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solve all the two paragraph question
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Asked by sarveshvibrantacademy | 18 May, 2019, 11:52: AM
answered-by-expert Expert Answer
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 Thiyagarajan K | 19 May, 2019, 15:39: PM

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