Question

A ring and a disc roll on a horizontal surface without slipping with same linear velocity. If both have same mass and total kinetic energy of the ring is \(4\,\text{J}\), then total kinetic energy of the disc is

• A. \(2\,\text{J}\)
• B. \(6\,\text{J}\)
• C. \(8\,\text{J}\)
• D. \(3\,\text{J}\)

Hint:

For rolling bodies with same mass and speed, total kinetic energy depends on the moment of inertia.

Answer 1

The Correct Option is D

Step 1: Write expression for total kinetic energy.
For rolling motion, total kinetic energy is given by:
\[ K = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 \]
Step 2: Kinetic energy of the ring.
For a ring, \(I = mR^2\) and \(\omega = \frac{v}{R}\).
\[ K_{\text{ring}} = \frac{1}{2}mv^2 + \frac{1}{2}mR^2\left(\frac{v}{R}\right)^2 = mv^2 \]
Step 3: Given kinetic energy of ring.
\[ mv^2 = 4\,\text{J} \]
Step 4: Kinetic energy of the disc.
For a disc, \(I = \frac{1}{2}mR^2\).
\[ K_{\text{disc}} = \frac{1}{2}mv^2 + \frac{1}{2}\cdot\frac{1}{2}mR^2\left(\frac{v}{R}\right)^2 = \frac{3}{4}mv^2 \]
Step 5: Substitute value.
\[ K_{\text{disc}} = \frac{3}{4}\times 4 = 3\,\text{J} \]
Step 6: Conclusion.
The total kinetic energy of the disc is \(3\,\text{J}\).