Question

A ring is hung on a nail. It can oscillate, without slipping or sliding
(i) in its plane with a time period \(T_1\) and,
(ii) back and forth in a direction perpendicular to its plane, with a period \(T_2\). The ratio \(\frac{T_1}{T_2}\) will be :

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

Hint:

Treat the ring as a physical pendulum, \(T = 2\pi\sqrt{\frac{I}{mgR}}\), with I taken about the nail axis (R = distance from the nail to the centre).

Answer 1

The Correct Option is B

Ring of mass m and radius R, pivot (nail) on the rim, so the centre of mass is at distance R from the pivot. For a physical pendulum \(T = 2\pi\sqrt{\frac{I}{mgR}}\).
(i) Oscillation in its plane: the axis is perpendicular to the plane through the rim: \(I_1 = mR^2 + mR^2 = 2mR^2\). \(T_1 = 2\pi\sqrt{\frac{2mR^2}{mgR}} = 2\pi\sqrt{\frac{2R}{g}}\).
(ii) Oscillation perpendicular to the plane: the axis is the tangent to the ring at the nail (in the plane): \(I_2 = \frac{1}{2}mR^2 + mR^2 = \frac{3}{2}mR^2\). \(T_2 = 2\pi\sqrt{\frac{3R}{2g}}\).
\(\frac{T_1}{T_2} = \sqrt{\frac{2}{3/2}} = \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}}\).
Answer: (2) \(\frac{2}{\sqrt{3}}\)