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Question

A thin circular ring of mass M and radius r is rotating about its axis with an angular speed ω. Two particles having mass m each are now attached at diametrically opposite points. The angular speed of the ring will become:

A
ωMM+2m
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B
ωMM+m
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C
ωM+2mM
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D
ωM2mM+2m
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Solution

The correct option is A ωMM+2m

External torque is zero on the system. τnet=0, so angular momentum is conserved.

By angular momentum conservation:
Iiωi=Ifωf

(MR2)ω=(MR2+2mR2)ωf

ωf=(MR2)ωMR2+2mR2=MωM+2m

ωf=MωM+2m

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