Question

A thin homogeneous stick of mass m and length L may rotate in the vertical plane around a horizontal axle pivoted at one end of the stick. A small ball of mass m and charge Q is attached to the opposite end of this stick. The whole system is placed in a constant horizontal electric field of magnitude $E=\frac{mg}{2Q}$. The stick is held horizontally at the beginning.
The acceleration of the small ball at the instance of releasing the stick is :

• A. $\frac{3g}{2}$
• B. $\frac{3g}{4}$
• C. $\frac{9g}{8}$
• D. None

Answer 1

The correct option is C $\frac{9g}{8}$

We do the torque balance
$\tau =I\alpha$
The force acting about the pivot are weight of the rod at a distance of l/2 and weight of the ball at a distance of l. Note the the electrostatic force wont have any moment about the pivot at the instant of the release of the rod. Thus we get
$(mgl/2+mgl)=\alpha \times (m{l}^2/3+m{l}^2)$
(here $m{l}^2/3$ is the moment of inertia of the rod about one end and $m{l}^2$ is the moment of inertia of the ball about the pivot).
Thus we get
$3/2mgl=\alpha \times 4/3m{l}^2\Rightarrow \alpha =9/8g/l$
$a=\alpha r=9/8g/l\times l=9/8g$