Q. A rigid massless rod of length 3l has two masses attached at each end as shown in the figure. The rod is pivoted at point P on the horizontal axis (see figure). When released from initial horizontal position, its instantaneous angular acceleration will be :

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Solution:

Applying torque equation about point P.
$2M_{0} \left(2l\right)-5 M_{0} gl = I\alpha $
$ I = 2M_{0}\left(2l\right)^{2} + 5M_{0} l^{2} = 13 M_{0} l^{2}d$
$ \therefore \alpha = - \frac{M_{0}g \ell}{13M_{0} \ell^{2}} \Rightarrow \alpha = - \frac{g}{13\ell} $
$ \therefore \alpha = \frac{g}{13\ell} $ anticlockwise

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