Question

A slender uniform rod of mass M and length l is pivoted at one end so that it can rotate in a vertical plane (see the figure).

Answer 1

Solution :

Taking torque about pivot T=lα

mgsinθl2=ml23α

α=3g2lsinθ

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Answer 2

Answer:

$\alpha = 3g \2l \sin(?)$

Explanation:

T=F*r

$t = i \alpha$

$mg sin() \times l \2$

$= mt ^2 \3 \times \alpha$

$\alpha = 3g \sin( ) \2l$