Question

A uniform rod mass m and length l is attached to smooth hinge at end A and to a string at end B as shown in figure. It is at rest . The angular acceleration of the rod just after the string is cut is:

Answer 1

When the string is cut, let angular acceleration of rod be $\alpha$.

The Torque due to in own weight (mg) provides the needed angular acceleration $\left(\alpha \right)$, given by :-

$\tau =I_A\alpha \to$ (1)

Where,

$\tau =$ Torque on rod

$=mg\frac{L}{2}cos\theta$ [As Torque = $\overrightarrow{r}\times \overrightarrow{F}$ $=r\perp .F$

and, $I_A$ = moment of Inertia about axis through A (end of rod)

$I_A=\frac{M{L}^2}{3}$

Thus, from (1),

$\Rightarrow mg\frac{L}{2}cos\theta =\frac{L}{3}\alpha$

$\Rightarrow \alpha =\frac{3}{2}gcos\theta$