Question

A uniform bar AB of mass $m$ and a ball of the same mass are released from rest from the same horizontal position. The bar hinged at end A. There is gravity downwards. What is the distance of the point from point B that has the same acceleration as that of a ball, immediately after release.?

Answer 1

Hint: Torque is given by the equation $\tau =I\alpha$. Using this equation find the value of $\alpha$. Then use the equation $a=\alpha r$ to find the distance

Complete step by step answer:
Step 1:
Torque is given by the equation $\tau =I\alpha$. From this equation we get,
$\alpha ={\displaystyle \frac{\tau }{I}}$ …………………………...(1)
We know that moment of inertia of the rod is given as $I={\displaystyle \frac{1}{3}}m{L}^2$.
Torque can also be found by the equation $\tau =F\times r$. Here $F$ is the force due to gravity that is,$F=mg$. It acts at the centre of gravity which is at the midpoint of the rod. Hence torque will be $\tau =mg\times {\displaystyle \frac{L}{2}}$
Substituting these values in equation (1), we get
$$\begin{gathered} \Rightarrow \alpha ={\displaystyle \frac{\tau }{I}} \\ \Rightarrow \alpha ={\displaystyle \frac{mg\times {\displaystyle \frac{L}{2}}}{{\displaystyle \frac{1}{3}}m{L}^2}} \\ \Rightarrow \alpha ={\displaystyle \frac{3g}{2L}} \end{gathered}$$
Step 2:
The equation connecting the linear acceleration $a$ and angular acceleration $\alpha$ is
 $a=\alpha r$ ……………….. (2)
We want to find the point at which linear acceleration is the acceleration due to gravity since the acceleration of the ball is the acceleration due to gravity. Let $r$ be the distance from the hinge point at which $a$ becomes equal to $g$. Thus equation (2) becomes $g=\alpha r$.
Now rearrange this equation to find $r$. Then we get $r={\displaystyle \frac{g}{\alpha }}$
We already got the value $\alpha ={\displaystyle \frac{3g}{2L}}$ in step 1. Substitute this in the last equation. Then we get,
$$\begin{gathered} \Rightarrow r={\displaystyle \frac{g}{{\displaystyle \frac{3g}{2L}}}} \\ \Rightarrow r={\displaystyle \frac{2L}{3}} \end{gathered}$$
This is the distance from the hinge. But we need to find the distance from B. Therefore the distance from B will be $L-{\displaystyle \frac{2L}{3}}$. Hence the distance of the point from point B that has the same acceleration like that of a ball, immediately after release is ${\displaystyle \frac{L}{3}}$.

Note:
Torque is the product of force and the perpendicular distance of the line of action of the force from the axis of rotation. Here hinge acts as the axis of rotation and the line of action of the force is at the centre of gravity. For a uniform rod, the centre of gravity will be at its midpoint. Hence, we should take the perpendicular distance as half the total length of the rod.