Question

A grandfather clock has a pendulum that consists of a thin brass disk of radius $$r=15.00cm$$ and mass $$1.000kg$$ that is attached to a long thin rod of negligible mass. The pendulum swings freely about an axis perpendicular to the rod and through the end of the rod opposite the disk, as shown in above figure. If the pendulum is to have a period of $$2.000s$$ for small oscillations at a place where $$g=9.800m/s^{2}$$, what must be the rod length $$L$$ to the nearest tenth of a millimeter?

Answer 1

The period formula, Eq. $$T=2\pi \sqrt{I/mgh}$$, requires knowing the distance $$h$$ from the axis of rotation and the center of mass of the system. We also need the rotational inertia $$I$$ about the axis of rotation. From the figure, we see $$h = L + R$$ where $$R = 0.15m$$. Using the

parallel-axis theorem, we find

$$I=\dfrac{1}{2}MR^{2}+M\left ( L+R \right )^{2}$$.

where $$M =1.0 kg$$. Thus, $$T = 2.0s$$, leads to,

$$2.0=2\pi\sqrt{\dfrac{\dfrac{1}{2}MR^{2}+M\left ( L+R \right )^{2}}{Mg\left ( L+R \right )}}$$.

which leads to $$L = 0.8315 m$$.