Question

If torques of equal magnitudes are applied to a hollow cylinder and a solid sphere both having the same mass and radius. The cylinder is free to rotate about its standard axis of symmetry and the sphere is free to rotate about an axis passing through its center. Which of the two will acquire a greater angular speed after a given time?

• A. \( \omega_1 > \omega_2 \)
• B. \( \omega_1 = \omega_2 \)
• C. \( \omega_2 > \omega_1 \)
• D. None of these

Hint:

For the same torque, an object with a smaller moment of inertia will acquire a greater angular speed.

Answer 1

The Correct Option is C

The moment of inertia for a solid sphere is \( I_{\text{sphere}} = \frac{2}{5} m r^2 \) and for a hollow cylinder, it is \( I_{\text{cylinder}} = m r^2 \).
When torques of equal magnitudes are applied, the angular acceleration \( \alpha \) is given by: \[ \alpha = \frac{\tau}{I} \] Since the cylinder has a larger moment of inertia compared to the solid sphere, the angular acceleration of the solid sphere will be greater, leading to a greater angular speed \( \omega_2 \) for the sphere.
Thus, the correct answer is (c).