Question

A charge particle of charge q and mass m is clamped rigidly at the circumference of a ring with mass m and radius R. Initially ring is in vertical plane resting on a sufficiently rough horizontal surface with charge q at the same horizontal level as that of the centre of the ring. There exist uniform horizontal electric fields as shown. At $t=0$ the system is let free. Given that $(qE=mg)$. Using the above information answer the following. The magnitude of the friction force at $t=0$ :
(Assume there is no sliding at point of contact at any moment of time during motion)

• A. $mg$
• B. $\sqrt{2}mg$
• C. $Zero$
• D. $\frac{mg}{2}$

Answer 1

Answer: (C)

$Zero$

Consider the force diagram shown above.

Let the point of contact with the ground be $A$.

The moment of inertia of the system about $A$ is $I_A=m{R}^2+m{R}^2+2m{R}^2=4m{R}^2$

Net torque at $A$ due to forces acting on the system is $\tau =qER+mgR=2mgR$

But, $\tau =I_A{\alpha }_A\Rightarrow 2mgR=4m{R}^2{\alpha }_A$

But, for the system to not slip, $a=R{\alpha }_A$

Thus, $a=g/2$

$2ma=qE+f\Rightarrow mg=mg+f\Rightarrow f=0$

Thus, static friction acting at $t=0$ is $0$