Question

A thin circular ring and a circular disc of equal mass are rolling without sliding. If their linear velocities are equal and the total kinetic energy of the disc is 6 J, then the total kinetic energy of the ring is:

• A. \( 6 \) J
• B. \( 3 \) J
• C. \( 8 \) J
• D. \( 4 \) J

Hint:

For rolling motion, remember: \[ K_{\text{total}} = \frac{1}{2} m v^2 + \frac{1}{2} I \omega^2 \] where \( v = r \omega \).

Answer 1

The Correct Option is D

Step 1: Kinetic Energy in Rolling Motion The total kinetic energy \( K \) of a rolling object is the sum of translational and rotational kinetic energies: \[ K = K_{\text{translational}} + K_{\text{rotational}} \] \[ K = \frac{1}{2} m v^2 + \frac{1}{2} I \omega^2 \] Since \( v = r \omega \), we substitute \( \omega \). Step 2: Comparing a Disc and a Ring For a solid disc: \[ I_{\text{disc}} = \frac{1}{2} m r^2 \] \[ K_{\text{disc}} = \frac{1}{2} m v^2 + \frac{1}{2} \times \frac{1}{2} m r^2 \times \frac{v^2}{r^2} \] \[ = \frac{1}{2} m v^2 + \frac{1}{4} m v^2 \] \[ = \frac{3}{4} m v^2 \] Given \( K_{\text{disc}} = 6 \) J, \[ \frac{3}{4} m v^2 = 6 \] For a thin ring: \[ I_{\text{ring}} = m r^2 \] \[ K_{\text{ring}} = \frac{1}{2} m v^2 + \frac{1}{2} m r^2 \times \frac{v^2}{r^2} \] \[ = \frac{1}{2} m v^2 + \frac{1}{2} m v^2 \] \[ = m v^2 \] Using \( \frac{3}{4} m v^2 = 6 \), solving for \( m v^2 \), \[ m v^2 = \frac{4}{3} \times 6 = 8 \] \[ K_{\text{ring}} = \frac{2}{3} \times 6 = 4 \text{ J} \] Conclusion Thus, the correct answer is: \[ 4 \text{ J} \]