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A solid sphere is rolling on a frictionl...

A solid sphere is rolling on a frictionless surface, shown in figure with a translational velocity `upsilon m//s`. If it is to climb the inclined surface then `upsilon` should be:

A

`ge sqrt(10//7 gh)`

B

`ge sqrt(2 gh)`

C

`2gh`

D

`10//7 gh`

Text Solution

Verified by Experts

The correct Answer is:
A
From conservation of energy Potential energy = translational kinetic energy + rotational kinetic energy
`mgh=(1)/(2)mv^(2)+(1)/((2)/(5))mR^(2) cdot (v^(2))/(R^(2))`
`rArr" "(7)/(10) mv^(2)=mgh rArr v ge sqrt((10)/(7))gh.`
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