Laws of Motion 4 Question 3

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 $rad / s$ ) is

(2011)

(a) 9

(b) 18

(c) 27

(d) 36

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Answer:

Correct Answer: 3. (d)

Solution:

$T \cos \theta$ component will cancel $m g$.

$T \sin \theta$ component will provide necessary centripetal force to the ball towards centre $C$.

$$ \begin{aligned} & \therefore & T \sin \theta & =m r \omega^{2}=m(l \sin \theta) \omega^{2} \text { or } T=m l \omega^{2} \\ & \therefore & \omega & =\sqrt{\frac{T}{m l}} \\ & \text { or } & \omega _{\max } & =\sqrt{\frac{T _{\max }}{m l}}=\sqrt{\frac{324}{0.5 \times 0.5}}=36 rad / s \end{aligned} $$



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