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Question

A solid cylinder of mass M and radius R rolls down an inclined plane of height h. The angular velocity of the cylinder when it reaches the bottom of the plane will be

A
2Rgh
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B
2Rgh2
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C
2Rgh3
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D
12Rgh
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Solution

The correct option is C 2Rgh3

[since there is no relative motion at point of contact, work done by friction Wf=0 in pure rolling]
Applying mechanical energy conservation,

Loss in PE=Gain in KETrans+Gain in KERot
or Mgh=12Mv2+12Iω2
Mgh=12MR2ω2+12(12MR2)ω2
[v=Rω for pure rolling]
Mgh=34MR2ω2
ω2=4gh3R2
ω=2Rgh3

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