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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. When released from the initial horizontal position, its instantaneous angular acceleration will be:
Moment of inertia about pivot $$P$$: $$I = \sum m_i r_i^2$$
$$I = (5M_0)(l)^2 + (2M_0)(2l)^2 = 5M_0l^2 + 8M_0l^2 = 13M_0l^2$$
$$\tau_{\text{net}} = \sum \tau_i = \tau_1 - \tau_2$$
$$\tau_{\text{net}} = (5M_0g)(l) - (2M_0g)(2l) = 5M_0gl - 4M_0gl = M_0gl$$
$$\tau_{\text{net}} = I\alpha$$
$$M_0gl = (13M_0l^2)\alpha \implies \alpha = \frac{M_0gl}{13M_0l^2} = \frac{g}{13l}$$
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