The figure shows a uniform rod (length L = 1.0 m,mass = 2.0 kg)suspended from a pivot a distance d = 0.25 m above its center of mass.The angular frequency (in rad/s)for small oscillations is approximately
A)1.0
B)2.5
C)1.5
D)4.1
E)3.5

Question

The figure shows a uniform rod (length L = 1.0 m,mass = 2.0 kg)suspended from a pivot a distance d = 0.25 m above its center of mass.The angular frequency (in rad/s)for small oscillations is approximately 
A)1.0 
B)2.5 
C)1.5 
D)4.1 
E)3.5

Quizplus · Accepted Answer

The angular frequency for small oscillations of a simple pendulum is given by:

ω = (g/L)^(1/2)

Where g is the acceleration due to gravity and L is the length of the pendulum. However, the given system is not a simple pendulum, but a physical pendulum. For a physical pendulum, the angular frequency for small oscillations is given by:

ω = (Mgd/I)^(1/2)

Where M is the mass of the rod, d is the distance between the pivot and the center of mass, and I is the moment of inertia about the pivot.

For a uniform rod rotating about its center of mass, the moment of inertia is given by:

I = (1/12)ML^2

Substituting the given values, we get:

ω = (2*9.81*0.25/[(1/12)*2*1^2])^(1/2) ≈ 4.1 rad/s

Therefore, the best choice is D.

The Figure Shows a Uniform Rod (Length L = 1