Question

A uniform rod of length 1.82 m is attached to a frictionless pivot at one end. It is released from rest from an angle θ = 29.8 ° above the horizontal.

Find the magnitude of the initial angular acceleration of the rod.

Find the magnitude of the initial linear acceleration of a point at the center of the rod.

Answer 1

The basic equation to be applied is:
Torque = (Moment of inertia) times (Angular acceleration).

The torque results from the weight of the rod, but only the component
perpendicular to the rod will exert a torque. We can treat the rod's
center of mass as the "lever arm". So the torque is
Tau = M g (L/2) cos(25 deg)

The moment of inertia of a uniform rod rotating about one of its ends is
(1/3) ML^2.

Thus, the initial angular acceleration is
[ M g (L/2) cos(29.8 deg) ] / [ (1/3) ML^2 ]
= [ 3g cos (29.8 deg) ] / 2L
= 3 (9.8 m/s^2) (0.8677) / (3.64 m) = 7.0083 (radians)/ s^2...........Ans.

To find the linear acceleration of the rod's center of mass,
note that the radius from the pivot to the center of mass is 0.91 m;
thus the initial acceleration of the center of mass is
(7.0083) (0.91 m) /s^2 = 6.37753 m/s^2..............Ans.