In the figure below, a hoop (radius R = 1.0 m, mass = 2.0 kg) having four spokes of negligible mass is suspended from a pivot a distance d = .25 m above its center of mass. The angular frequency (in rad/s) for small oscillations is approximately
A) 4.0
B) 2.5
C) 1.5
D) 1.0
E) 0.5

Question

In the figure below, a hoop (radius R = 1.0 m, mass = 2.0 kg) having four spokes of negligible mass is suspended from a pivot a distance d = .25 m above its center of mass. The angular frequency (in rad/s) for small oscillations is approximately 
A) 4.0 
B) 2.5 
C) 1.5 
D) 1.0 
E) 0.5

Quizplus · Accepted Answer

To find the angular frequency of small oscillations for the hoop, we use the equation:

ω = √(g*h/I)

where g is the acceleration due to gravity, h is the distance between the pivot and the center of mass, and I is the moment of inertia of the hoop about an axis passing through its center of mass.

To find I, we can use the parallel axis theorem:

I = MR^2 + 2MR^2 = 3MR^2

where M is the mass of the hoop.

Plugging in the given values, we get:

ω = √(9.8 m/s^2 * 0.25 m / 3 * 2.0 kg * 1.0 m^2) = 1.5 rad/s

Therefore, the best choice is C) 1.5.

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