Question

A rod of length l is standing vertically frictionless surface. It is disturbed slightly from this position. Let $\omega$ and $\alpha$ be the angular speed and angular acceleration of the rod, when the rod turns through an angle $\theta$ with the vertical, then find the value of acceleration of centre of mass of the rod.

Answer 1

Suppose mass of rod be M, and at a certain time, it makes angle $\theta$ with the vertical.

The only force acting on the rod is its own weight, which provides a torque about point O,

$\tau =Mg\frac{L}{2}\sin \theta$

Moment of inertia of rod about I is:

$I_C=\frac{M{L}^2}{3}$

During rotation, we have:-

$\tau =I\alpha$, and for point O:

$Mg\frac{L}{2}\sin \theta =\frac{M{L}^2}{3}\alpha$

$\Rightarrow \alpha =\frac{3g\sin \theta }{2L}$

Hence centre of mass has an acceleration:

$a_{CM}=R\alpha$, where $R=\frac{L}{2}$ gives $a_{CM}=\frac{3}{4}g\sin \theta$.