Consider a physical pendulum constructed by attaching a thin, massless rod to the center of a uniform disc of mass 500 g and radius 4.00 cm. The disc has a moment of inertia of ½ MR² about an axis perpendicular to the plane of the disc, through its center. With the rod hung from one end (the pivot), the plane of the disc (attached to the other end) is vertical. If the rod has a length of 50.0 cm, then the period of (small amplitude) oscillations of the pendulum will be A) 1.24 sec. B) 1.42 sec. C) 1.74 sec. D) 1.90 sec. E) 2.25 sec.

Question

Quizplus · Accepted Answer

The period of a physical pendulum is given by:
$$T=2\pi\sqrt{\frac{I}{mgh}}$$
where I is the moment of inertia about the pivot point, m is the mass, g is the acceleration due to gravity, and h is the distance from the pivot to the center of mass.
In this case, the pivot is at the end of the rod, so h = L/2 = 25.0 cm. The moment of inertia of the disc about its center is given as ½ MR², but since we want the moment of inertia about the pivot, we need to use the parallel axis theorem:
$$I_p=I_{cm}+md^2$$
where $I_{cm}=1/2MR^2$, $m$ is the mass of the disc, and $d$ is the distance from the center of mass to the pivot. Since the pivot is at the end of the rod and the disc is attached to the other end, the distance $d=L=50.0$ cm.
Substituting these values into the equation for the period, we get:
$$T=2\pi\sqrt{\frac{I_p}{mgh}}=2\pi\sqrt{\frac{(1/2)MR^2+ML^2}{mgh}}$$
$$=2\pi\sqrt{\frac{(1/2)(0.5	ext{ kg})(0.04	ext{ m})^2+(0.5	ext{ kg})(0.5	ext{ m})^2}{(0.5	ext{ kg})(9.81	ext{ m/s}^2)(0.25	ext{ m})}}$$
$$\approx 1.42	ext{ s}$$
Therefore, the best choice is B.

Consider a Physical Pendulum Constructed by Attaching a Thin, Massless