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 centre 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 centre 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 simple pendulum is given by:

ω = √(g/L)

where g is the acceleration due to gravity and L is the length of the pendulum.

For this rod, we can calculate the moment of inertia about the pivot point as:

I = (1/12)ML^2 + (d^2)M

where M is the mass of the rod.

The torque about the pivot point due to a small angular displacement θ is:

τ = -Mgdθ

Applying Newton's second law for rotational motion, we get:

Iα = τ

where α is the angular acceleration.

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

α = (-3g/2L)θ

Comparing this with the equation for simple harmonic motion:

α = -ω^2θ

we can identify:

ω = √(3g/2L)

Substituting the given values, we get:

ω = √(3 x 9.81 / 2 x 1) ≈ 4.1 rad/s

Therefore, the best choice is D.

The Figure Shows a Uniform Rod (Length L = 1