Question

The uniform rod of mass $20kg$ and length $1.6m$ 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 as shown in the figure is:

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

Answer 1

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

Torque on weight $mg=I\propto$

$mg\frac{l}{2}=\frac{m{l}^2}{3}\propto$

$\propto =\frac{3g}{2l}=\frac{3g}{2\times 1.6}=\frac{15}{16}g$