Question

A uniform bar of length l is hinged at one end is released from horizontal position so as to fall under gravity.The initial angular acceleration of the rod will be

Answer 1

Initial angular acceleration of the rod is  $\frac{3g}{2l}$

Step-by-step explanation:

The moment of inertia of a uniform rod about an axis via its end and perpendicular to its length, $I = \frac{ml^2}{3}$

where,

m denotes the mass of a rod and

l denotes the length of a rod

Torque applied at the center of the rod, $t = I\alpha$

where,

I refers to the moment of inertia and

$\alpha$  refers to angular acceleration

Torque on the rod= Moment of inertia of the rod about point P,

As we know, $t = I\alpha$

$t = mg\frac{l}{2}$

$I\alpha =mg\frac{l}{2}$

Put the value of I

$\frac{ml^2}{3}\alpha =mg\frac{l}{2}$

$\alpha =\frac{3g}{2l}$

Thus, initial angular acceleration of the rod is  $\frac{3g}{2l}$