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 (in rad/s) for small oscillations is given by 
ω = (g/R)^(1/2)
where g is the acceleration due to gravity. 
In this case, since the disk is suspended from a pivot a distance d = 0.25 m above its center of mass, the effective radius of rotation is R' = R^2/d = 4 m. 
So, 
ω = (g/R')^(1/2)
plugging g = 9.8 m/s^2 and R' = 4 m, we get
ω = (9.8/4)^(1/2) = 2.2 rad/s
which is closest to choice B, 2.1 rad/s.

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