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)

36

Answer 2

$T \cos\theta$ component will cancel $mg$.
$T \sin \theta$ component will provide necessary centripetal force to the ball towards centre $C$
$\therefore\, T \sin \theta =mr \omega^2=m(l \sin \theta)\omega^2\,$ or $T=m l \omega^2$
$\therefore 15mm \omega=\sqrt{\frac{T}{ml}}$
or $15mm \omega_{max}=\sqrt{\frac{T_{max}}{ml}}=\sqrt{\frac{324}{0.5\times 0.5}}$
$25mm = 36\, rad/s$
$\therefore$ Correct option is (d)