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Question

A round uniform body of radius R, mass M and moment of inertia I, rolls down (without slipping) an inclined plane making an angle θ with the horizontal. Then its acceleration is

A
gsinθ1+IMR2
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B
gsinθ1+MR2I
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C
gsinθ1IMR2
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D
gsinθ1MR2I
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Solution

The correct option is A gsinθ1+IMR2
Assume that no energy is used up against friction, the loss in potential energy is equal to the total gain in the kinetic energy.
ie, Mgh=12I(v2R2)+12Mv2
or 12v2(M+IR2)=Mgh
or v2=2MghM+IR2=2gh1+IMR2
If s be the distance covered along the plane, then
h=s sinθ
v2=2gssinθ1+IMR2
Now, V2=2as
2as=2gssinθ1+IMR2
or a=gsinθ1+IMR2


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