. A solid cylinder is released from rest. If string does not slip relative to cylinder then find relative acceleration of A w.r.t B just after it is released.

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Vikas TU · Answer

Dear student

The moment of inertia of the cylinder I=12mr2Let us consider the velocity of the block after it falls through height h is v. Then equatingkinetic energy and potential energy we get12Iω2+12mv2=mgh⇒12mr22(vr)2+12mv2=mgh⇒32mv2
v = sqrt(2gh^3)

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. A solid cylinder is released from rest. If string does not slip relative to cylinder then find relative acceleration of A w.r.t B just after it is released.

. A solid cylinder is released from rest. If string does not slip relative to cylinder then find relative acceleration of A w.r.t B just after it is released.

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. A solid cylinder is released from rest. If string does not slip relative to cylinder then find relative acceleration of A w.r.t B just after it is released.

. A solid cylinder is released from rest. If string does not slip relative to cylinder then find relative acceleration of A w.r.t B just after it is released.

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