Question

A rod PQ of mass M and length L is hinged at end P. The rod is kept horizontal by a massless string tied to point Q as shown in the figure. When the string is cut, the initial angular acceleration of the rod is

Answer 1

Step 1: Given data

1. The length of the rod is $L$
2. The mass of the rod is $M$

Diagram

Step 2: Formula Used

1. Torque is the product of force and distance. Torque is defined as, $\overrightarrow{\tau }=\overrightarrow{S}\times \overrightarrow{F}$, where, F is the force and S is the distance.
2. For a body rotating with acceleration $\alpha$, then the torque exerted on the body is $\tau =I\alpha$, where, I is the moment of inertia of the body.
3. The moment of inertia of a rod about an axis passing through the one end is $\frac{M{L}^2}{12}$ where M= mass of rod, L= length of rod

Step 3: Calculations

Finding the angular acceleration

When the string is cut, due to gravity a torque will be acted on the rod.

This torque is on the rod is $\overrightarrow{\tau }=\overrightarrow{L}\times \overrightarrow{F}$,

We know due to rotation torque is $\tau =I\alpha$. considering the balancing torque at point Q due to angular acceleration.

So,

$$\begin{gathered} FL=I\alpha \\ or\alpha =\frac{FL}{I}...............\left(1\right) \end{gathered}$$

We know, that the moment of inertia of a rod about an axis passing through the one end is $\frac{M{L}^2}{3}$.

From equation (1)

$$\begin{gathered} \alpha =\frac{FL}{I}=\frac{Mg{\displaystyle \frac{L}{2}}}{{\displaystyle \frac{M{L}^2}{12}}}=\frac{3g}{2L} \\ or\alpha =\frac{3g}{2L}. \end{gathered}$$

Therefore, the initial angular acceleration of the rod is $\frac{3g}{2L}$.