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Question

# A uniform ring of mass m, with the outside radius r2, is fitted tightly on a shaft of radius r1. The shaft is rotated about its axis with a constant angular acceleration β. Find the moment of elastic forces in the ring as a function of the distance r from the rotation axis.

A
N=12mβ(r22r21)(r4r41)
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B
N=12mβ(r22+r21)(r42r4)
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C
N=12mβ(r22+r21)(r42+r4)
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D
None of these
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Solution

## The correct option is A N=12mβ(r22−r21)(r4−r41)Moment applied by the elastic force due to inside and outside part on the elementary ring of radius r are N and N+dNdrdr respectively.∴ Net moment of force is (N+dNdrdr−N)=dNdrdr=dNGiven: Mass of the ring is mArea of cross-section of the ring A=πr22−πr21=π(r22−r21)Area of the cross-section of the elementary ring A′=2πrdr∴ Mass of the elementary ring dm=mAA′=mπ(r22−r21)2πrdrNow moment of inertia of the elementary part dI=r2dm=mπ(r22−r21)2πr3drMoment of force on the elementary part dN=dI×β ∴ dN=2mβ(r22−r21)r3drTotal moment of force on the ring N=2mβ(r22−r21)∫rr1r3dr N=2mβ(r22−r21)×r44∣∣∣rr1 N=mβ2(r22−r21)(r4−r41)

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