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 $rad{s}^{-1}$ ) 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 $Tsin\theta =mr{\left(\omega \right)}^2=m(lsin\theta ){\left(\omega \right)}^2$ or $T=ml{\omega }^2$
Therefore $\omega =\sqrt{\frac{T}{ml}}$
or ${\omega }_{max}$ $=\sqrt{\frac{T_{max}}{ml}}=\sqrt{\frac{324}{0.5}}=36rad{s}^{-1}$