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Finding angle of departure for rolling disk on semicyliner

Nasser M. Abbasi

December 4, 2017 compiled on — Monday December 04, 2017 at 01:43 AM [public]

A small sphere of mass m  starts to roll with no slip on top of semicylinder. The problem is to determine at what angle θ  the small sphere will depart the surface of the semicylinder.


The free body diagram for the sphere is


Resolving forces along the normal N  gives



       (                 )
N =  m  gcos θ − θ˙2(R + r)

To find when N =  0  , we need to find θ  . Taking moments around point p  where sphere is on contact with the cylinder (this way we do not have to solve for F  , the friction). Using anti-clock wise as positive then

mg sinθ = Icgαs + ma θr

Notice that we had to add ma θr  , which is the moment around p  due to inertia acceleration of the sphere, since the point we are taking moment about (point p  ) is not fixed and it is not the C.G. In (2) αs  is the angular acceleration of the sphere around its mass center. Not to confuse this with θ¨ of the whole sphere around the center of the semicylinder itself.

Now, since the sphere rolls without slip, then

aθ = rαs

And since Icg = 2mr2
     5   , then (2) becomes


But a = (R + r) ¨θ
 θ , therefore the above becomes

        7-       ¨
g sinθ = 5 (R + r)θ

Let ¨   d˙θ   d˙θdθ   ˙d˙θ
θ = dt = dθdt = θdθ  . Hence (3) becomes

           ˙7-        ˙
gsinθdθ = θ 5 (R + r)dθ

Integrating (The sphere starts rolling with zero initial velocity)


Using the above expression for  ˙2
θ   in (1) gives


This is zero when


The first solution for this is

θslip = 53.96810

This is the angle from the vertical when the sphere will depart the surface of the cylinder.