Q 11-06-206JEE MainJEE Main 2019 (10 Jan, Shift 2)Medium
A rigid massless rod of length $3l$ has two masses attached at its ends: $5M_0$ at one end and $2M_0$ at the other. The rod is pivoted at a point $P$ on the horizontal axis, at a distance $l$ from the $5M_0$ end. When released from the initial horizontal position, its instantaneous angular acceleration will be
Answer: (D) $\dfrac{g}{13l}$
Torques about $P$: $5M_0g\cdot l$ one way and $2M_0g\cdot2l$ the other, so the net torque is
$$\tau = 5M_0gl - 4M_0gl = M_0gl$$
$$I = 5M_0l^2 + 2M_0(2l)^2 = 13M_0l^2$$
$$\alpha = \frac{\tau}{I} = \frac{g}{13l}$$
Solution by Sreeraj P, M.Sc Physics