Question

A spherical bob of mass m and charge q suspended from a string of length l rotates about a fixed charge identical to that of the bob (Fig.). The angle between the string and the vertical is q. Find the angular velocity of uniform rotation of the bob. It is given that $\frac{kq^2}{ml^2}=g$

• A. $\sqrt{\frac{g}{l}(sec\theta - cosec^3\theta)}$
• B. $\sqrt{\frac{g}{l}sec\theta}$
• C. $\sqrt{\frac{g}{l}sec\theta}$
• D. $\sqrt{\frac{g}{l}}$

Answer 1

Answer: (A)

$\sqrt{\frac{g}{l}(sec\theta - cosec^3\theta)}$

Answer 2

To find the angular velocity of the bob, we analyze the forces acting on it: tension in the string, weight, and electrostatic force.

1. Vertical Equilibrium: The vertical component of tension balances the weight:

   $$T\cos\theta = mg \implies T = \frac{mg}{\cos\theta}$$

2. Horizontal Force Balance: The horizontal component of tension and the electrostatic force provide the centripetal force:

   $$T\sin\theta - F_e = m\omega^2 (l\sin\theta)$$

3. Electrostatic Force: Given $F_e = \frac{kq^2}{(l\sin\theta)^2}$ and $\frac{kq^2}{ml^2} = g$, we get:

   $$F_e = \frac{mg}{\sin^2\theta}$$

4. Substitute and Solve: Substituting $T$ and $F_e$ into the horizontal force balance equation:

   $$mg\tan\theta - \frac{mg}{\sin^2\theta} = m\omega^2 l\sin\theta$$

   Dividing by $ml\sin\theta$:

   $$\omega^2 = \frac{g}{l\sin\theta} \left(\tan\theta - \frac{1}{\sin^2\theta}\right)$$

5. Simplify: Expressing in terms of $\sec\theta$ and $\csc\theta$:

   $$\omega^2 = \frac{g}{l} (\sec\theta - \csc^3\theta)$$

   Thus,

   $$\omega = \sqrt{\frac{g}{l} (\sec\theta - \csc^3\theta)}$$