CBSE Class 11-science Answered
find
Asked by shekhar14351 | 23 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)
Regards
Team Topperlearning.com
Answered by | 23 Feb, 2010, 05:12: PM
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