Problems

In this question, \(\mathbf{i}\) and \(\mathbf{j}\) are perpendicular unit vectors and \(\mathbf{j}\) is vertically upwards. A smooth hemisphere of mass \(M\) and radius \(a\) rests on a smooth horizontal table with its plane face in contact with the table. The point \(A\) is at the top of the hemisphere and the point \(O\) is at the centre of its plane face. Initially, a particle \(P\) of mass \(m\) rests at \(A\). It is then given a small displacement in the positive \(\mathbf{i}\) direction. At a later time \(t\), when the particle is still in contact with the hemisphere, the hemisphere has been displaced by \(-s\mathbf{i}\) and \(\angle AOP = \theta\).

1. Let \(\mathbf{r}\) be the position vector of the particle at time \(t\) with respect to the initial position of \(O\). Write down an expression for \(\mathbf{r}\) in terms of \(a\), \(\theta\) and \(s\) and show that $$\dot{\mathbf{r}} = (a\dot{\theta} \cos \theta - \dot{s})\mathbf{i} - a\dot{\theta} \sin \theta \mathbf{j}.$$ Show also that $$\dot{s} = (1 - k)a\dot{\theta} \cos \theta,$$ where \(k = \frac{M}{m + M}\), and deduce that $$\dot{\mathbf{r}} = a\dot{\theta}(k \cos \theta \mathbf{i} - \sin \theta \mathbf{j}).$$
2. Show that $$a\dot{\theta}^2 \left(k \cos^2 \theta + \sin^2 \theta\right) = 2g(1 - \cos \theta).$$
3. At time \(T\), when \(\theta = \alpha\), the particle leaves the hemisphere. By considering the component of \(\ddot{\mathbf{r}}\) parallel to the vector \(\sin \theta \mathbf{i} + k \cos \theta \mathbf{j}\), or otherwise, show that at time \(T\) $$a\dot{\theta}^2 = g \cos \alpha.$$ Find a cubic equation for \(\cos \alpha\) and deduce that \(\cos \alpha > \frac{2}{3}\).

Show Solution

---

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 \(\cos \alpha > \frac23\)

A solid figure is composed of a uniform solid cylinder of density \(\rho\) and a uniform solid hemisphere of density \(3\rho\). The cylinder has circular cross-section, with radius \(r\), and height \(3r\), and the hemisphere has radius \(r\). The flat face of the hemisphere is joined to one end of the cylinder, so that their centres coincide. The figure is held in equilibrium by a force \(P\) so that one point of its flat base is in contact with a rough horizontal plane and its base is inclined at an angle \(\alpha\) to the horizontal. The force \(P\) is horizontal and acts through the highest point of the base. The coefficient of friction between the solid and the plane is \(\mu\). Show that \[\mu \ge \left\vert \tfrac98 -\tfrac12 \cot\alpha\right\vert\,. \]

Show Solution

---

Solution: The centre of mass of the sphere will be at \((0, \frac{3}{2}r)\) and the centre of mass of the hemisphere will be at \((0, 3r + \frac38r)\), their masses will be \(3\pi r^3 \cdot \rho \) and \(\frac23 \pi r^3 \cdot 3\rho \), meaning the center of mass will be \(\frac{\frac92r + \frac{27}{8} \cdot 2r}{3 + 2} = \frac{45/4}{5}r = \frac{9}{4}r\) above the center of the base.

+ - ↺

\begin{align*} \text{N2}(\uparrow): && R -mg &= 0 \\ \overset{\curvearrowright}{X}: && P\cdot 2r \sin \alpha + mg (r \cos \alpha -\tfrac94 r\sin \alpha) &= 0 \\ \Rightarrow && P &= mg(\tfrac98 - \tfrac12 \cot \alpha) \\ \text{N2}(\rightarrow): && |F| &= |P| \\ (|F| \leq \mu R): && mg|\tfrac98 - \tfrac12 \cot \alpha| & \leq \mu mg \\ \Rightarrow && |\tfrac98 - \tfrac12 \cot \alpha| &\leq \mu \end{align*}

The base of a non-uniform solid hemisphere, of mass \(M,\) has radius \(r.\) The distance of the centre of gravity, \(G\), of the hemisphere from the base is \(p\) and from the centre of the base is \(\sqrt{p^2 + q^2} \,\). The hemisphere rests in equilibrium with its curved surface on a horizontal plane. A particle of mass \(m,\,\) where \(m\) is small, is attached to \(A\,\), the lowest point of the circumference of the base. In the new position of equilibrium, find the angle, \(\alpha\), that the base makes with the horizontal. The particle is removed and attached to the point \(B\) of the base which is at the other end of the diameter through \(A\,\). In the new position of equilibrium the base makes an angle \({\beta}\) with the horizontal. Show that $$\tan(\alpha-\beta)= \frac{2mMrp} {M^2\left(p^2+q^2\right)-m^2r^2}\;.$$

Show Solution

---

Solution:

+ - ↺

In the coordinate system where \((0,0)\) is the centre base of the hemisphere, \(G\) is at \((p, q)\). Once the mass is attached at \(A\), the new centre of mass will satisfy: \(M \begin{pmatrix} p \\ q \end{pmatrix} + m \begin{pmatrix} r \\ 0 \end{pmatrix} = (M+m)\bar{x} \Rightarrow \bar{x} = \frac{1}{M+m} \begin{pmatrix} Mp+mr \\ Mq \end{pmatrix}\) The angle between the horizontal and \(AB\), \(\alpha\) will satisfy: $$\tan \alpha = \frac{Mp + mr}{Mq}$$ Similarly, when the mass is attached at \(B\), the new centre of mass will satisfy: \(M \begin{pmatrix} p \\ q \end{pmatrix} + m \begin{pmatrix} -r \\ 0 \end{pmatrix} = (M+m)\bar{x} \Rightarrow \bar{x} = \frac{1}{M+m} \begin{pmatrix} Mp-mr \\ Mq \end{pmatrix}\) The angle between the horizontal and \(AB\), \(\beta\) will satisfy: $$\tan \beta = \frac{Mp - mr}{Mq}$$ We are trying to find: \begin{align*} \tan \l \alpha - \beta \r &= \frac{\tan \alpha - \tan \beta}{1+ \tan \alpha \tan \beta} \\ &= \frac{\frac{Mp + mr}{Mq} - \frac{Mp - mr}{Mq}}{1 + \frac{Mp + mr}{Mq} \frac{Mp - mr}{Mq}} \\ &= \frac{(Mp + mr)Mq - (Mp - mr)Mq}{M^2q^2 + (Mp + mr)(Mp - mr)} \\ &= \frac{2Mmrp}{M^2(q^2+p^2) -m^2r^2} \\ \end{align*}

A child's toy consists of a solid cone of height \(\lambda a\) and a solid hemisphere of radius \(a\), made out of the same uniform material and fastened together so that their plane faces coincide. (Thus the diameter of the hemisphere is equal to that of the base of the cone.) Show that if \(\lambda < \sqrt{3}\) the toy will always move to an upright position if placed with the surface of the hemisphere on a horizontal table, but that if \(\lambda > \sqrt{3}\) the toy may overbalance. Show, however, that if the toy is placed with the surface of the cone touching the table it will remain there whatever the value of \(\lambda\). [The centre of gravity of a uniform solid cone of height \(h\) is a height \(h/4\) above its base. The centre of gravity of a uniform solid hemisphere of radius \(a\) is at distance \(3a/8\) from the centre of its base.]

Show Solution

---

Solution:

+ - ↺

By symmetry the centre of mass will lie on the main axis. Taking the plane faces as \(x = 0\) we have the following centers of mass: \begin{align*} && \text{COM} && \text{Mass} \\ \text{Hemisphere} && -\frac{3a}{8} && \frac{2\pi a^3}{3} \\ \text{Cone} && \frac{\lambda a}{4} && \frac{\lambda \pi a^3}{3} \\ \text{Toy} && \bar{x} && \frac{(\lambda + 2)\pi a^3}{3} \\ \end{align*} Therefore, \begin{align*} && \frac{(\lambda + 2)\pi a^3}{3} \cdot \bar{x} &= -\frac{3a}{8} \cdot \frac{2\pi a^3}{3} + \frac{\lambda a}{4} \cdot \frac{\lambda \pi a^3}{3} \\ \Rightarrow && (\lambda + 2) \bar{x} &= \frac{(\lambda^2 -3)a}{4} \end{align*} Therefore the centre of mass will be inside the hemisphere (and it will always move to an upright position) iff \(\bar{x} < 0 \Leftrightarrow \lambda < \sqrt{3}\).

+ - ↺

For the toy to topple from this position, \(\bar{x}\) must be longer than it would need to be to form a right-angled triangle with the vertical at the plane face. The angle at this point will be \(\theta\), so we need: \(\bar{x} > a\tan \theta = a \frac{a}{\lambda a} = \frac{a}{\lambda}\) \begin{align*} && \bar{x} &> \frac{a}{\lambda} \\ \Leftrightarrow && \frac{(3-\lambda^2)a}{4(\lambda + 2)} &> \frac{a}{\lambda} \\ \Leftrightarrow && {(3-\lambda^2)\lambda} &> {4(\lambda + 2)} \\ \Leftrightarrow && -\lambda^3-\lambda -8 &> 0 \\ \end{align*} Contradiction! Therefore it can never topple when laid on its side.