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Question

A stone tied to an inextensible string of length 1 m is released from horizontal position. The stone is free to revolve in a vertical plane around the other end of the string. Find the angular speed of the stone when the string makes an angle θ=30 with horizontal. Take g=10 m/s2.

A
ω=4 rad/ s
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B
ω=15 rad/ s
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C
ω=10 rad/ s
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D
ω=2.5 rad/ s
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Solution

The correct option is C ω=10 rad/ s

On applying energy conservation between intial and final position i.e A and B, taking datum for PE=0 at point A:
PEA+KEA=PEB+KEB
0+12mu2=mgh+12mv2
v2=u2+2gh
where h=lsinθ, u=0
v2=0+2g(lsinθ)
or v=2×10×1×sin30=10 m/s

Now angular speed of body at B:
ω=vr
radius of circular path about O is r=l=1 m
ω=101=10 rad/s

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