Question

The disc of mass m and radius R is confirmed to roll without slipping at A and B as shown

• A. The distance of instantaneous centre from point A is $\frac{2R}{3}$
• B. The angular velocity of the disc is $\frac{3V}{2R}$
• C. The velocity of the centre of mass of disc is $\frac{V}{2}$
• D. The total energy of the disc is $\frac{11}{16}mv^2$

Answer 1

Answer: (A), (B), (C), (D)

A, B, C, D

Answer 2

The problem describes a disc of mass $m$ and radius $R$ rolling without slipping between two horizontal surfaces. The top surface moves to the left with velocity $V$, and the bottom surface moves to the right with velocity $2V$. We need to evaluate the given statements.

Let $V_{CM}$ be the velocity of the center of mass of the disc and $\omega$ be its angular velocity. We adopt a coordinate system where the positive x-direction is to the right, and counter-clockwise rotation is positive $\omega$.

1. Rolling without slipping conditions:

   • For the top contact point (let's call it P_A), its velocity must match the velocity of the top surface. The top surface moves left with velocity $V$, so $V_{P_A} = -V$. The velocity of P_A on the disc is $V_{CM} - \omega R$ (assuming $V_{CM}$ is to the right and $\omega$ is counter-clockwise, the relative velocity of P_A with respect to CM is $\omega R$ to the left). So, we have the equation: $V_{CM} - \omega R = -V \quad (1)$

   • For the bottom contact point (let's call it P_B), its velocity must match the velocity of the bottom surface. The bottom surface moves right with velocity $2V$, so $V_{P_B} = 2V$. The velocity of P_B on the disc is $V_{CM} + \omega R$ (assuming $V_{CM}$ is to the right and $\omega$ is counter-clockwise, the relative velocity of P_B with respect to CM is $\omega R$ to the right). So, we have the equation: $V_{CM} + \omega R = 2V \quad (2)$

2. Solving for $V_{CM}$ and $\omega$: Add equation (1) and (2): $(V_{CM} - \omega R) + (V_{CM} + \omega R) = -V + 2V$ $2V_{CM} = V$ $V_{CM} = \frac{V}{2}$ (The center of mass moves to the right).

   Subtract equation (1) from (2): $(V_{CM} + \omega R) - (V_{CM} - \omega R) = 2V - (-V)$ $2\omega R = 3V$ $\omega = \frac{3V}{2R}$ (The disc rotates counter-clockwise).

3. Evaluating the options:

   • C) The velocity of the centre of mass of disc is $\frac{V}{2}$ From our calculation, $V_{CM} = \frac{V}{2}$. This statement is correct.

   • B) The angular velocity of the disc is $\frac{3V}{2R}$ From our calculation, $\omega = \frac{3V}{2R}$. This statement is correct.

   • A) The distance of instantaneous centre from point A is $\frac{2R}{3}$ The instantaneous center of rotation (ICR) is the point on the disc with zero velocity. Let the ICR be at a vertical distance $y_{ICR}$ from the center of mass (positive upwards). The velocity of a point at a distance $y$ from the CM is $V_y = V_{CM} - \omega y$. For the ICR, $V_y = 0$: $0 = V_{CM} - \omega y_{ICR}$ $y_{ICR} = \frac{V_{CM}}{\omega} = \frac{V/2}{3V/(2R)} = \frac{V}{2} \times \frac{2R}{3V} = \frac{R}{3}$. So, the ICR is located at a distance $R/3$ above the center of mass. Point A is at a distance $R$ above the center of mass. The distance of the ICR from point A is $R - y_{ICR} = R - \frac{R}{3} = \frac{2R}{3}$. This statement is correct.

   • D) The total energy of the disc is $\frac{11}{16}mv^2$ The total kinetic energy of a rolling disc is the sum of its translational and rotational kinetic energies: $KE = \frac{1}{2}m V_{CM}^2 + \frac{1}{2}I_{CM}\omega^2$ For a disc, the moment of inertia about its center of mass is $I_{CM} = \frac{1}{2}mR^2$. Substitute the values of $V_{CM}$ and $\omega$: $KE = \frac{1}{2}m \left(\frac{V}{2}\right)^2 + \frac{1}{2}\left(\frac{1}{2}mR^2\right)\left(\frac{3V}{2R}\right)^2$ $KE = \frac{1}{2}m \frac{V^2}{4} + \frac{1}{4}mR^2 \frac{9V^2}{4R^2}$ $KE = \frac{1}{8}mV^2 + \frac{9}{16}mV^2$ $KE = \frac{2}{16}mV^2 + \frac{9}{16}mV^2$ $KE = \frac{11}{16}mV^2$. This statement is correct.

All four options A, B, C, and D are correct.