product: a new way to multiply vectors
Rotation needs a product of two vectors that gives a vector. The vector (cross) product $a \times b$ has magnitude $ab\sin\theta$ and points perpendicular to the plane of a and b, in the direction given by the right-hand rule (curl your fingers from a to b and the thumb points along a × b).
| Property | Scalar product a · b | Vector product a × b |
|---|---|---|
| Result | Scalar: $ab\cos\theta$ | Vector: $ab\sin\theta$, perpendicular to both |
| Reversing the order | Unchanged (commutative) | Changes sign: $a \times b = -b \times a$ |
| Parallel vectors ($\theta = 0$) | Maximum, ab | Zero |
| Perpendicular vectors | Zero | Maximum, ab |
| Same vector | $a \cdot a = a^2$ | $a \times a = 0$ |
| Unit vectors | $\hat{i} \cdot \hat{i} = 1$, $\hat{i} \cdot \hat{j} = 0$ | $\hat{i} \times \hat{j} = \hat{k}$, $\hat{j} \times \hat{k} = \hat{i}$, $\hat{k} \times \hat{i} = \hat{j}$ |
For vectors in the x–y plane, the z-component of the cross product is the quick formula ($a_x b_y - a_y b_x$).
Angular velocity and the link to linear velocity
For a particle going around a fixed axis, angular velocity $\omega = d\theta/dt$ is the rate of sweeping angle (rad s⁻¹), and angular acceleration is $\alpha = d\omega/dt$. For a rigid body, every particle has the same ω, but the speed of a particle grows with its distance r from the axis:
$v = \omega r$, or in vector form $v = \omega \times r$
The vector ω lies along the axis of rotation (right-hand rule). For uniform rotation, $\omega = 2\pi \times$ (revolutions per second). Always convert rpm to rad s⁻¹ with $\omega = 2\pi \times (\text{rpm}/60)$ before using any formula.
Torque: the turning effect of a force
Open a door by pushing near the hinge and it is hard. Push at the handle and it is easy. Same force, different effect. Torque (moment of force) about a point is
$\tau = r \times F$, with magnitude $\tau = rF\sin\theta = r_\perp F = rF_\perp$
where r runs from the axis to the point where the force acts, and $r_\perp$ is the perpendicular distance of the line of force from the axis. Torque is a vector perpendicular to the plane of r and F. Its SI unit is the newton metre (N m) and its dimensions are $[M\,L^2\,T^{-2}]$.
Beyond the textbook. For the exam: torque and work have the same dimensions, $ML^2T^{-2}$. In reality: a torque of 1 N m is not 1 joule. Torque is a vector that causes angular acceleration. Energy is a scalar. They are linked by $W = \tau\theta$, where $\theta$ is the angle in radians turned. That is why the unit of torque is written as N m, never J, even though the dimensions match.
Angular momentum: momentum of rotation
The angular momentum of a particle about a point is the moment of its momentum:
$l = r \times p$, with magnitude $l = rp\sin\theta = r_\perp p$
It is perpendicular to the plane of r and p. For a particle moving in a straight line past a point at perpendicular distance d, l = m v d, and it stays constant even though r keeps changing. If the line of motion passes through the point (v parallel to r), then l = 0. In components, for motion in the x–y plane, $l(z) = x\,p_y - y\,p_x$.
Differentiating gives the rotational form of Newton's second law, $dl/dt = \tau$. For a system of particles, the internal torques cancel in pairs (when internal forces act along the line joining the particles), so
$dL/dt = \tau(\text{external})$, and if $\tau(\text{external}) = 0$ then $L = \text{constant}$
This is the law of conservation of angular momentum, applied separately to each component.
Equilibrium, couples and levers
A rigid body is in mechanical equilibrium when both its translational and rotational motion stay unchanged:
$\sum F = 0$ and $\sum \tau = 0$
The two conditions are independent. A couple is a pair of equal and opposite forces with different lines of action. The net force is zero, so the CM does not accelerate, but the net torque is not zero, so the body rotates. Turning a tap, a steering wheel or a key uses a couple.
A lever is in equilibrium when the torques about the fulcrum balance. This is the principle of moments: load × load arm = effort × effort arm, $d_1F_1 = d_2F_2$. The mechanical advantage is $F_1/F_2 = d_2/d_1$, the ratio of the arms.
Worked example: a loaded beam. A uniform 4 m beam of mass 20 kg rests on supports at its two ends. A 60 kg load is placed 1 m from the left end. Find the reaction at each support (g = 10 m s⁻²).
| Step | What we do | Result |
|---|---|---|
| 1 | Weights: beam 200 N acts at its centre (2 m); load 600 N at 1 m | Total downward force = 800 N |
| 2 | Take torques about the left support (this removes the left reaction from the equation) | R(B) × 4 = 200 × 2 + 600 × 1 |
| 3 | Solve | R(B) = 1000/4 = 250 N |
| 4 | Translational equilibrium: R(A) + R(B) = 800 | R(A) = 550 N |
Common wrong step. Splitting the total weight equally between the supports (400 N each). The load sits nearer the left end, so the left support carries more.
Centre of gravity. The centre of gravity (CG) is the point about which the total gravitational torque on the body is zero. If g is uniform over the body, the CG and the CM are the same point. The CM is defined by the mass distribution alone, the CG also depends on gravity. A body is balanced when supported at its CG.
Moment of inertia: rotational mass
In translation, mass resists a change in velocity. In rotation, a particle of mass m at distance r moves at $v = \omega r$, so its kinetic energy is $1/2\,m(\omega r)^2$. Adding over all particles:
$$K = \dfrac{1}{2} I \omega^2, \qquad \text{where } I = \sum m_i r_i^2$$
I is the moment of inertia (SI unit kg m², dimensions $ML^2$). It plays the role of mass, but it depends on how the mass is spread about the axis. Mass far from the axis counts for much more, because of the r². The radius of gyration k is defined by $I = Mk^2$: the distance at which the whole mass would have to sit to give the same I.
| Body (mass M, radius R) | Axis | I | Radius of gyration k |
|---|---|---|---|
| Thin ring | Through the centre, perpendicular to the plane | $MR^2$ | R |
| Thin ring | A diameter | $1/2\,MR^2$ | $R/\sqrt{2}$ |
| Thin rod (length L) | Perpendicular, through the midpoint | $ML^2/12$ | $L/\sqrt{12}$ |
| Disc | Through the centre, perpendicular to the plane | $1/2\,MR^2$ | $R/\sqrt{2}$ |
| Disc | A diameter | $1/4\,MR^2$ | $R/2$ |
| Hollow cylinder | Its own axis | $MR^2$ | R |
| Solid cylinder | Its own axis | $1/2\,MR^2$ | $R/\sqrt{2}$ |
| Solid sphere | A diameter | $2/5\,MR^2$ | $R\sqrt{2/5}$ |
Why the numbers differ. In a ring every bit of mass is at distance R, so $I = MR^2$ exactly. In a disc of the same mass and radius much of the mass lies closer in, so I falls to half. A solid sphere has even more mass near the axis, so $I = 0.4\,MR^2$. For equal M and R about the central axis, the order is ring (1) > disc ($1/2$) > solid sphere ($2/5$). Also, the same disc has $I = 1/2\,MR^2$ about its central axis but only $1/4\,MR^2$ about a diameter. The axis always matters.
Rotational kinematics and dynamics: the translation dictionary
For constant angular acceleration, the equations look exactly like those of straight-line motion:
$$\omega = \omega_0 + \alpha t, \qquad \theta = \omega_0 t + \dfrac{1}{2}\alpha t^2, \qquad \omega^2 = \omega_0^2 + 2\alpha\theta$$
The dynamics follow the same pattern. Newton's second law for a fixed axis is $\tau = I\alpha$. A torque turning the body through dθ does work $dW = \tau\,d\theta$, so the power is $P = \tau\omega$. The angular momentum about the axis is $L = I\omega$, and this connects back to $\tau = dL/dt = I\alpha$ when I is constant.
| Linear motion | Rotational motion |
|---|---|
| Displacement x | Angle θ |
| Velocity v | Angular velocity ω |
| Acceleration a | Angular acceleration α |
| Mass m | Moment of inertia I |
| Force F | Torque τ |
| Momentum $p = mv$ | Angular momentum $L = I\omega$ |
| $F = ma$ | $\tau = I\alpha$ |
| Kinetic energy $1/2\,mv^2$ | Kinetic energy $1/2\,I\omega^2$ |
| Work $F\,dx$ | Work $\tau\,d\theta$ |
| Power $Fv$ | Power $\tau\omega$ |
Worked example: a driven wheel. A uniform disc of mass 10 kg and radius 0.5 m is free to rotate about its central axis. A tangential force of 10 N is applied at the rim for 5 s, starting from rest. Find α, the final ω and the kinetic energy gained.
| Step | What we do | Result |
|---|---|---|
| 1 | Torque: $\tau = FR$ | 10 × 0.5 = 5 N m |
| 2 | Moment of inertia of a disc: $1/2\,MR^2$ | $1/2 \times 10 \times 0.25 = 1.25$ kg m² |
| 3 | $\alpha = \tau/I$ | 5/1.25 = 4 rad s⁻² |
| 4 | $\omega = \alpha t$ | 4 × 5 = 20 rad s⁻¹ |
| 5 | $K = 1/2\,I\omega^2$ | $1/2 \times 1.25 \times 400 = 250$ J |
| 6 | Cross-check: angle turned $\theta = 1/2\,\alpha t^2 = 50$ rad, and $W = \tau\theta$ | 5 × 50 = 250 J |
Both routes give 250 J, the work-energy idea in rotational form. Common wrong step. Using $I = MR^2$ (the ring value), which gives $I = 2.5$ kg m² and halves the acceleration. Always match the formula to the shape.
Worked example: a slowing fan. A ceiling fan running at 600 rpm is switched off and comes to rest uniformly in 20 s. How many revolutions does it make while stopping?
| Step | What we do | Result |
|---|---|---|
| 1 | Convert: $\omega_0 = 2\pi \times 600/60$ | $20\pi$ rad s⁻¹ |
| 2 | Uniform deceleration: $\alpha = (0 - \omega_0)/t$ | $-\pi$ rad s⁻² |
| 3 | $\theta = $ average $\omega \times t = (20\pi/2) \times 20$ | $200\pi$ rad |
| 4 | Revolutions = $\theta/2\pi$ | 100 |
Common wrong step. Putting 600 straight into the equations. The rotation formulas need rad s⁻¹, and using rpm throws the answer off by a factor of $2\pi/60$ in ω.
Conservation of angular momentum: the skater
For rotation about a fixed axis, if the external torque about that axis is zero then $L = I\omega$ stays constant: $I_1\omega_1 = I_2\omega_2$. When the skater pulls her arms in, her moment of inertia falls, so her angular velocity must rise to keep Iω fixed. Divers, acrobats and ballet dancers use the same trick, and the same law explains why a swivel-chair experiment works with outstretched weights.
Worked example: the spinning skater. A skater spins at 2 rad s⁻¹ with arms out (I = 4 kg m²). She pulls her arms in so that I becomes 1.6 kg m². Find the new angular velocity and compare the kinetic energies.
| Step | What we do | Result |
|---|---|---|
| 1 | No external torque about the vertical axis, so $I\omega$ is constant | $4 \times 2 = 1.6 \times \omega_2$ |
| 2 | Solve | $\omega_2 = 5$ rad s⁻¹ |
| 3 | Kinetic energy before: $1/2 \times 4 \times 2^2$ | 8 J |
| 4 | Kinetic energy after: $1/2 \times 1.6 \times 5^2$ | 20 J |
Common wrong step. Keeping ω constant, or keeping kinetic energy constant. The conserved quantity is $L = I\omega$, not $\omega$ and not K.
Pause and predict: in this example the kinetic energy rose from 8 J to 20 J, but there was no external torque. Where did the extra 12 J come from?
Check your answer: from the skater herself. Her muscles exert an inward force on her arms while the arms move inward, so they do positive work. Angular momentum is conserved, but kinetic energy is not.
Beyond the textbook. For the exam: $L = I\omega$ is conserved when the external torque is zero. In reality: the Earth–Moon system conserves angular momentum on a grand scale. Ocean tides act as a brake on the Earth's spin, and the lost spin angular momentum goes to the Moon's orbit, so the Moon drifts away by a few centimetres every year (measured by laser ranging) and the Earth's day grows longer by a few thousandths of a second per century.
Exam traps
| The trap | Why it's tempting | The truth |
|---|---|---|
| "The CM must be inside the body." | It usually is | A ring, a hollow shell or a bent rod has its CM in empty space |
| "CM equals centroid." | True for equal masses and for uniform plates | For unequal point masses the CM shifts toward the heavier mass |
| "When the skater pulls her arms in, ω is conserved." | Both are angular quantities | $L = I\omega$ is conserved. I falls, so ω rises |
| "A couple has zero torque because net force is zero." | $\sum F = 0$ sounds like equilibrium | A couple has zero net force but a non-zero torque |
| "Torque is measured in joules." | Same dimensions as energy | Torque is a vector, written in N m. Joule is for scalar energy |
| "I of a disc is $1/2\,MR^2$ about any axis." | One formula for one shape | $1/2\,MR^2$ is for the central axis. About a diameter it is $1/4\,MR^2$ |
| "Angular momentum of a moving particle is always $rp$." | Looks like a simple product | $l = rp\sin\theta$. It is zero if the velocity points along r |
| "$a \times b = b \times a$." | The scalar product is commutative | The vector product changes sign when the order is reversed |
Formula sheet
| Quantity | Formula | SI unit | When to use |
|---|---|---|---|
| CM (n particles) | $R = \sum m_i r_i/M$ | m | Point masses, composite bodies |
| Motion of CM | $MA = F(\text{external})$ | N | Explosions, boats, system problems |
| Link between v and ω | $v = \omega r$ | m s⁻¹ | Any point on a rotating body |
| Torque | $\tau = rF\sin\theta = r_\perp F$ | N m | Turning effect of a force |
| Angular momentum (particle) | $l = rp\sin\theta$ | kg m² s⁻¹ | Particle about a point |
| Newton's law (system) | $dL/dt = \tau(\text{external})$ | N m | General rotation |
| Equilibrium | $\sum F = 0$, $\sum \tau = 0$ | Beams, levers, ladders | |
| Principle of moments | $d_1F_1 = d_2F_2$ | Lever, seesaw | |
| Moment of inertia | $I = \sum m_i r_i^2$, $I = Mk^2$ | kg m² | Rotational inertia |
| Rotational KE | $K = 1/2\,I\omega^2$ | J | Energy of spinning bodies |
| Fixed-axis dynamics | $\tau = I\alpha$, $P = \tau\omega$, $L = I\omega$ | N m, W, kg m² s⁻¹ | Wheels, pulleys, flywheels |
| Kinematics (constant α) | $\omega = \omega_0 + \alpha t$, $\theta = \omega_0 t + 1/2\,\alpha t^2$ | rad s⁻¹, rad | Spin-up and slowing down |
| Conservation | $I_1\omega_1 = I_2\omega_2$ | No external torque (skater, diver) |
60-second revision
| Topic | Lock this in |
|---|---|
| Rigid body | Fixed inter-particle distances; pure translation, pure rotation or both |
| Centre of mass | $R = \sum m_i r_i/M$; at the geometric centre for uniform symmetric bodies |
| CM motion | $MA = F(\text{external})$; internal forces never matter |
| Vector product | $ab\sin\theta$; perpendicular to both; $a \times b = -b \times a$; $\hat{i} \times \hat{j} = \hat{k}$ |
| Torque | $r \times F$; N m; same dimensions as work but a vector |
| Angular momentum | $r \times p$; $dL/dt = \tau(\text{external})$; constant if $\tau = 0$ |
| Equilibrium | $\sum F = 0$ and $\sum \tau = 0$; a couple breaks the second condition only |
| Moment of inertia | Ring $MR^2$, disc $1/2\,MR^2$, rod $ML^2/12$, solid sphere $2/5\,MR^2$ |
| Rotational laws | $\tau = I\alpha$, $K = 1/2\,I\omega^2$, $L = I\omega$, $P = \tau\omega$ |
| Skater | I falls, ω rises, L constant, KE rises |
Where this chapter connects
- Previous: Work, Energy and Power gives the energy and power ideas that become $K = 1/2\,I\omega^2$ and $P = \tau\omega$ here.
- Next: Gravitation uses conservation of angular momentum to derive Kepler's law of areas.
- Class XII link: Atoms quantises angular momentum in the Bohr model ($mvr = nh/2\pi$), and Moving Charges and Magnetism uses torque on a current loop.
🎯 Practice MCQs
First 3 questions free · create a free account to unlock the rest — answers & explanations included, no payment needed
Q1. A 1 kg particle is at x = 0 and a 4 kg particle is at x = 5 m on the x-axis. The x-coordinate of the centre of mass is
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Answer: B
$X = (1 \times 0 + 4 \times 5)/(1 + 4) = 20/5 = 4.0$ m, closer to the heavier particle. Option (A) is the midpoint, which would be right only for equal masses. Option (C) is the distance of the CM from the heavier particle, not from the origin.
Q2. A shell moving along a parabolic path explodes in mid-air into several fragments, and all the fragments are still in flight. The centre of mass of the fragments will
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Answer: B
The explosion forces are internal, and internal forces cannot change the motion of the CM. Gravity is the only external force and it is unchanged, so the CM continues along the same parabola until a fragment lands and an external force from the ground begins to act.
Q3. For two non-zero vectors a and b, which statement about the vector product is correct?
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Answer: C
The vector product is perpendicular to the plane of a and b, hence to both. Its magnitude is $ab\sin\theta$, it changes sign on reversing the order, and $a \times a$ is the zero vector. The result $a^2$ belongs to the scalar product $a \cdot a$.
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Q4. A uniform thin rod has mass 1.2 kg and length 1.0 m. Its moment of inertia about an axis perpendicular to the rod through its midpoint is
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Answer: A
$I = ML^2/12 = 1.2 \times 1^2/12 = 0.10$ kg m². Option (B) is $ML^2/3$, the value about an axis through one end. Option (C) is $ML^2/6$, and option (D) is $ML^2$, the value for a point mass at distance L.
Q5. The radius of gyration of a uniform disc of radius R about a diameter is
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Answer: D
About a diameter, $I = 1/4\,MR^2$. Setting $I = Mk^2$ gives $k^2 = R^2/4$, so $k = R/2$. Option (A) is the radius of gyration about the central axis perpendicular to the plane (where $I = 1/2\,MR^2$).
Q6. A solid cylinder of mass 20 kg and radius 0.25 m rotates about its own axis at 100 rad s⁻¹. Its rotational kinetic energy is
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Answer: C
$I = 1/2\,MR^2 = 1/2 \times 20 \times (0.25)^2 = 0.625$ kg m². $K = 1/2\,I\omega^2 = 1/2 \times 0.625 \times 100^2 = 3125$ J. Option (B) drops the ½ in I, and option (D) uses $I = MR^2$ (the hollow-cylinder value).
Q7. Which pair of physical quantities has the same dimensions?
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Answer: D
Torque is force × distance, $ML^2T^{-2}$, the same as work. Angular momentum is $ML^2T^{-1}$, so it differs from torque by a factor of T. Linear momentum is $MLT^{-1}$ and moment of inertia is $ML^2$, neither of which matches the other member of its pair.
Q8. A skater spinning with angular velocity ω pulls in her arms so that her moment of inertia becomes one third of its initial value. There is no external torque. Her new angular velocity and rotational kinetic energy compared with the initial values are
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Answer: A
$I\omega$ is conserved, so $I/3 \times \omega' = I\omega$ gives $\omega' = 3\omega$. Kinetic energy $K = 1/2\,I\omega^2$, so $K' = 1/2\,(I/3)(9\omega^2) = 3 \times (1/2\,I\omega^2)$, which is 3 times the original. The extra energy comes from the work done by her muscles. Option (B) wrongly assumes kinetic energy is also conserved.
Q9. Match the rotational quantities in List I with their translational analogues in List II. List I: (P) moment of inertia (Q) torque (R) angular momentum (S) angular velocity. List II: (1) force (2) mass (3) linear momentum (4) velocity
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Answer: C
Moment of inertia plays the part of mass, torque the part of force, angular momentum the part of linear momentum and angular velocity the part of velocity. Option (A) swaps mass and force, and option (B) swaps momentum and velocity.
Q10. Assertion (A): A couple acting on a rigid body has zero net force but a non-zero net torque. Reason (R): If the total external force on a body is zero, the total external torque is automatically zero.
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Answer: C
The two forces of a couple are equal and opposite, so their sum is zero, but their lines of action differ and they produce a torque. So A is true. The reason is false: vanishing net force and vanishing net torque are independent conditions, and the couple is the counter-example.
Q11. The angular speed of a motor wheel is increased uniformly from 1200 rpm to 3120 rpm in 16 s. The angular acceleration of the wheel is
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Answer: C
$\omega_0 = 2\pi \times 1200/60 = 40\pi$ rad s⁻¹ and $\omega = 2\pi \times 3120/60 = 104\pi$ rad s⁻¹. $\alpha = (104\pi - 40\pi)/16 = 64\pi/16 = 4\pi$ rad s⁻². Options (A) and (B) come from slips of a factor of 2 in the conversion from rpm, and option (D) from a slip of a factor of 4.
Q12. A cord of negligible mass is wound on the rim of a flywheel of mass 20 kg and radius 0.20 m, and a steady pull of 25 N is applied to the cord. Treating the flywheel as a uniform disc, its angular acceleration is
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Answer: D
Torque $\tau = FR = 25 \times 0.20 = 5.0$ N m. For a disc, $I = 1/2\,MR^2 = 1/2 \times 20 \times 0.04 = 0.40$ kg m². So $\alpha = \tau/I = 5.0/0.40 = 12.5$ rad s⁻². Option (A) uses $I = MR^2$ (0.8 kg m²) and gets 6.25. Option (B) is the torque itself, not divided by I. Option (C) uses $I = 1/4\,MR^2$, the value about a diameter.
Q13. The moment of inertia of a solid sphere of mass M and radius R about a diameter is
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Answer: D
For a solid sphere $I = 2/5\,MR^2$. Option (A) is the ring value and option (B) the disc or solid-cylinder value. Option (C) is the value for a thin hollow spherical shell.
Q14. A rotating flywheel has angular velocity ω and kinetic energy K. No external torque acts on it, but its moment of inertia suddenly doubles as masses slide outward. The new angular velocity and kinetic energy are
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Answer: A
$I\omega$ is conserved, so $\omega$ falls to $\omega/2$. The kinetic energy is $K = 1/2\,I\omega^2 = L^2/(2I)$, so doubling I with fixed L halves K. Option (B) wrongly assumes K is conserved, and option (C) takes the energy as proportional to $\omega^2$ only.
Q15. A particle at position $r = (2\hat{i} + 3\hat{j})$ m has linear momentum $p = (4\hat{i} + 5\hat{j})$ kg m s⁻¹. Its angular momentum about the origin is
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Answer: D
$l = r \times p = (xp_y - yp_x)\hat{k} = (2 \times 5 - 3 \times 4)\hat{k} = (10 - 12)\hat{k} = -2\hat{k}$ kg m² s⁻¹. Option (A) has the wrong sign (the order of the cross product reversed). Option (B) adds the products, and option (C) adds the products and reverses the sign.
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