A uniform rod is 2.0 m long. The rod is pivoted about a horizontal, frictionless pin through one end. The rod is released from rest at an angle of 30° above the horizontal. What is the angular acceleration of the rod at the instant it is released? A) 4.7 rad/s² B) 6.9 rad/s² C) 6.4 rad/s² D) 5.6 rad/s² E) 4.2 rad/s²

Question

Quizplus · Accepted Answer

The angular acceleration of the rod can be found using the equation for torque:
τ = Iα

Where τ is the torque, I is the moment of inertia, and α is the angular acceleration.

Since the rod is pivoted about one end, the moment of inertia can be found using the formula for a thin rod rotating about one end:

I = (1/3)ML^2

Where M is the mass of the rod and L is its length.

Substituting in the given values, we get:

I = (1/3)(m)(L^2) = (1/3)(m)(2^2) = (4/3)m

The torque acting on the rod is due to the gravitational force, which creates a torque of:

τ = mgL/2 sin θ

Where θ is the angle the rod makes with the horizontal.

Substituting in the given values, we get:

τ = (mgL/2)sin30° = (mgL/4)

So we have:

τ = Iα
(mgL/4) = (4/3)mα
α = (3/16)gL sin θ

Substituting in the given values, we get:

α = (3/16)(9.81 m/s^2)(2 m) sin 30° = 6.35 rad/s^2

Rounding to two significant figures gets us answer choice C: 6.4 rad/s^2.

A Uniform Rod Is 2