In the figure below, a disk (radius R = 1.0 m, mass = 2.0 kg) is suspended from a pivot a distance d = 0.25 m above its center of mass. For a circular disk, . The angular frequency (in rad/s) for small oscillations is approximately
A) 4.2
B) 2.1
C) 1.5
D) 1.0
E) 3.8

Question

In the figure below, a disk (radius R = 1.0 m, mass = 2.0 kg) is suspended from a pivot a distance d = 0.25 m above its center of mass. For a circular disk,  . The angular frequency (in rad/s) for small oscillations is approximately 
A) 4.2 
B) 2.1 
C) 1.5 
D) 1.0 
E) 3.8

Quizplus · Accepted Answer

The angular frequency for small oscillations of a simple pendulum is given by:
ω = √(g/L)
where g is the acceleration due to gravity and L is the length of the pendulum.
For a physical pendulum, the angular frequency is given by:
ω = √(mgh/I)
where m is the mass of the object, h is the distance between the pivot point and the center of mass, and I is the moment of inertia of the object about the pivot point.
Using the given values, we have:
m = 2.0 kg
h = 0.25 m
I = (1/2)mr^2 = (1/2)(2.0 kg)(1.0 m)^2 = 1.0 kg·m^2
g = 9.81 m/s^2
Substituting these values into the equation above, we get:
ω = √(mgh/I) = √[(2.0 kg)(9.81 m/s^2)(0.25 m)/1.0 kg·m^2] ≈ 2.1 rad/s
Therefore, the best choice is B.

In the Figure Below, a Disk (Radius R = 1