How Does a Rolling Sphere Overcome a Block Without Slipping?

[Mr.Darcy]

A uniform solid sphere of radius R and massM rolls without slipping on
a horizontal plane. The centre of mass of the sphere moves initially with
velocity ~v directed perpendicularly to a fixed block of height h < R.
The sphere hits the block and the collision is such that the point P on
the sphere which touches the block does not slip and remains in contact
with the block until the sphere starts rolling on top of the block. See
figure belowa) Compute the initial kinetic energy of the sphere and its angular
momentum with respect to the point P.
[Notice: The angular momentum is a vector, so it is characterised
by its length and direction.]
b) Determine the minimum value, vmin, of the initial speed of the
centre of mass of the sphere, which allows it to climb over the block
(express the result in terms of the parameters of the problem and
the gravitational acceleration g).
c) Assuming that the initial velocity of the centre of mass is greater
than the vmin found in (b), compute the amount of mechanical
energy dissipated in the process.
Why is energy not conserved?

Hi guys.

So I got the first bit. I think K is just equal to sum of translational and rotational energies- that is K=1/2Mv^2+2/5MR^2. and then using the P-A theorem find the angular momentum to be 7/5MVR.

I'm not so sure how to go about the second part though. I got v=gsqrth^2-2hr but don't think this is right.

any help appreciated!

 

[BigJDubs]

a) The initial kinetic energy of the sphere is K = 1/2Mv^2 + 2/5MR^2 and its angular momentum is L = 7/5MVR.b) The minimum speed, vmin, of the centre of mass of the sphere required for it to climb over the block is given by vmin = √gh(h - 2R).c) The amount of mechanical energy dissipated in the process can be calculated as E = K + (total change in angular momentum). Energy is not conserved because some of it is dissipated due to friction between the sphere and the block.