Classical Mechanics - Moment Inertia

The period of physical pendulum is 2π√(I/mgd), where I is the moment of inertia about the pivot point and d is the distance from the pivot to the center of mass. A circular hoop hangs from a nail on a barn wall. The mass of the hoop is 3 kilograms and its radius is 20 centimeters. If it is displaced slightly by a passing breeze, what is the period of the resulting oscillations?

A. 0.63 s
B. 1.0 s
C. 1.3 s
D. 1.8 s
E. 2.1 s

(GR9677 #21)

Solution:

T = 2π√(I/mgd)
m = 3 kg
r = 20 cm = 0.2 m
In this case d = r

To find total I:
Parallel axis theorem: I = ml² + I_CM
I_CM  = I_loop = mr² 

In this case l = r
→ I = mr² + mr² = 2mr²

T = 2π√(2mr² / mgr) 
= 2π√(2r/g) 
= 2π√[2(0.2)/(10)] 
= 2π(0.2)
= 1.25 s

Answer: C