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 a vertical axis. The maximum tension that the string can bear is 324 N. The maximum possible value of angular velocity of the ball (in radian/s) is [2011]

Question

Smart Achievers · Accepted Answer

(4)

$T \sin θ = m R ω^{2}$ ...(i)

$T \cos θ = m g$ ...(ii)

Dividing (ii) by (i), we get

$\tan θ = \frac{ω^{2}}{R g} \Rightarrow ω = \sqrt{R g \tan θ}$

Clearly, $ω$ is maximum when $\tan θ$ is maximum, i.e., $θ = 90 °$

So, $T \sin 90 ° = m R ω^{2}$

$T = m L ω^{2} [Here, R = L]$

$\Rightarrow ω = \sqrt{\frac{T}{m L}} = \sqrt{\frac{324}{0.5 \times 0.5}}$

$= \frac{18}{0.5} = 36 rad/s$

Solution

Q.
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 a vertical axis. The maximum tension that the string can bear is 324 N. The maximum possible value of angular velocity of the ball (in radian/s) is [2011]

1 9

2 18

3 27

4 36

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