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$\omega$ .Find the maximum $\omega$ with which the ring can be rotated if the strings can withstand a maximum tension of $3T_0/2$

Answer 1

In equilibrium,
$2T_0=mg$
or $T_0=\frac{mg}{2}$ ...........(i)
Magnetic moment,$M=iA=\left(\frac{\omega }{2\pi }Q\right)(\pi R^2)$
Let $T_{1}and{T}_2$ be the tensions in the two strings when magnetic
field is switched on $(T_1>T2)$.
For translational equilibrium,
$T_1+T_2=mg.........(ii)$
For rotational equilibrium,
$(T_1-T_2)\frac{D}{2}=\tau =\frac{\omega BQ{R}^2}{2}$
or $T_1-T_2=\frac{\omega BQ{R}^2}{2}$ ............(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, can be $\frac{3T_0}{2}$we have
$\frac{3T_0}{2}=T_0+\frac{{\omega }_{max}BQ{R}^2}{2D}(\frac{mg}{2}=T_0)$
$\therefore {\omega }_{max}=\frac{D{T}_0}{BQ{R}^2}$