Asked by shekhar14351 | 23rd Feb, 2010, 01:34: PM

Expert Answer:

Dear Student

 Initial angular velocity = Ai=0
Final angular velocity= Af
Torque acting  T=Rmg
Angular acceleratio =  alpha
T=I  X alpha  ==>  alpha= T/I= Rmg/I
When the chorg falls down by h distance  wheel rotates h distance in the perimeter.==> Angular displacement=2pi X   h/(2 pi R) =h/R  radians

Angualr displacement  s= Ai t+(1/2) alpha t^2 = (1/2) alpha t^2  ==>  t=sqrt (2s/alpha)
Af=Ai+alpha t
= alpha X sqrt (2s/alpha)
= Sqrt(2s alpha)
= sqrt (2sRmg/I)
=sqrt(2hmg/I)             (since s=h/r)

the angular velocity of the wheel after falling through a distance 'h'.=Af= sqrt(2hmg/I)            

Answered by  | 23rd Feb, 2010, 05:12: PM

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