Question

A uniform rod of mass M and length L is free to rotate about a frictionless pivot located L/3 from one end. The roc
1s released from rest incrementally away from being perfectly vertical, resulting in the rod rotating clockwise abou
the pivot. When the rod is horizontal, what is the magnitude of the tangential acceleration of its center of mass?​

Answer 1

Given: A uniform rod of mass M and length L is free to rotate about a friction less pivot located L/3 from one end.

To find: The magnitude of the tangential acceleration of its center of mass when the rod is horizontal.

Solution:

• Now we have given that the rod  is free to rotate about a friction less pivot located L/3 from one end.

• Also The rod 1st released from rest incrementally away from being perfectly vertical, resulting in the rod rotating clockwise about the pivot.

• So equating Torque about P, when the rod becomes horizontal is:

            M x g x (l/6) = ( M x l² / 9 ) x α

            α = 3 x g / 2 x l

• Now α(tangential) = α x l/6

            α(tangential) =  3 x g / 2 x l x l / 6

            α(tangential) = g / 4

Answer:

                 The magnitude of the tangential acceleration of its center of mass when the rod is horizontal is g / 4.