Question

A ring and a disc of same mass and same diameter are rolling without slipping. Their linear velocities are same, then the ratio of their kinetic energy is:

• A. \( 0.75 \)
• B. \( 1.33 \)
• C. \( 0.5 \)
• D. \( 2.66 \)

Hint:

For rolling motion, the total kinetic energy is the sum of translational and rotational components. The moment of inertia plays a key role in determining the distribution of kinetic energy between these components.

Answer 1

The Correct Option is B

Step 1: Understanding kinetic energy in rolling motion
The total kinetic energy of a rolling body consists of translational and rotational kinetic energy. It is given by:
\[ KE = KE_{\text{translational}} + KE_{\text{rotational}} \] For a rolling body, we use: \[ KE = \frac{1}{2} M v^2 + \frac{1}{2} I \omega^2 \] where \( M \) is the mass, \( v \) is the linear velocity, \( I \) is the moment of inertia, and \( \omega \) is the angular velocity. Since rolling without slipping implies \( v = R\omega \), we substitute \( \omega = \frac{v}{R} \).
Step 2: Kinetic energy for a ring
The moment of inertia of a ring about its central axis is: \[ I_{\text{ring}} = MR^2 \] \[ KE_{\text{ring}} = \frac{1}{2} M v^2 + \frac{1}{2} M R^2 \left(\frac{v^2}{R^2}\right) \] \[ KE_{\text{ring}} = \frac{1}{2} M v^2 + \frac{1}{2} M v^2 = M v^2 \] Step 3: Kinetic energy for a disc
The moment of inertia of a disc about its central axis is: \[ I_{\text{disc}} = \frac{1}{2} M R^2 \] \[ KE_{\text{disc}} = \frac{1}{2} M v^2 + \frac{1}{2} \left(\frac{1}{2} M R^2\right) \left(\frac{v^2}{R^2}\right) \] \[ KE_{\text{disc}} = \frac{1}{2} M v^2 + \frac{1}{4} M v^2 = \frac{3}{4} M v^2 \] Step 4: Ratio of kinetic energies
The ratio of the kinetic energy of the ring to the disc is: \[ \frac{KE_{\text{ring}}}{KE_{\text{disc}}} = \frac{M v^2}{\frac{3}{4} M v^2} = \frac{4}{3} = 1.33 \] Thus, the ratio of their kinetic energy is: \[ \mathbf{1.33} \]