Question

A ball of mass (m) 0.5 kg is attached to
the end of a string having length (l)
0.5 m. The ball is rotated on a horizontal
circular path about vertical axis. The
maximum tension that the string can bear
is 324 N. The maximum possible value
of angular velocity of ball (in rad/s) is

• A. 9
• B. 18
• C. 27
• D. 36

Answer 1

Answer: (D)

$r = l sin \theta $
$ Tcos \theta $ component will cancel mg.
$ Tsin \theta $ component will provide necessary centripetal force
to the ball towards centre C.
$\therefore \, \, \, Tsin \theta = m r {\omega}^2 = m (l sin \theta ) {\omega}^2 \, \, \, or \, \, \, T = m l {\omega}^2$
$\therefore \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \omega = \sqrt{ \frac{T}{ml}}$
or $ \, \, \, \, \, \, \, \, \, \, \, {\omega}_{max } = \sqrt{\frac{T_{max}}{ml}} = \sqrt{\frac{ 324}{ 0.5 \times 0.5}} $
$ \, \, \, \, \, \, \, = 36 rad/s$
$\therefore $ Correct option is (d).