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: Use angular momentum form:
KE $= \frac{L^2}{2I}$, but here it is easier to use $\omega$ directly.

Step 2: Plug numbers:
Let $MR^2 = 1$. Then $I_A = 0.5$ and $I_B = 1.5$. Equal KE gives $0.5\omega_A^2 = 1.5\omega_B^2$.

Step 3: Solve:
$\omega_B^2 = \frac{\omega_A^2}{3}$, so $\frac{\omega_B}{\omega_A} = \frac{1}{\sqrt3}$.

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