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Finding angle of departure for rolling disk on semicyliner

Nasser M. Abbasi

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.

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The free body diagram for the sphere is

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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\begin{equation} N=m\left ( g\cos \theta -\dot{\theta }^{2}\left ( R+r\right ) \right ) \tag{1} \end{equation} 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\begin{equation} mg\sin \theta =I_{cg}\alpha _{s}+ma_{\theta }r \tag{2} \end{equation} 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\begin{equation} g\sin \theta =\frac{7}{5}\left ( R+r\right ) \ddot{\theta }\tag{3} \end{equation} 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.