Question

I have nearly completed this problem, but I am stuck on one of the final steps in my derivation. I would like to see a solution to compare with my own work

3. A sphere of radius a and mass m is placed on top of a horizontal cylinder of radius R. Let P denote the initial position of contact and O the center of the cylinder and suppose that the cylinder is held fixed Suppose that the sphere starts to roll without slipping and falls off the cylinder when its center of mass is at an angle 0 with respect to the vertical axis OP Show that 2 cos I 3+ ma2 Here I is the moment of inertia of the sphere about an axis through its center and all motion takes place in a plane transverse to the cylinder axis. Conclude that the sphere loses contact with the cylinder at a larger angle than a point particle of the same mass, and explain why. The surprising fact is that the result does not depend on R. To understand this look at the figure below: GRA LYZAB REG LEZAB

Answer 1

^{a d- mg lese (R-A) For Sthere (R-h) RCore hf(-0s From, At tot tesition. considers 2m8h 2 mg h 22h 1 I mar mar at point B. Neto the sphere mg Cos N Dses Contact then N=0 cos R 23R (1-C0s0) (t CAS8 mar 2. Oaticle ee Cos 3tna aD}
angle of contact of a particle is greater because for it's case the angle is equal to cos^{_​​​​​​​​ -1}(2/3) which is less than that of sphere !!!... Because there is no (1/2 I w^{​​​​​​ 2} )term involved in case of evaluating velocity term...

But in case of sphere it has that term so that the angle of contact of a sphere is greater than that of a particle!.