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A uniform rod of mass M and length L is ...


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

A

g/6

B

g/2

C

g/4

D

2g/3

Text Solution

Verified by Experts

The correct Answer is:
C

Torque about P
`Mg((l)/(6))=(Ml^(2))/(9)alpha`
`implies=(3)/(2)(g)/(l)`
`impliesa_(t)=alpha(l)/(6)=(g)/(4)`
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