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 of small oscillations for a physical pendulum with a uniform rod is given by:
ω = √(mgd/I)

where m is the mass of the rod, g is the acceleration due to gravity, d is the distance between the pivot and the center of mass, and I is the moment of inertia of the rod about its center of mass.

Using the parallel axis theorem to calculate I:
I = (1/12) ml^2 + md^2

where l is the length of the rod.

Substituting the given values, we get:
I = (1/12)(2.0 kg)(1.0 m)^2 + (2.0 kg)(0.25 m)^2 = 0.34375 kg m^2

Substituting into the formula for ω and solving, we get:
ω = √[(2.0 kg)(9.81 m/s^2)(0.25 m)/0.34375 kg m^2] ≈ 4.1 rad/s

Therefore, the best choice is D, 4.1.

The Figure Shows a Uniform Rod (Length L = 1