Laws Of Motion Ques 45
- 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
(2011)
9
18
27
36
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Answer:
Correct Answer: 45.(d)
Solution:
Formula:
$T \cos \theta$ component will cancel $m g$.
$T \sin \theta$ component will provide necessary centripetal force to the ball towards centre $C$.
$$ \begin{aligned} & \therefore & T \sin \theta & =m r \omega^{2}=m(l \sin \theta) \omega^{2} \text { or } T=m l \omega^{2} \\ & \therefore & \omega & =\sqrt{\frac{T}{m l}} \\ & \text { or } & \omega _{\max } & =\sqrt{\frac{T _{\max }}{m l}}=\sqrt{\frac{324}{0.5 \times 0.5}}=36 rad / s \end{aligned} $$
BETA
