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Question

Find the speed of uniform solid sphere after rolling down (without sliding) an inclined plane of vertical height h=0.14 m from rest is (Take g=9.8 m/s2)

A
1.4 m/s
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B
1.2 m/s
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C
1 m/s
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D
1.3 m/s
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Solution

The correct option is A 1.4 m/s
For pure rolling, ω=vr ...(i)
where, v=vCM
Applying mechanical energy conservation:
Loss in PEgrav=Gain in KETrans+Gain in KERot
mgh=(12mv20)+(12Iω20) ...(ii)
Initially KETrans=0, KERot=0
For solid sphere,
ICMI=25mr2
From Eq (i) & (ii),
mgh=12mv2+12×25mr2×(vr)2
mgh=12mv2+15mv2
mgh=710mv2
v=10gh7
v=10×9.8×0.147=1.4 m/s

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