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# solve all the two paragraph question

when yo-yo rolls with angular velocity ω without slipping, velocity is zero at the point of contact at bottom and
velocity is 2ωR at top most point as marked in figure. Where R is major radius of yo-yo.

Hence velocity of centre of mass  = ωR ..................(1)

if we linearly interpolate, at a point where the yo-yo is pulled by unwinding the string attached,  velocity is ω(R+r).

But at this point yo-yo is pulled with speed v,  hence   v = ω(R+r) ....................(2)

if we substitute for ω in eqn.(1), using eqn.(2), then we have, velocity of centre of mass = v R/(R+r)  .............................(3)

If θ is the angle subtended by bar with horizontal surface, then we have (refer figure),  tan(θ/2) = y/x  = R/x   .......................(4)

if we differentiate equation (4),   sec2(θ/2) (1/2) (dθ/dt)  = -R/x2 (dx/dt)  .....................(5)

By substituting x using eqn.(4) and substituting (dx/dt) as velocity of centre of mass from eqn.(3),
we rewrite eqn.(5) as     sec2(θ/2) (1/2) (dθ/dt)  = - tan2(θ/2) (1/R) [ v R/(R+r) ]

Hence after simplification of above equation, we get angular velocity of bar,  (dθ/dt) = -2 sin2(θ/2) [ v / (R+r) ]

(-ve sign of angular velocity indicates, direction of rotation is clockwise )
Answered by Expert 19th May 2019, 3:39 PM
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