A uniform solid cylinder of radius `R=15cm` rolls over a horizontal plane passing into an inclined plane forming an angle `alpha=30^@` with the horizontal(figure). Find the maximum value of the velocity `v_0` which still permits the cylinder to roll onto the inclined plane section without a jump. The sliding is assumed to be absent.

Explanation

Since the cylinder moves without sliding, the centre of the cylinder rotates about the point O, while passing through the common edge of the planes. In other words, the point O becomes the foot of the instantaneous axis of rotation of the cylinder.
It at any instant during this motion the velocity of the C.M. is `v_1` when the angle (shown in the figure) is `beta`, we have
`(mv_1^2)/(R)=mgcosbeta-N`,
where N is the normal reaction of the edge
or, `v_1^2=gRcosbeta-(NR)/(m)` (1)
From the energy conservation law,
`1/2I_0(v_1^2)/(R^2)-1/2I_0(v_0^2)/(R^2)=mgR(1-cosbeta)`
But `I_0=(mR^2)/(2)+mR^2=3/2mR^2`,
(from the parallel axis theorem)
Thus, `v_1^2=v_0^2+4/3gR(1-cosbeta)` (2)
From (1) and (2)
`v_0^2=(gR)/(3)(7cosbeta-4)-(NR)/(m)`
The angle `beta` in this equation is clearly smaller than or equal to `alpha` so putting `beta=alpha` we get
`v_0^2=(gR)/(3)(7cosalpha-4)-(N_0R)/(M)`
where `N_0` is the corresponding reaction. Note that `NgeN_0`. No jumping occurs during this turning if `N_0gt0`. Hence, `v_0` must be less than
`v_(max)=sqrt((gR)/(3)(7cosalpha-4))`