A solid sphere is set into rotation at an angular velocity and it is then placed on a rough horizontal surface. The ratio of distances covered by rotational and translational motions up to the start of the pure rolling is (Assume uniformly accelerated motion up to start of pure rolling):
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a
5/2
b
7/2
c
7/4
d
9/2
answer is D.
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Detailed Solution
Angular momentum about A is conserved.L1=25mR2ω0L2=25mR2ω+mR2ω=75mR2ωL1=L2ω=27ω0ω0>ω and v>v0Sphere accelerates forward and rotation decelerates.ω=ω0−αtα=ω0−ωt .......(i)ω2=ω02−2αθθ=ω0−ωω+ω02α=αt⋅ω+72ω2α=94ωtRθ=94ωRt ....(ii)v=v0+at v0=0a=ωR/tv2=v02+2aSS=v⋅v2a=(ωR)(v⋅t)2⋅ωR=vt2 .......(iii) Ratio of distance =9/2