JEE Advance - Physics (2011 - Paper 1 Offline - No. 3)
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 radian/s) is
Explicación
Resolve T along the horizontal and the vertical directions. As the particle moves in a horizontal plane, net vertical fore on it is zero. Net horizontal force on it provides the necessary centripetal acceleration for circular motion. Apply Netwon's second law in the horizontal direction to get
$$T\sin \theta = m{\omega ^2}r = m{\omega ^2}(l\sin \theta )$$
$$T = m{\omega ^2}l$$
$$\omega = \sqrt {{T \over {ml}}} $$
Substituting the given values, we get
$$\omega = \sqrt {{{324} \over {0.5 \times 0.5}}} = 36$$ rad/s