A uniform solid cyinder of radius `R=15 cm` rolls over a horizontal plane passing into an inclined plane forming an ange `alpha=30^@` with the horizontal. Find the maximum value of the velocity `v_(0)` which still permits the cylinder to roll on the inclined plane section without a jump. (The sliding is asumed to be absent).

Explanation

Initial Energy
`E_(1)=1/2mv_(0)^(2)+1/2I_(cm)omega^(2)+mgR`

For rolling `(v_(0))/R=omega`
`implies E_(1)=1/2mv_(0)^(2)+1/2.1/2mR^(2)v_(0)^(2).R^(2)+mgR`
`=3/4mv_(0)^(2)+mgR`
As it reaches the edge
`E_(2)=1/2I_(P)omega^('2)+mgRcosalpha`
`E_(2)=1/2mv^(2)+1/2I_(cm)omega^('2)+mgRcosalpha`
Here `omega'=v/R`
`=3/4mv^(2)=mgRcosalpha`
From `COE, 3/4 mv^(2)+mgRcosalpha=3/4mv_(0)^(2)+mgR`
`implies mv^(2)=mv_(0)^(2)+4/3mgR(1-cosalpha)`.........i
Consider the `FBD` of the cyider when it is at the edge. Centre of mass of the cylinder describes circular motion about `P`. Hence, `mgcosalpha-N=mv^(2)//R`
`implies=mgcosalpha-mv^(2)//R`
`=mgcosalpha- (mv_(0)^(2))/R-4/3mg+4/3mgcosalpha`, from i.
For no jumping `Nge0`
`implies 7/3 mgcosalpha-4/3mg-(mv_(0)^(2))/Rge0`
`implies v_(0)lesqrt((7gR)/3cosalpha-4/3gR)`