A ring of mass m and radius R rotates about an axis passing through its centre and perpendicular to its plane with angular velocity ω. Find the velocity of transverse pulse in the ring.
A
Rω2
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B
2Rω
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C
32Rω
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D
Rω
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Solution
The correct option is DRω Let T be the tension produce in the ring.
Take a elementary part of ring having mass dm, subtending an angle dθ at centre
Here, T sin (dθ2)+T sin (dθ2)=2T sin (dθ2) will be responsible for coertripetal force
So 2T sin(dθ2)=dmRω2 ⇒2Tdθ2=m2πRRdθ.Rω2 ⇒T=mRω22π
and μ=m2πR [dθ2is very smalland dm=m2πRRdθ]
So, velocity of transverse pulse v=√Tμ=
⎷(mRω22π)(m2πR)=Rω