SATHEE: Laws of Motion 4 Question 3

Laws of Motion 4 Question 3

3. A ball of mass $(m) 0.5 k g$ 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 $r a d / s$ ) is

(2011)

(a) 9

(b) 18

(d) 36

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Answer:

Correct Answer: 3. (d)

Solution:

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

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

$\begin{aligned} ∴ & T \sin θ & = m r ω^{2} = m (l \sin θ) ω^{2} or T = m l ω^{2} \\ ∴ & ω & = \sqrt{\frac{T}{m l}} \\ or & ω_{max} & = \sqrt{\frac{T_{max}}{m l}} = \sqrt{\frac{324}{0.5 \times 0.5}} = 36 r a d / s \end{aligned}$

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Laws of Motion 4 Question 3

3. A ball of mass (m)0.5kg is attached to the end of a string having length (l)0.5m. The ball is rotated on a horizontal circular path about vertical axis. The maximum tension that the string can bear is 324N. The maximum possible value of angular velocity of ball (in rad/s ) is

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3. A ball of mass $(m) 0.5 k g$ 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 $r a d / s$ ) is