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Question

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 the vertical axis, as shown in figure. The maximum tension that the string can bear is 324 N. Then, maximum possible value of angular speed of ball (in rad/s) will be:


A
9 rad/s
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B
18 rad/s
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C
27 rad/s
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D
36 rad/s
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Solution

The correct option is D 36 rad/s

Radius of the horizontal circular path is:
r=lsinθ

Tcosθ component will balance the weight mg, since particle's acceleration in y-direction ay=0

Tcosθ=mg

Tsinθ component is directed towards the center of the circle and will provide necessary centripetal force to the ball.
The ball is rotating with angular speed ω

Tsinθ=m×ac=mω2r
Tsinθ=mω2(lsinθ)

The maximum value of tension T, will be achieved when sinθ=1
i.e Tmax=mlω2max
or ωmax=Tmaxml

Given: maximum tension which string can withhold is 324 N
Thus, ωmax=Tmaxml=3240.5×0.5=180.5=36 rad/s

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