Question

A uniform rod of length $l=1m$ is kept as shown in the figure. $H$ is a horizontal smooth surface and $W$ is a vertical smooth well. The rod is release from this position. What is the angular angular acceleration of the rod just after the released?

• A. $\frac{6g\cos \theta }{l}$
• B. $\frac{3g\cos \theta }{l}$
• C. $6g\cos \theta$
• D. $\frac{2g\cos \theta }{l}$

Answer 1

The correct option is A $\frac{6g\cos \theta }{l}$

By balancing forces along $H$ and $W$.

$N_1=mg$

$N_2=0$

Torque about COM

$\frac{l}{2}\times N_1\cos \theta \frac{l}{2}\times mg.\frac{3}{5}=\frac{3mgl}{10}$

About COM

$\frac{l}{12}\times m{l}^2\therefore \alpha =\frac{z}{I}=\frac{3mgl}{10m{l}^2}\times 12=\frac{36g}{10l}=35.28$