Question

Two identical rings A and B of same mass and radius are revolving, ring A arounds its own diameter and ring B about tangential axis in its own plane. Both the rings A and B have same rotational kinetic energy. The ratio of the angular velocity of ring B to that of ring A is

• A. \(1:3\)
• B. \(1:\sqrt{3}\)
• C. \(3:2\)
• D. \(\sqrt{3}:2\)

Hint:

Find the moment of inertia of a ring about a diameter and about an in-plane tangent.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Rotational KE is \(\frac12I\omega^2\). With equal KE, \(I_A\omega_A^2 = I_B\omega_B^2\), so \(\frac{\omega_B}{\omega_A} = \sqrt{\frac{I_A}{I_B}}\).

Step 2: Moments of inertia:
Ring A about its diameter: \(I_A = \frac12MR^2\).
Ring B about a tangent in its plane: use the parallel axis theorem, \(I_B = \frac12MR^2 + MR^2 = \frac32MR^2\).

Step 3: Ratio:
\[ \frac{\omega_B}{\omega_A} = \sqrt{\frac{\frac12MR^2}{\frac32MR^2}} = \sqrt{\frac13} = \frac{1}{\sqrt3} \]
So \(\omega_B : \omega_A = 1 : \sqrt3\).

Step 4: Why the other options are wrong.
The ratio 1:3 comes from forgetting the square root. Options (C) and (D) involve 2, as if a different moment of inertia such as \(MR^2\) was used for one of the rings.

Final Answer:
The ratio is \(1 : \sqrt3\), option (B). \[ \boxed{1:\sqrt{3}} \]