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Question

A sphere of mass M and radius r shown in the figure, slips on rough horizontal plane. At some instant it has translational velocity v0 and rotational velocity about centre v0/2r. The percentage change of translational velocity after the sphere start pure rolling:

160356_27af5de4d3cd4af79bb70d2765f7fdef.png

A
14.28%
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B
7.14%
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C
21.42%
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D
None
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Solution

The correct option is A 14.28%
Initially the velocity of COM of sphere is vo and angular velocity is wi.
When the pure rolling starts, let the velocity of COM of sphere be v and angular velocity be wf.
Now f=Maa=fM
v=v0atv=vofMt ..............(1)
Also, τ=Iα
fr=25Mr2αα=5f2Mr
wf=wi+αt
wf=v02r+5ft2Mr25rwf=vo2×25+ftM ..............(2)
Adding (1) and (2), v+25rwf=vo5+vo
Also v=rwf (pure rolling)
v=67vo

Now vovvo×100=17×100=14.28%

447338_160356_ans_3909302bfd0e4aea8a0de477a5a28f57.png

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