Question

A ring of radius R having uniformly distributed charge Q is mounted on a rod suspended by two identical strings. The tension in strings in equilibrium is $T_0$. Now a vertical magnetic field is switched on and ring is rotated at constant angular velocity $\omega$. If the maximum $\omega$ with which the ring can be rotated if the strings can withstand a maximum tension of $3T_0/2$ is $omeg{a}_{max}=\frac{x\times D{T}_0}{BQ{R}^2}$ Find x.

Answer 1

In equilibrium, $2T_0=mg$ or $T_0=\frac{mg}{2}$ (i)
Magnetic moment, $M=iA=\left(\frac{\omega }{2\pi }Q\right)\left(\pi R^2\right)$
$\tau =MBsin{90}^o=\frac{\omega BQ{R}^2}{2}$
Let $T_1$ and $T_2$ be the tensions in the two strings when magnetic field is switched on $(T_1>T_2)$

For translational equilibrium: $T_1+T_2=mg$ (ii)
For rotational equilibrium: $(T_1-T_2)\frac{D}{2}=\tau =\frac{\omega BQ{R}^2}{2}$
$T_1-T_2=\frac{\omega BQ{R}^2}{D}$ (iii)

Solving Eqs. (ii) and (iii) we have
$T_1=\frac{mg}{2}+\frac{\omega BQ{R}^2}{2D}$
As $T_1>T_2$ and maximum values of $T_1$ can be $3T_0/2$, we have
$\frac{3T_0}{2}=T_0+\frac{{\omega }_{max}BQ{R}^2}{2D}(\frac{mg}{2}=T_0)$
$\therefore {\omega }_{max}=\frac{D{T}_0}{BQ{R}^2}$