Question

A uniform rod $AB$ of mass $M$ and length $L$ is hinged at end $A$. It is released from rest at a horizontal position. The linear acceleration of the centre of mass immediately after release is

• A. g
• B. $\frac{g}{2}$
• C. $\frac{g}{4}$
• D. $\frac{3g}{4}$

Answer 1

The correct option is D $\frac{3g}{4}$


Torque about $A$ immediately after release
$\tau =Mg\times \frac{L}{2}\Rightarrow I\alpha =\frac{MgL}{2}$
or $\alpha =\frac{MgL}{2I}=\frac{3g}{2L}(\because I=\frac{M{L}^2}{3})$

Therefore, linear acceleration of the centre of mass, immediately after release is
$a_{CM}=\frac{L}{2}\times \alpha =\frac{L}{2}\times \frac{3g}{2L}=\frac{3g}{4}$