Question

The uniform rod of mass $20\, kg$ and length of $1.6 \,m$ 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

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

Answer 1

Answer: (A)

Initial velocity of each point on the rod is zero so angular velocity of rod is zero
$\text { Torque about } O$ $\,\,\,\, T = I \alpha$
$ 20 g (0.8)=\frac{m \ell^{2}}{3} \alpha$
$\Rightarrow 20g (0.8)=\frac{20(1.6)^{2}}{3} \alpha$
$ \Rightarrow \frac{3 g}{3.2}=\alpha=\text { angular acceleration} $