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

Classical Mechanics - Rotational Motion

Seven pennies are arranged in a hexagonal, planar pattern so as to touch each neighbor, as shown in the figure. Each penny is a uniform disk of mass m and radius r. What is the moment of inertia of the system of seven pennies about an axis that passes through the center of the central penny and is normal to the plane of the pennies?

A. (7/2) mr²
B. (13/2) mr²
C. (29/2) mr²
D. (49/2) mr²
E. (55/2) mr²

(GR0177 #25)

Solution:

Parallel axis theorem: I = ml² + I_CM

Moment inertia of each penny (a uniform disk): I = ½ mr²

with l = 2r,

I = m(2r)² + ½ mr² = (9/2) mr²

I_total =  I_{6 outer pennies} + I_{1 central penny}

I_total = 6 × (9/2) mr²  + ½ mr² = (55/2) mr²

Answer: E