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Question

A solid cylinder of mass m and radius r starts rolling down on inclined plane of inclination θ. Friction is sufficient enough to prevent slipping. Find the speed of centre of mass of solid cylinder, when it's centre of mass has fallen through a height of h=9 m. (Take g=10 m/s2)

A
120 m/s
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B
125 m/s
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C
180 m/s
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D
90 m/s
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Solution

The correct option is A 120 m/s
In case of pure rolling, total mechanical energy remains conserved as work done by dissipative forces are zero i.e Wf=0, mechanical energy is conserved.

Since sphere is starting from rest ωi=0, vi=0
KETrans=12mv2CM
KERot=12ICMω2
Total mechanical Energy at position (1):
E1=PE+KETrans+KERot
E1=mgh+0+0=mgh
Taking reference of PE=0 at position (2)
Total mechanical Energy at position (2):
E2=PE+KETrans+KERot
E2=0+12mv2CM+12ICMω2
vCMr=ω and ICM=mr22
E2=12mv2CM+12(mr22)(v2CMr2)
E2=34mv2CM
From conservation of total energy,
E1=E2
mgh=34mv2CM
vCM=43gh
vCM=4×10×93=120 m/s
vCM=120 m/s

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