10 Essential Rotational Motion MCQs for JEE Mains Physics Success

Rotational Motion MCQs for JEE Mains

1. A uniform rod of length \( l \) and mass \( m \) is pivoted at one end. It is held horizontal and then released. What is the angular acceleration of the rod just after it is released?

(A) \(\frac{3g}{2l}\) (B) \(\frac{2g}{3l}\) (C) \(\frac{g}{l}\) (D) \(\frac{3g}{l}\)

Correct Answer: (A) \(\frac{3g}{2l}\)

Explanation: Torque due to gravity acts at the center of mass (at \( l/2 \)), so \(\tau = mg \cdot \frac{l}{2}\). Moment of inertia of a rod about one end is \( I = \frac{1}{3} m l^2 \). Using \(\tau = I \alpha\), we get \( mg \cdot \frac{l}{2} = \frac{1}{3} m l^2 \cdot \alpha \), solving to \(\alpha = \frac{3g}{2l}\).

2. A solid cylinder of radius \( r \) and mass \( m \) is attached to a thin cylindrical shell of radius \( 2r \) and mass \( 2m \), both sharing the same axis. What is the moment of inertia of the system about their common axis?

(A) \(\frac{1}{2} m r^2 + 2 m (2r)^2\) (B) \(\frac{1}{2} m r^2 + \frac{1}{2} (2m) (2r)^2\) (C) \( m r^2 + 2 m (2r)^2\) (D) \(\frac{1}{2} m r^2 + \frac{2}{3} (2m) (2r)^2\)

Correct Answer: (A) \(\frac{1}{2} m r^2 + 2 m (2r)^2\)

Explanation: For a solid cylinder, \( I = \frac{1}{2} m r^2 \). For a thin cylindrical shell, \( I = m R^2 \), so for mass \( 2m \) and radius \( 2r \), \( I = 2m (2r)^2 = 8 m r^2 \). Total \( I = \frac{1}{2} m r^2 + 8 m r^2 \).

3. A disk of mass \( m \) and radius \( r \) is spinning with angular velocity \( \omega \) when a point mass \( m \) is dropped onto it at a distance \( \frac{r}{2} \) from the center. If the point mass sticks to the disk, what is the new angular velocity?

(A) \(\frac{\omega}{1 + \frac{m r^2}{2 I_{\text{disk}}}}\) (B) \(\frac{\omega}{1 + \frac{m \left(\frac{r}{2}\right)^2}{I_{\text{disk}}}}\) (C) \(\omega \cdot \frac{I_{\text{disk}}}{I_{\text{disk}} + m \left(\frac{r}{2}\right)^2}\) (D) \(\omega \cdot \frac{I_{\text{disk}} + m r^2}{I_{\text{disk}}}\)

Correct Answer: (C) \(\omega \cdot \frac{I_{\text{disk}}}{I_{\text{disk}} + m \left(\frac{r}{2}\right)^2}\)

Explanation: Angular momentum is conserved. Initial \( L = I_{\text{disk}} \omega \), where \( I_{\text{disk}} = \frac{1}{2} m r^2 \). Final \( I = \frac{1}{2} m r^2 + m \left(\frac{r}{2}\right)^2 = \frac{3}{4} m r^2 \). So \( \omega' = \frac{I_{\text{disk}} \omega}{I_{\text{disk}} + m \left(\frac{r}{2}\right)^2} \).

4. A solid sphere of radius \( r \) and mass \( m \) rolls without slipping down an inclined plane of angle \( \theta \). What is the linear acceleration of the sphere's center of mass?

(A) \(\frac{5}{7} g \sin \theta\) (B) \(\frac{2}{5} g \sin \theta\) (C) \( g \sin \theta\) (D) \(\frac{7}{5} g \sin \theta\)

Correct Answer: (A) \(\frac{5}{7} g \sin \theta\)

Explanation: For rolling, \( a = \frac{g \sin \theta}{1 + \frac{I}{m r^2}} \). For a solid sphere, \( I = \frac{2}{5} m r^2 \), so \( a = \frac{g \sin \theta}{1 + \frac{2}{5}} = \frac{5}{7} g \sin \theta \).

5. A rigid body is subjected to two forces: \( F_1 = 10 \, \text{N} \) acting along the positive x-axis at point \( (1, 0, 0) \) and \( F_2 = -10 \, \text{N} \) acting along the negative x-axis at point \( (0, 1, 0) \), with the origin as the pivot. What is the net torque on the body?

(A) \( 10 \, \hat{k} \, \text{Nm} \) (B) \( -10 \, \hat{k} \, \text{Nm} \) (C) \( 10 \, \hat{j} \, \text{Nm} \) (D) \( 0 \, \text{Nm} \)

Correct Answer: (A) \( 10 \, \hat{k} \, \text{Nm} \)

Explanation: Torque \( \tau = \vec{r} \times \vec{F} \). For \( F_1 \), \( \vec{r} = \langle 1, 0, 0 \rangle \), \( \vec{F}_1 = \langle 10, 0, 0 \rangle \), so \( \tau_1 = 0 \). For \( F_2 \), \( \vec{r} = \langle 0, 1, 0 \rangle \), \( \vec{F}_2 = \langle -10, 0, 0 \rangle \), so \( \tau_2 = \langle 0, 0, 10 \rangle = 10 \, \hat{k} \, \text{Nm} \). Net torque = \( 10 \, \hat{k} \, \text{Nm} \).

6. Using the parallel axis theorem, find the moment of inertia of a thin rod of length \( l \) and mass \( m \) about an axis perpendicular to the rod and passing through a point at a distance \( \frac{l}{4} \) from one end.

(A) \(\frac{7}{48} m l^2\) (B) \(\frac{1}{12} m l^2\) (C) \(\frac{1}{3} m l^2\) (D) \(\frac{13}{48} m l^2\)

Correct Answer: (A) \(\frac{7}{48} m l^2\)

Explanation: \( I_{\text{cm}} = \frac{1}{12} m l^2 \) (about center). Distance from center to \( \frac{l}{4} \) from end is \( \frac{l}{2} - \frac{l}{4} = \frac{l}{4} \). Using parallel axis theorem, \( I = I_{\text{cm}} + m d^2 = \frac{1}{12} m l^2 + m \left(\frac{l}{4}\right)^2 = \frac{1}{12} m l^2 + \frac{1}{16} m l^2 = \frac{7}{48} m l^2 \).

7. A pulley with moment of inertia \( I \) and radius \( r \) has a light string wrapped around it, with a mass \( m \) attached to the string. The pulley is fixed but can rotate about its center. When the mass is released, what is the linear acceleration of the mass?

(A) \(\frac{m g}{m + \frac{I}{r^2}}\) (B) \(\frac{m g r}{I}\) (C) \(\frac{m g}{m + I}\) (D) \( g \)

Correct Answer: (A) \(\frac{m g}{m + \frac{I}{r^2}}\)

Explanation: \( m g - T = m a \), \( T r = I \alpha \), and \( a = r \alpha \). So \( T = \frac{I a}{r^2} \). Then \( m g = m a + \frac{I a}{r^2} \), so \( a = \frac{m g}{m + \frac{I}{r^2}} \).

8. If a rigid body is rotating about a fixed axis, the linear velocity of a point on the body is perpendicular to:

(A) The axis of rotation (B) The radius vector from the axis to the point (C) Both (A) and (B) (D) Neither (A) nor (B)

Correct Answer: (C) Both (A) and (B)

Explanation: \( \vec{v} = \vec{\omega} \times \vec{r} \), so \( \vec{v} \) is perpendicular to both \( \vec{\omega} \) (axis) and \( \vec{r} \) (radius vector).

9. A disk is rotating with angular velocity \( \omega \). If a constant torque \( \tau \) is applied to it for time \( t \), what is the work done by the torque?

(A) \( \tau \omega t \) (B) \(\frac{1}{2} I \omega^2\) (C) \( \tau \theta \), where \( \theta \) is the angle rotated (D) Both (A) and (C)

Correct Answer: (C) \( \tau \theta \), where \( \theta \) is the angle rotated

Explanation: Work done by constant torque is \( W = \int \tau \, d\theta = \tau \theta \), where \( \theta \) is the total angular displacement.

10. A uniform disk of radius \( 0.2 \, \text{m} \) and mass \( 5 \, \text{kg} \) is rotating about its axis at \( 10 \, \text{rad/s} \). What is its rotational kinetic energy?

(A) \( 5 \, \text{J} \) (B) \( 10 \, \text{J} \) (C) \( 25 \, \text{J} \) (D) \( 50 \, \text{J} \)

Correct Answer: (A) \( 5 \, \text{J} \)

Explanation: \( I = \frac{1}{2} m r^2 = \frac{1}{2} \cdot 5 \cdot (0.2)^2 = 0.1 \, \text{kg m}^2 \). \( K = \frac{1}{2} I \omega^2 = \frac{1}{2} \cdot 0.1 \cdot 10^2 = 5 \, \text{J} \).