Question 334: Centre of Mass and Rotational Motion

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 figure). There is negligible friction at the
pivot. The free end is held vertically above the pivot and then released. The
angular acceleration of the rod when it makes an angle $	heta$ with the vertical is
![](https://app-content.cdn.examgoal.net/fly/@width/image/1l7ym8h4g/7880f427-9eac-4dd7-b039-8bc9a1ea1ebd/bc3b7b00-3284-11ed-8893-19b23ee4c66d/file-1l7ym8h4h.png?format=png)

Accepted Answer

Forces acting on the rod are shown below.
Torque about pivot point O due to force N
x
and N
y
are zero.
Mgcos$	heta $ is passing through O, so torque will be zero due to this force.
So torque about the point O is
$	au  = Mg\sin 	heta  	imes {l \over 2}$
We know, $	au  = I\alpha $
$	herefore$ $I\alpha  = Mg\sin 	heta  	imes {l \over 2}$
Moment of inertia of rod about point O, $I$ = ${{M{l^2}} \over 3}$
$	herefore$ ${{M{l^2}} \over 3} 	imes \alpha  = Mg\sin 	heta  	imes {l \over 2}$
$ \Rightarrow \alpha  = {3 \over 2}{g \over l}\sin 	heta $
![](https://imagex.cdn.examgoal.net/INxxN2Rul5s4JpZ8P/3MhfX4bpIwQMKSggWi7Y9ysIYVXh6/6fb0NHVAbSAh7UsagJjT55/image.png)

Answer

${{2g} \over {3l}}\cos 	heta $

Answer

${{3g} \over {2l}}\sin 	heta $

Answer

${{2g} \over {3l}}\sin 	heta $

Answer

${{3g} \over {3l}}\sin 	heta $

Centre of Mass and Rotational Motion

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