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Question

A uniform disc of mass M and radius R is mounted on an axle supported in frictionless bearings. A light cord is wrapped around the rim of the disc as shown in figure below. If we hang a body of mass m with the cord, it has acceleration a as shown in figure. Find the angular acceleration of the disc (α).



A
2m(a+g)MR
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B
2M(ga)mR
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C
2m(ga)MR
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D
2M(a+g)mR
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Solution

The correct option is C 2m(ga)MR
Let acceleration of the mass m=a .
Therefore tangential acceleration of disc at is also equal to a due to string constraints.

F.B.D

Force balance on mass m
mgT=maT=m(ga) ...(i)

Torque balance on disc,
τ=R×T=Iα ...(ii)
where, I=MR22 for uniform disc.
From equation (i) and (ii)
R×m(ga)=(MR22)α
So, angular acceleration α=R×m(ga)(MR22)=2m(ga)MR

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