Question

The uniform rod of mass $20\text{kg}$ and length $1.6\text{m}$ is pivoted at one of its ends and is held in horizontal position, as shown in the diagram. What will be its initial angular acceleration, if it is just released from this position?

• A. $\frac{15g}{16}$
• B. $\frac{17g}{16}$
• C. $\frac{19g}{16}$
• D. $\frac{21g}{16}$

Answer 1

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

Torque about the pivoted end,

$\tau =I\alpha$

$\Rightarrow mg\left(l/2\right)=[\frac{m{l}^2}{12}+m(l/2{)}^2]\alpha$

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