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## Finding angle of departure for rolling disk on semicyliner

December 4, 2017   Compiled on May 19, 2020 at 4:31am  [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 $$\theta$$ 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\begin{align*} N-mg\cos \theta & =ma_{r}\\ & =-m\dot{\theta }^{2}\left ( R+r\right ) \end{align*}

Hence$$N=m\left ( g\cos \theta -\dot{\theta }^{2}\left ( R+r\right ) \right ) \tag{1}$$ To ﬁnd when $$N=0$$, we need to ﬁnd $$\theta$$. 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$$mg\sin \theta =I_{cg}\alpha _{s}+ma_{\theta }r \tag{2}$$ Notice that we had to add $$ma_{\theta }r$$, 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) $$\alpha _{s}$$ is the angular acceleration of the sphere around its mass center. Not to confuse this with $$\ddot{\theta }$$ of the whole sphere around the center of the semicylinder itself.

Now, since the sphere rolls without slip, then$a_{\theta }=r\alpha _{s}$ And since $$I_{cg}=\frac{2}{5}mr^{2}$$, then (2) becomes\begin{align*} rmg\sin \theta & =\frac{2}{5}mr^{2}\frac{a_{\theta }}{r}+ma_{\theta }r\\ rg\sin \theta & =\frac{2}{5}ra_{\theta }+a_{\theta }r\\ g\sin \theta & =\frac{7}{5}a_{\theta } \end{align*}

But $$a_{\theta }=\left ( R+r\right ) \ddot{\theta }$$, therefore the above becomes$$g\sin \theta =\frac{7}{5}\left ( R+r\right ) \ddot{\theta }\tag{3}$$ Let $$\ddot{\theta }=\frac{d\dot{\theta }}{dt}=\frac{d\dot{\theta }}{d\theta }\frac{d\theta }{dt}=\dot{\theta }\frac{d\dot{\theta }}{d\theta }$$. Hence (3) becomes$g\sin \theta d\theta =\dot{\theta }\frac{7}{5}\left ( R+r\right ) d\dot{\theta }$ Integrating (The sphere starts rolling with zero initial velocity)\begin{align*} \int _{0}^{\theta _{slip}}g\sin \theta d\theta & =\int _{0}^{\dot{\theta }_{slip}}\dot{\theta }\frac{7}{5}\left ( R+r\right ) d\dot{\theta }\\ -g\left ( \cos \theta \right ) _{0}^{\theta _{slip}} & =\frac{7}{10}\left ( R+r\right ) \dot{\theta }_{slip}^{2}\\ g\left ( 1-\cos \theta _{slip}\right ) & =\frac{7}{10}\left ( R+r\right ) \dot{\theta }_{slip}^{2}\\ \dot{\theta }_{slip}^{2} & =\frac{10}{7}\frac{g\left ( 1-\cos \theta _{slip}\right ) }{\left ( R+r\right ) } \end{align*}

Using the above expression for $$\dot{\theta }^{2}$$ in (1) gives\begin{align*} N & =m\left ( g\cos \theta -\left ( \frac{10}{7}\frac{g\left ( 1-\cos \theta \right ) }{\left ( R+r\right ) }\right ) \left ( R+r\right ) \right ) \\ & =m\left ( g\cos \theta -\frac{10}{7}g\left ( 1-\cos \theta \right ) \right ) \end{align*}

This is zero when \begin{align*} \cos \theta -\frac{10}{7}\left ( 1-\cos \theta \right ) & =0\\ \cos \theta -\frac{10}{7}+\frac{10}{7}\cos \theta & =0\\ \frac{17}{7}\cos \theta & =\frac{10}{7}\\ \cos \theta & =\frac{10}{17} \end{align*}

The ﬁrst solution for this is $\theta _{slip}=53.9681^{0}$ This is the angle from the vertical when the sphere will depart the surface of the cylinder.