Question: The Silently Ringing Bell. A large, 34.0-kg bell is hung from a wooden beam so it can swing back and forth with negligible friction. The bell’s center of mass is 0.60 m below the pivot. The bell’s moment of inertia about an axis at the pivot is \({\bf{18}}.{\bf{0}}{\rm{ }}{\bf{kg}} \cdot {{\bf{m}}^2}\). The clapper is a small, 1.8-kg mass attached to one end of a slender rod of length L and negligible mass. The other end of the rod is attached to the inside of the bell; the rod can swing freely about the same axis as the bell. What should be the length L of the clapper rod for the bell to ring silently—that is, for the period of oscillation for the bell to equal that of the clapper?

\(\begin{array}{c}{\rm{mass}}\;{\rm{of}}\;{\rm{the}}\;{\rm{bell}} = 34\;{\rm{kg }}\\{\rm{distnace}}\;{\rm{of}}\;{\rm{the}}\;{\rm{center}}\;{\rm{of}}\;{\rm{mass}}\;{\rm{from}}\;{\rm{the}}\;{\rm{pivot}};\;d = 0.6\;{\rm{m}}\\{\rm{moment}}\;{\rm{of}}\;{\rm{inertia}}\;{\rm{of}}\;{\rm{the}}\;{\rm{bell, }}\;{\rm{I}} = 18\;{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}\end{array}\)

If the motion of the body has no resistance because of the smoothness surface, then it leads to negligible of friction.

By considering the clapper as a simple pendulum its period of oscillation is,

\({T_c} = 2\pi \sqrt {\frac{L}{g}} \)

The bell oscillates as a physical pendulum and its period of oscillation is,

\({T_b} = 2\pi \sqrt {\frac{I}{{{m_b}gd}}} \)

The bell must ring silently, therefore the period of oscillation of the bell equals to the period of oscillation of the clapper

\(\begin{array}{c}{T_c} = {T_b}\\2\pi \sqrt {\frac{L}{g}} = 2\pi \sqrt {\frac{I}{{{m_b}gd}}} \\\sqrt {\frac{L}{g}} = \sqrt {\frac{I}{{{m_b}gd}}} \\\frac{L}{g} = \frac{I}{{{m_b}gd}}\\L = \frac{I}{{{m_b}d}}\\ = \frac{{18}}{{34 \times 0.6}}\\ = 0.88\;{\rm{m}}\end{array}\)

Hence the length of the clapper rod is 0.88 m.