Solution to 1992 Problem 74

The period of a physical pendulum is
\begin{align*}T = 2 \pi\sqrt{\frac{I}{m g d}}\end{align*}
where I is the moment of inertia about the pivot point and d is the distance from the pivot to the center of mass. Let M be the mass of hoop Y and let R be the radius of hoop Y. Then

\begin{align*}I_X &= 2 \cdot 4M \cdot (4R)^2 = 128 M R^2 \\I_Y &= 2 M R^2 \\d_X &= 4 R \\d_Y &= R\end{align*}
\begin{align*}\frac{T_Y}{T_X} = \frac{\displaystyle  2 \pi\sqrt{\frac{I_Y}{m_Y g d_Y}}}{ \displaystyle 2 \pi\sqrt{\frac{I_X}{...
In other words,
\begin{align*}T_Y = \frac{T_X}{2} = \boxed{\frac{T}{2}}\end{align*}
Therefore, answer (B) is correct.

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