It's time I disclosed my reason for posting this problem.
The major reason why I did so, is that the problem is one of the simplest that highlights the intricacies involved in the standard definition of "rolling".
The basic solution technique is, as rayjohn pointed out, that of using energy conservation.
This can be regarded as the "trivial" part of the problem.
The "hard" part is to state the rolling condition correctly!
(I call this "hard", since most textbooks I've seen either skips it entirely, or formulates it in an ambigouous way that might easily lead to false conclusions, because the examples typically used may, in a subtle manner, lead to false generalizations. More about this later on.)
1. The basic contact point condition in energy conservation:
Let's first "forget" the rolling aspect, and instead focus on answering the question:
What must the contact point condition be, in order for the system cylinder+sphere to be energy-conservative?
Clearly, it must be: The relative velocity between the contact point on the cylinder and the contact point on the sphere must be zero!
(Otherwise, the internal force acting between them would do work on the entire system)
Or, differently stated, the contact point velocities must be equal.
Let us assume that the cylinder (radius R) rotates with angular velocity
\omega_{c}
Since the C.M. of the sphere (radius r) moves in a circular orbit about the cylinder axis, we assign to it an angular velocity \omega_{C.M}
The sphere itself has an angular velocity attached to it, which we write as:
\omega_{s}=\omega_{C.M.}+\omega_{rel}
Hence, we have, for equal contact point velocities:
R\omega_{c}=(R+r)\omega_{C.M}-r\omega_{s}
Or:
R\omega_{c}=R\omega_{C.M}-r\omega_{rel} (1)
This is the correct contact point condition for energy conservation.
I believe eq. (1) raises some eyebrows, since the sphere's (total) angular velocity is not an explicit parameter!
I will now state the standard "rolling" condition, and show that this, correctly interpreted, is, in fact, equivalent to (1)
2. The rolling condition:
The classical rolling condition, is that given objects 1 and 2, they are said to roll on each other, if the "contact-point arclength" (denoted either as "CPA" or "s") as traced out on object 1 equals the CPA as traced out on object 2.
Or, in terms, of velocities, we have:
\dot{s}_{1}=\dot{s}_{2} (2)
Equation (2) is the classical rolling condition.
Let us rewrite eq. (1) in the form:
R(\omega_{C.M}-\omega_{c})=r\omega_{rel} (3)
Now, the left-hand side of (3) is easy to interpret:
R\omega_{C.M} is the velocity by which the contact point slides down the cylinder surface, because the contact point always lies at distance R on the line connecting the cylinder axis and the sphere's C.M. (and that line, clearly, rotates with the C.M s angular velocity).
But, since the cylinder itself rotates, the actual CPA-velocity is not given by
R\omega_{C.M} !
Because a material point on the cylinder moves with velocity R\omega_{c}
the CPA-velocity on the cylinder must be given by their difference, namely the left-hand side of (3)
In order therefore, that the rolling condition is equivalent with the energy conservation demand, I must show that r\omega_{rel} is, in fact, the correct expression for the CPA-velocity as traced out on the sphere.
(I'll do that later..)
