Q 11-06-039NEETJEE MainMedium
A uniform rod of length $L$, pivoted at one end, is held horizontal and then released. Its angular acceleration at the moment of release is
Answer: (A) $\dfrac{3g}{2L}$
Torque of gravity about the pivot: $\tau = mg\dfrac{L}{2}$. $I = \dfrac{mL^2}{3}$.
$$\alpha = \frac{\tau}{I} = \frac{mgL/2}{mL^2/3} = \frac{3g}{2L}$$
(The free end then starts with acceleration $\alpha L = 1.5g$, more than $g$.)
Solution by Sreeraj P, M.Sc Physics