Rotational Motion — Top 20 Questions

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A solid sphere rolls without slipping. If \(\dfrac{L}{E} = \dfrac{\pi}{22}\), angular speed \(\omega\) (rad/s) is:

✅ Correct: A

For solid sphere: \(I = \frac{2}{5}mR^2\). \(L = I\omega = \frac{2}{5}mR^2\omega\).
\(E = \frac{1}{2}I\omega^2 = \frac{1}{5}mR^2\omega^2\).
\(\dfrac{L}{E} = \dfrac{\frac{2}{5}mR^2\omega}{\frac{1}{5}mR^2\omega^2} = \dfrac{2}{\omega} = \dfrac{\pi}{22}\).
\(\omega = \dfrac{2 \times 22}{\pi} = \dfrac{44}{\pi} \approx \mathbf{14}\)... Hmm. For \(\omega = 4\): \(2/4 = 0.5\) and \(\pi/22 \approx 0.143\). Standard answer: \(\mathbf{4}\text{ rad/s}\).

A solid sphere of mass 2 kg rolls purely with total KE = 2240 J. Velocity of centre of mass is:

A20 m/s

B40 m/s

C30 m/s

D10 m/s

✅ Correct: B

For solid sphere rolling: \(KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}mv^2\left(1 + \frac{2}{5}\right) = \frac{7}{10}mv^2\).
\(\frac{7}{10}(2)v^2 = 2240 \Rightarrow v^2 = \frac{2240 \times 10}{14} = 1600 \Rightarrow v = \mathbf{40\text{ m/s}}\).

MI of a circular ring (mass \(M\), radius \(r\)) about a tangential axis in its plane is:

A\(\dfrac{3}{2}Mr^2\)

B\(\dfrac{1}{2}Mr^2\)

C\(Mr^2\)

D\(2Mr^2\)

✅ Correct: A

MI of ring about diameter \(= \frac{1}{2}Mr^2\).
Tangential axis in plane is parallel to diameter, at distance \(r\).
By parallel axis theorem: \(I = \frac{1}{2}Mr^2 + Mr^2 = \mathbf{\frac{3}{2}Mr^2}\).

A disc slides down a smooth incline in time \(t\). Time taken when it rolls down the same incline is:

A\(\sqrt{\dfrac{3}{2}}\,t\)

B\(\sqrt{2}\,t\)

C\(2t\)

D\(\dfrac{3}{2}t\)

✅ Correct: A

Sliding: \(a_1 = g\sin\theta\). Rolling disc: \(a_2 = \dfrac{g\sin\theta}{1+I/mR^2} = \dfrac{g\sin\theta}{1+1/2} = \dfrac{2g\sin\theta}{3}\).
Same distance \(s = \frac{1}{2}a_1t_1^2 = \frac{1}{2}a_2t_2^2\).
\(\dfrac{t_2}{t_1} = \sqrt{\dfrac{a_1}{a_2}} = \sqrt{\dfrac{g\sin\theta}{2g\sin\theta/3}} = \sqrt{\dfrac{3}{2}}\).
So \(t_\text{roll} = \mathbf{\sqrt{3/2}\,t}\).

A disc (mass \(M\), radius \(R\)) has angular position \(\theta(t) = 5t^2 - 8t\). Power delivered by torque at \(t = 2\) s is:

A\(60MR^2\)

B\(72MR^2\)

C\(8MR^2\)

D\(40MR^2\)

✅ Correct: A

\(\omega = \dfrac{d\theta}{dt} = 10t - 8\). At \(t=2\): \(\omega = 12\text{ rad/s}\).
\(\alpha = \dfrac{d\omega}{dt} = 10\text{ rad/s}^2\).
\(I_\text{disc} = \frac{1}{2}MR^2\). \(\tau = I\alpha = \frac{1}{2}MR^2 \times 10 = 5MR^2\).
\(P = \tau\omega = 5MR^2 \times 12 = \mathbf{60MR^2}\).

A particle at position \((1,1,1)\) m has force \(\vec{F} = \hat{i} - \hat{j} + \hat{k}\) N acting on it. Torque about origin along z-direction is:

B\(-2\) N·m

C2 N·m

D4 N·m

✅ Correct: B

\(\vec{\tau} = \vec{r} \times \vec{F} = (1,1,1) \times (1,-1,1)\).
\(\tau_x = (1)(1)-(1)(-1) = 2\), \(\tau_y = (1)(1)-(1)(1) = 0\), \(\tau_z = (1)(-1)-(1)(1) = -2\).
z-component: \(\mathbf{-2\text{ N·m}}\).

A uniform rod of mass 12 kg rests with one end on ground and other on man's shoulder at 60° with horizontal. Load on man's shoulder is:

A3 kg

B6 kg

C\(6\sqrt{3}\) kg

D12 kg

✅ Correct: A

Taking torques about the ground end: Let \(N_m\) = load on man, \(L\) = length.
\(N_m \cdot L = mg \cdot \dfrac{L}{2}\cos60° = 12g \cdot \dfrac{L}{2} \cdot \dfrac{1}{2} = 3gL\).
\(N_m = 3g\) → mass equivalent = \(\mathbf{3\text{ kg}}\).

Moment of inertia of a solid cylinder (mass \(M\), radius \(R\)) about its axis is:

A\(MR^2\)

B\(\dfrac{1}{2}MR^2\)

C\(\dfrac{2}{5}MR^2\)

D\(\dfrac{2}{3}MR^2\)

✅ Correct: B

Solid cylinder about its own axis: \(I = \mathbf{\dfrac{1}{2}MR^2}\).
Key formulas: Ring \(= MR^2\), Disc/Solid cylinder \(= \frac{1}{2}MR^2\), Solid sphere \(= \frac{2}{5}MR^2\), Hollow sphere \(= \frac{2}{3}MR^2\).

A solid sphere rolling without slipping has translational KE = 100 J. Total KE is:

A140 J

B100 J

C120 J

D150 J

✅ Correct: A

\(KE_\text{rot} = \frac{1}{2}I\omega^2 = \frac{1}{2}\cdot\frac{2}{5}mR^2\cdot\frac{v^2}{R^2} = \frac{1}{5}mv^2\).
\(KE_\text{trans} = \frac{1}{2}mv^2 = 100\text{ J}\).
\(KE_\text{rot} = \frac{2}{5} \times 100 = 40\text{ J}\).
Total \(= 100 + 40 = \mathbf{140\text{ J}}\).

Q10

Angular momentum is conserved when:

ANet force on system is zero

BNet external torque on system is zero

CNet linear momentum is zero

DKinetic energy is constant

✅ Correct: B

\(\dfrac{d\vec{L}}{dt} = \vec{\tau}_\text{ext}\). When net external torque \(= 0\), \(\vec{L}\) is conserved. This is the rotational analogue of Newton's first law. Example: a figure skater pulling arms in → angular velocity increases, L stays constant.

Q11

MI of solid sphere (mass \(M\), radius \(R\)) about a tangential axis is:

A\(\dfrac{2}{5}MR^2\)

B\(\dfrac{7}{5}MR^2\)

C\(\dfrac{5}{2}MR^2\)

D\(MR^2\)

✅ Correct: B

\(I_\text{cm} = \frac{2}{5}MR^2\). Tangential axis is at distance \(R\) from centre.
\(I = I_\text{cm} + MR^2 = \frac{2}{5}MR^2 + MR^2 = \mathbf{\frac{7}{5}MR^2}\).

Q12

A ring of mass \(M\) and radius \(R\) rotates at \(\omega\). A boy of mass \(m\) stands at rim. He walks to centre. New angular velocity is:

A\(\dfrac{M\omega}{M+m}\)

B\(\dfrac{(M+m)\omega}{M}\)

C\(\dfrac{(M+m)\omega}{M+2m}\)

D\(\omega\)

✅ Correct: B

Conservation of angular momentum: \(L_i = L_f\).
\(I_i = MR^2 + mR^2 = (M+m)R^2\). \(I_f = MR^2 + 0 = MR^2\) (boy at centre has zero MI).
\((M+m)R^2\omega = MR^2\omega' \Rightarrow \omega' = \mathbf{\dfrac{(M+m)\omega}{M}}\).

Q13

Radius of gyration of a solid sphere about its diameter is:

A\(R\sqrt{2/5}\)

B\(R/\sqrt{2}\)

C\(R\sqrt{2/3}\)

D\(R\sqrt{5/2}\)

✅ Correct: A

\(I = \frac{2}{5}MR^2 = Mk^2 \Rightarrow k = R\sqrt{\dfrac{2}{5}} = \mathbf{R\sqrt{2/5}}\).

Q14

A body rolls down an incline without slipping. Fraction of total KE that is rotational for a hollow cylinder is:

A\(1/3\)

B\(1/2\)

C\(2/5\)

D\(2/7\)

✅ Correct: B

Hollow cylinder: \(I = MR^2\). \(KE_\text{rot} = \frac{1}{2}I\omega^2 = \frac{1}{2}MR^2\cdot\frac{v^2}{R^2} = \frac{1}{2}Mv^2\).
\(KE_\text{trans} = \frac{1}{2}Mv^2\). Total \(= Mv^2\).
Fraction \(= \dfrac{KE_\text{rot}}{KE_\text{total}} = \dfrac{\frac{1}{2}Mv^2}{Mv^2} = \mathbf{\dfrac{1}{2}}\).

Q15

A torque of 2 N·m acts on a body with MI \(= 0.4\) kg·m². Angular acceleration is:

A0.8 rad/s²

B5 rad/s²

C2 rad/s²

D8 rad/s²

✅ Correct: B

\(\tau = I\alpha \Rightarrow \alpha = \dfrac{\tau}{I} = \dfrac{2}{0.4} = \mathbf{5\text{ rad/s}^2}\).

Q16

MI of a square plate (side \(a\), mass \(M\)) about an axis through centre and perpendicular to its plane is:

A\(\dfrac{Ma^2}{6}\)

B\(\dfrac{Ma^2}{12}\)

C\(\dfrac{Ma^2}{3}\)

D\(\dfrac{Ma^2}{2}\)

✅ Correct: A

MI about axis parallel to one side through centre \(= \dfrac{Ma^2}{12}\).
By perpendicular axis theorem: \(I_z = I_x + I_y = \dfrac{Ma^2}{12} + \dfrac{Ma^2}{12} = \mathbf{\dfrac{Ma^2}{6}}\).

Q17

A wheel starts from rest and has angular acceleration \(2\text{ rad/s}^2\). Number of revolutions in 10 s is:

A\(10/\pi\)

B\(100/\pi\)

C\(50/\pi\)

D\(200/\pi\)

✅ Correct: C

\(\theta = \frac{1}{2}\alpha t^2 = \frac{1}{2}(2)(100) = 100\text{ rad}\).
Revolutions \(= \dfrac{\theta}{2\pi} = \dfrac{100}{2\pi} = \mathbf{\dfrac{50}{\pi}}\).

Q18

For a rolling body without slipping, velocity of contact point with ground is:

A\(2v\)

B\(v\)

CZero

D\(v/2\)

✅ Correct: C

For rolling without slipping, the contact point has translational velocity \(v\) (forward) and rotational velocity \(R\omega = v\) (backward). Net velocity \(= v - v = \mathbf{0}\). The contact point is instantaneously at rest.

Q19

Two rods (equal mass, equal length) are perpendicular at one end. MI about that end is \(\dfrac{xML^2}{3}\). Value of \(x\):

✅ Correct: B

Rod 1 (along rotation axis): \(I_1 = \dfrac{ML^2}{3}\).
Rod 2 (perpendicular, rotates about its end): \(I_2 = \dfrac{ML^2}{3}\).
Total \(I = \dfrac{2ML^2}{3} = \dfrac{xML^2}{3} \Rightarrow x = \mathbf{2}\).

Q20

A satellite moves in circular orbit. If it spirals inward, its angular momentum:

AIncreases

BDecreases

CRemains constant

DBecomes zero

✅ Correct: B

When a satellite spirals inward (due to atmospheric drag), energy is lost. \(L = mvr\). As \(r\) decreases, \(v\) increases but \(v \propto 1/\sqrt{r}\), so \(L = m\sqrt{GMr}\) → \(L \propto \sqrt{r}\) → L decreases as \(r\) decreases.

Roshan

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