Problems

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Solution:

1. \(\mathbf{r} = (a \sin \theta - s) \mathbf{i}+a\cos \theta\mathbf{j}\), so \begin{align*} && \dot{\mathbf{r}} &=(a \dot{\theta} \cos \theta - \dot{s}) \mathbf{i}- a\dot{\theta} \sin \theta \mathbf{j}\\ \\ \text{COM}(\rightarrow): && 0 &= M(-\dot{s}) + m(a \dot{\theta} \cos \theta - \dot{s}) \\ \Rightarrow && \dot{s} &= \frac{ma \dot{\theta} \cos \theta}{m+M} \\ &&&= \left ( 1- \frac{M}{m+M} \right) a\dot{\theta} \cos \theta \\ &&&= (1 - k) a\dot{\theta} \cos \theta \\ \\ \Rightarrow && \dot{\mathbf{r}} &=(a \dot{\theta} \cos \theta - \dot{s}) \mathbf{i}- a\dot{\theta} \sin \theta \mathbf{j} \\ &&&= (a \dot{\theta} \cos \theta - (1 - k) a\dot{\theta} \cos \theta) \mathbf{i}- a\dot{\theta} \sin \theta \mathbf{j} \\ &&&= a\dot{\theta} \left ( k \cos \theta \mathbf{i} - \sin \theta \mathbf{j} \right) \end{align*}
2. \(\,\) \begin{align*} COE: &&\underbrace{0}_{\text{k.e.}}+ \underbrace{mga}_{\text{GPE}} &= \underbrace{\frac12 m \mathbf{\dot{r}}\cdot\mathbf{\dot{r}}}_{\text{k.e. }P} + \underbrace{mg a\cos \theta}_{\text{GPE}} + \underbrace{\frac12 M \dot{s}^2}_{\text{k.e. hemisphere}} \\ \Rightarrow && 2amg(1-\cos \theta) &= a^2m \dot{\theta}^2(k^2 \cos^2 \theta + \sin^2 \theta)+ M(1 - k)^2 a^2\dot{\theta}^2 \cos^2 \theta \\ \Rightarrow && 2mg(1-\cos \theta) &= a \dot{\theta}^2 \left (m\sin^2 \theta + (mk^2 + M(1-k)^2)\cos^2 \theta \right) \\ &&&= a \dot{\theta}^2 \left (m\sin^2 \theta + mk\cos^2 \theta \right) \\ \Rightarrow && 2g(1-\cos \theta) &= a \dot{\theta}^2 \left (\sin^2 \theta + k\cos^2 \theta \right) \\ \end{align*}
3. The equation of motion is \(m \ddot{\mathbf{r}} = \mathbf{R} - mg\mathbf{j}\) and the particle will leave the surface when \(\mathbf{R} = 0\). If we take the component in the directions suggested: \begin{align*} && \ddot{\mathbf{r}} &= a\ddot{\theta}(k \cos \theta \mathbf{i}- \sin \theta \mathbf{j}) + a \dot{\theta}(-k\dot{\theta} \sin \theta \mathbf{i}- \dot{\theta} \cos \theta \mathbf{j}) \\ &&&= ak (\ddot{\theta} \cos \theta - \dot{\theta}^2 \sin \theta) \mathbf{i} -a(\ddot{\theta} \sin \theta + \dot{\theta}^2 \cos \theta) \mathbf{j} \\ \Rightarrow && \mathbf{\ddot{r}} \cdot (\sin \theta \mathbf{i} + k \cos \theta \mathbf{j}) &= ak (\ddot{\theta} \cos \theta - \dot{\theta}^2 \sin \theta) \sin \theta -ak(\ddot{\theta} \sin \theta + \dot{\theta}^2 \cos \theta)\cos \theta \\ &&&= - ak \dot{\theta}^2 \\ && (-g\mathbf{j}) \cdot (\sin \theta \mathbf{i} + k \cos \theta \mathbf{j}) &= -gk \cos \theta \\ \mathbf{R} = 0: && gk \cos \theta &= ak \dot{\theta}^2 \\ \Rightarrow && g \cos \theta &= a \dot{\theta}^2 \end{align*}
4. \(\,\) \begin{align*} && 2g(1-\cos \theta) &= a \dot{\theta}^2(k \cos^2 \theta + \sin^2 \theta) \\ && a \dot{\theta}^2 &= g \cos \alpha \\ \Rightarrow && 2g(1-\cos \alpha) &= g \cos \alpha(k \cos^2 \alpha + (1-\cos^2 \alpha)) \\ \Rightarrow && 0 &= g(k-1)c^3+3gc-2g \\ \Rightarrow && 0 &= (k-1)c^3+3c - 2 \end{align*} When \(c =1, f(c) = k > 0\) when \(c = \frac23, f(c) = k-1 < 0\). Therefore there is a root with \(\cos \alpha > \frac23\)

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Solution:

Whilst the particle is on the surface of the sphere consider the energy. Letting the height of centre of the sphere by our \(0\) GPE level, the initial energy is \(mgr\) (assuming that the initial speed is so close to \(0\) as to make no difference). When it makes an angle \(\theta\) with the horizontal it's energy will be \(mgr \sin \theta + \frac12 m v^2\). By conservation of energy: \(mgr \sin \theta + \frac12 m v^2 = mgr \Rightarrow v^2 = 2gr(1-\sin \theta)\) \begin{align*} \text{N2}(\text{radially}): && mg \sin \theta - R &= m \frac{v^2}{r} \\ \Rightarrow && R &= mg\sin \theta - \frac{m}{r} 2gr(1-\sin \theta) \\ &&&=mg \l 3\sin \theta - 2 \r \end{align*} Since \(R\) must be positive whilst the particle is in contact with the sphere, the angle \(\theta\) makes with the horizontal when it leaves the sphere is \(\sin^{-1} \frac{2}{3}\). At this point \(v^2 = 2gr(1-\sin \theta) = \frac{2}{3}gr\) Again, considering energy, the initial energy is \(mgr\). The final energy is \(-mgr + \frac12mu_x^2 + \frac12mu_y^2\) When the particle leaves the surface it has speed \(v= \frac23 gr\), so the component \(u_x = \sqrt{v}\sin \theta\). By conservation of energy therefore: \begin{align*} && mgr &= -mgr + \frac12mu_x^2 + \frac12mu_y^2 \\ \Rightarrow && \frac12 u_y^2 &= 2gr - \frac12 u_x^2 \\ &&&= 2gr - \frac12 (\sqrt{v} \sin \theta)^2 \\ &&&= 2gr - \frac12 \frac23gr \sin^2 \theta \\ &&&= 2gr - \frac13gr \frac{4}{9} \\ &&&= \frac{50}{27}gr \\ \Rightarrow && u_y &= \frac{10}{3\sqrt{3}}\sqrt{gr} \end{align*} Therefore vertical component of momentum is \(\displaystyle \frac{10}{3\sqrt{3}}\sqrt{gr}m\)