Question

A thin disc and a thin ring, both have mass M and radius R. Both rotate about axis through their center of mass and are perpendicular to their surfaces at the same angular velocity. Which of the following is true?

• A. The ring has higher kinetic energy
• B. The disc has higher kinetic energy
• C. The ring and the disc have the same kinetic energy
• D. Kinetic energies of both the bodies are zero since they are not in linear motion

Answer 1

Answer: (A)

The ring has higher kinetic energy

Answer 2

The rotational kinetic energy is $K = \frac{1}{2} I \omega^2$. For a thin disc, $I_{disc} = \frac{1}{2} M R^2$. For a thin ring, $I_{ring} = M R^2$. Since both have the same $M$, $R$, and $\omega$:

$K_{disc} = \frac{1}{2} \left( \frac{1}{2} M R^2 \right) \omega^2 = \frac{1}{4} M R^2 \omega^2$

$K_{ring} = \frac{1}{2} \left( M R^2 \right) \omega^2 = \frac{1}{2} M R^2 \omega^2$

Comparing the two, $K_{ring} = 2 \times K_{disc}$, which means the ring has higher kinetic energy.