Rotational motion question

Question

A long rod of length L is pivoted about its left end. It is released from an angle 𝜃 above the horizontal. What is the magnitude of the angular acceleration of the rod about the pivot when the rod is released?

Answer 1

Let

g = 9.81 m/s² be gravitational acceleration,

m be mass of the rod (unknown),

I be the moment of inertia of the rod around the pivot,

τ be the torque around the pivot caused by the weight of the rod,

α be the angular acceleration to be computed.

You can solve this problem by applying Newton's Second Law for circular motion, which is

α = τ/I

For that we need to compute τ and I.

For that we need to know the distribution of the mass of the rod.

In the absence of any information, we assume that the mass of the rod is distributed uniformly.

Under than assumption, the rod center of mass is in its center, that is, distance L/2 from the pivot.

Therefore the torque

τ = mgLcos(θ)/2

Also under the assumption of uniform mass

I = mL²/3.

Then

α = τ/I = (mgLcos(θ)/2)/(mL²/3) = 3gcos(θ)/2L