Question

A disc of mass $ M $ and radius $ R $ is rolling without slipping with linear velocity $ v $. What is its total kinetic energy?

• A. \( \frac{1}{2} Mv^2 \)
• B. \( \frac{3}{4} Mv^2 \)
• C. \( \frac{1}{4} Mv^2 \)
• D. \( \frac{5}{6} Mv^2 \)

Hint:

For rolling objects, always add both translational and rotational kinetic energy. Use the relation \( v = \omega R \) when rolling without slipping. For a disc, the moment of inertia is \( \frac{1}{2}MR^2 \).

Answer 1

The Correct Option is B

When a disc rolls without slipping, it has two forms of kinetic energy:
1. Translational Kinetic Energy: \[ K_{\text{trans}} = \frac{1}{2} M v^2 \] 2. Rotational Kinetic Energy: \[ K_{\text{rot}} = \frac{1}{2} I \omega^2 \] For a solid disc, the moment of inertia about its center is: \[ I = \frac{1}{2} M R^2 \] Since the disc is rolling without slipping, we use the rolling condition: \[ v = \omega R \quad \Rightarrow \quad \omega = \frac{v}{R} \] Substitute into rotational energy: \[ K_{\text{rot}} = \frac{1}{2} \cdot \frac{1}{2} M R^2 \cdot \left( \frac{v}{R} \right)^2 = \frac{1}{4} M v^2 \] Now, total kinetic energy is the sum of both: \[ K_{\text{total}} = K_{\text{trans}} + K_{\text{rot}} = \frac{1}{2} M v^2 + \frac{1}{4} M v^2 = \frac{3}{4} M v^2 \]