Question

A uniform rod of mass 20 kg and length 1.6 m is pivoted at its end and swings freely in the vertical plane. Angular acceleration of rod just after the rod is released from rest in the horizontal position is:

• A. $\frac{15g}{16}$
• B. $\frac{17g}{16}$
• C. $\frac{16g}{15}$
• D. $\frac{g}{15}$

Answer 1

The correct option is A $\frac{15g}{16}$

Just after the rod is released,

Torque of weight = $I\alpha$

$mg\times \frac{l}{2}=\frac{m{l}^2}{3}\times \alpha$,

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

$\Rightarrow \alpha =\frac{3\times g}{2\times 1.6}$

$\Rightarrow \alpha =\frac{15g}{16}$