Question

The total energy of a rolling ring of mass \( M \), velocity \( V \), and radius \( R \) is -

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

Hint:

For a ring rolling without slipping, both translational and rotational energies are equal, so total energy is \( MV^2 \).

Answer 1

The Correct Option is C

For a rolling ring, the total mechanical energy is the sum of translational and rotational kinetic energies.
Moment of inertia of a ring about its center is \( I = MR^2 \), and angular velocity \( \omega = \frac{V}{R} \).
\[ \text{Translational K.
E.
} = \frac{1}{2}MV^2, \quad \text{Rotational K.
E.
} = \frac{1}{2}I\omega^2 = \frac{1}{2}MR^2 \cdot \left(\frac{V}{R}\right)^2 = \frac{1}{2}MV^2 \] \[ \text{Total energy} = \frac{1}{2}MV^2 + \frac{1}{2}MV^2 = MV^2 \]