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 $3 T_0/2$

Answer 1

In equilibrium,
$ \, \, \, \, \, \, \, \, \, \, \, 2 T_0 = mg $
or $ \, \, \, \, \, \, \, \, \, \, \, T_0 = \frac{mg}{2} $ ...........(i)
Magnetic moment,$ M=iA = \bigg( \frac{\omega }{2 \pi } Q \bigg) (\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 BQR^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 BQR^2}{2D}$
As $T_1 > T_2$ and maximum values of T, can be $\frac{ 3 T_0}{2}$we have
$ \, \, \, \, \, \, \, \, \, \, \, \frac{3 T_0}{2} =T_0 + \frac{ {\omega}_{ max} BQR^2}{2D} \bigg( \frac{mg}{2 }= T_0 \bigg) $
$\therefore \, \, \, \, \, \, {\omega}_{max} = \frac{DT_0}{BQR^2}$