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Question

A heavy solid sphere is thrown on a horizontal rough surface with initial velocity u without rolling. What will be its speed, when it starts pure rolling motion

Joshi sir comment

for translational motion

0-μMg = Ma so a = -μg

so after time t velocity v = u-μgt (1)

for rotational motion

about centre, μMgr = Iα = 2Mr²α/5

so α = 5μg/2r

so ω = 0 + αt

so ω = 5μgt/2r (2)

for pure rolling

v = rω implies that u-μgt = r5μgt/2r

implies that u = 7μgt/2

so t = 2u/7μg

now put t in (1) for getting v

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