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

For small oscillations about the equilibrium position, the restoring torque on the hoop can be approximated as:

τ = -mgdθ

where m is the mass of the hoop, g is the acceleration due to gravity, d is the distance from the pivot to the center of mass of the hoop, and θ is the angular displacement from the equilibrium position.

However, the moment of inertia of the hoop about its center of mass is given by:

I = MR^2

where R is the radius of the hoop.

Using the parallel axis theorem, the moment of inertia of the hoop about the pivot point is given by:

I' = I + md^2

Substituting the expressions for I and d, we get:

I' = MR^2 + md^2 = M(R^2 + d^2)

The angular frequency of small oscillations is given by:

ω = √(τ/I')

Substituting the expressions for τ and I', we get:

ω = √(mgd/M(R^2 + d^2))

Substituting the given values, we get:

ω = √(2(9.81)(0.25)/(2(1^2 + 0.25^2))) ≈ 1.5 rad/s

Thus, the best choice is option C) 1.5.

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