Q: A ball of mass (m) 0.5 kg is attached to the end of a string having (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

Ans: (d)

Sol: Let T = tension in the string

θ = angle made by string with vertical

$\large T cos\theta = mg$ …(i)

$\large T sin\theta = m \omega^2 r $

$\large T sin\theta = m \omega^2 (l sin\theta) $

$\large \omega = \sqrt{\frac{T}{m l}}$

$\large \omega_{max} = \sqrt{\frac{T_{max}}{m l}}$

$\large \omega_{max} = \sqrt{\frac{324}{0.5 \times 0.5}}$

= 36 rad/s