Question

A ball of mass ( $m$ ) $0.5kg$ is attached to the end of a string having a length ( $L$ ) $0.5m$ . The ball is rotated on a horizontal circular path about the vertical axis. The maximum tension that the string can bear is $324N$ . The maximum possible value of the angular velocity of ball (in $rads^{- 1}$ ) is

• A. $9$
• B. $18$
• C. $27$
• D. $36$

Answer 1

Answer: (D)

$r=lsin\theta Tcos\theta$ component will cancel $mg$ .
T sin θ component will provide the necessary centripetal force to the ball towards centre C.
Therefore $Tsin\theta =mr\left(\omega \right)^{2}=m\left(\right.lsin\theta \left.\right)\left(\omega \right)^{2}$ or $T=ml\omega ^{2}$
Therefore $\omega = \sqrt{\frac{T}{m l}}$
or $\omega _{max}$ $=\sqrt{\frac{T_{max}}{m l}}=\sqrt{\frac{324}{0.5}}=36rads^{- 1}$