# Cylinder Rotational Motion Question

A sphere of radius a rests on top of a fixed rough circular cylinder of radius R which is lying with its principal axis horizontal. The sphere is disturbed and rolls, without any slipping, around the surface of the cylinder. Show from energy considerations that, if theta is the angle to the vertical made by the line joining the centre of the sphere directly to the axis of the cylinder, then

(w)^2 = [10g(1 - cos theta)] / [7(R+a)]

where (w)^2=angular velocity squared, g=acceleration due to gravity

Hence, show that the sphere will leave the surfacve of the cylinder at

theta = arccos (10/17)

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Welcome to PF!

Hi kongieieie! Welcome to PF! (have a theta: θ and an omega: ω and try using the X2 tag just above the Reply box )
Show from energy considerations …

Hence, show that the sphere will leave the surfacve of the cylinder …
Use KE + PE = constant, and v = rω, and remember the sphere will lose contact when the normal force is zero. Show us how far you get, and where you're stuck, and then we'll know how to help! Problem solved. Thanks for the hint. :)