Problems

The points \(A\) and \(B\) have position vectors \(\bf i +j+k\) and \(5{\bf i} - {\bf j} -{\bf k}\), respectively, relative to the origin \(O\). Find \(\cos2\alpha\), where \(2\alpha\) is the angle \(\angle AOB\).

1. The line \(L _1\) has equation \({\bf r} =\lambda(m{\bf i}+n {\bf j} + p{\bf k})\). Given that \(L _1\) is inclined equally to \(OA\) and to \(OB\), determine a relationship between \(m\), \(n\) and~\(p\). Find also values of \(m\), \(n\) and~\(p\) for which \(L _1\) is the angle bisector of \(\angle AOB\).
2. The line \(L _2\) has equation \({\bf r} =\mu(u{\bf i}+v {\bf j} + w{\bf k})\). Given that \( L _2\) is inclined at an angle \(\alpha\) to \(OA\), where \(2\alpha = \angle AOB\), determine a relationship between \(u\), \(v\) and \(w\). Hence describe the surface with Cartesian equation \(x^2+y^2+z^2 =2(yz+zx+xy)\).

A right circular cone has base radius \(r\), height \(h\) and slant length \(\ell\). Its volume \(V\), and the area \(A\) of its curved surface, are given by \[ V= \tfrac13 \pi r^2 h \,, \ \ \ \ \ \ \ A = \pi r\ell\,. \]

1. Given that \(A\) is fixed and \(r\) is chosen so that \(V\) is at its stationary value, show that \(A^2 = 3\pi^2r^4\) and that \(\ell =\sqrt3\,r\).
2. Given, instead, that \(V\) is fixed and \(r\) is chosen so that \(A\) is at its stationary value, find \(h\) in terms of \(r\).

A solid right circular cone, of mass \(M\), has semi-vertical angle \(\alpha\) and smooth surfaces. It stands with its base on a smooth horizontal table. A particle of mass \(m\) is projected so that it strikes the curved surface of the cone at speed \(u\). The coefficient of restitution between the particle and the cone is \(e\). The impact has no rotational effect on the cone and the cone has no vertical velocity after the impact.

1. The particle strikes the cone in the direction of the normal at the point of impact. Explain why the trajectory of the particle immediately after the impact is parallel to the normal to the surface of the cone. Find an expression, in terms of \(M\), \(m\), \(\alpha\), \(e\) and \(u\), for the speed at which the cone slides along the table immediately after impact.
2. If instead the particle falls vertically onto the cone, show that the speed \(w\) at which the cone slides along the table immediately after impact is given by \[ w= \frac{mu(1+e)\sin\alpha\cos\alpha}{M+m\cos^2\alpha}\,. \] Show also that the value of \(\alpha\) for which \(w\) is greatest is given by \[ \cos \alpha = \sqrt{ \frac{M}{2M+m}}\ . \]

A uniform right circular cone of mass \(m\) has base of radius \(a\) and perpendicular height \(h\) from base to apex. Show that its moment of inertia about its axis is \({3\over 10} ma^2\), and calculate its moment of inertia about an axis through its apex parallel to its base. \newline[{\em Any theorems used should be stated clearly.}] The cone is now suspended from its apex and allowed to perform small oscillations. Show that their period is $$ 2\pi\sqrt{ 4h^2 + a^2\over 5gh} \,. $$ \newline[{\em You may assume that the centre of mass of the cone is a distance \({3\over 4}h\) from its apex.}]

The diagram shows a coffee filter consisting of an inverted hollow right circular cone of height \(H\) cm and base radius \(a\) cm. \noindent

Prove that the area of the zone of the surface of a sphere between two parallel planes cutting the sphere is given by \[ 2\pi\times(\mbox{radius of sphere})\times(\mbox{perpendicular distance between the planes}). \] A tangent from the origin \(O\) to the curve with cartesian equation \[ (x-c)^{2}+y^{2}=a^{2}, \] where \(a\) and \(c\) are positive constants with \(c>a,\) touches the curve at \(P\). The \(x\)-axis cuts the curve at \(Q\) and \(R\), the points lying in the order \(OQR\) on the axis. The line \(OP\) and the arc \(PR\) are rotated through \(2\pi\) radians about the line \(OQR\) to form a surface. Find the area of this surface.

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Solution:

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We can choose a coordinate frame where the parallel planes are parallel to the \(y-z\) axis. Then we can compute the surface area as an integral of the surface of revolution for \(x^2 + y^2 = r^2\). Using \(y = r \sin t, x = r \cos t\) we have: \begin{align*} S &= 2\pi\int_{\cos^{-1}a}^{\cos^{-1}b}y \sqrt{\left ( \frac{\d x}{\d t} \right)^2+\left ( \frac{\d y}{\d t} \right)^2} \d t \\ &=2\pi\int_{\cos^{-1}a}^{\cos^{-1}b} r^2 \sin t \d t \\ &= 2\pi \cdot r^2 \cdot (a - b) \\ &= 2 \pi \cdot r \cdot (ra-rb) \\ &= 2\pi\times(\mbox{radius of sphere})\times(\mbox{perpendicular distance between the planes}). \end{align*}

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We can view this surface as a sphere missing a cap of height \(XQ\) and adding a cone of slant height \(OP\) and radius \(PX\) The centre of the circle is at \((c,0)\) and \(OP^2 + a^2 = c^2 \Rightarrow OP = \sqrt{c^2-a^2}\) Since \(OPC \sim OXP\) we must have that \(\frac{OX}{OP} = \frac{OP}{OC} \Rightarrow OX = \frac{c^2-a^2}{c}\) and \(\frac{PX}{OP} = \frac{CP}{OC} \Rightarrow PX = \frac{a}{c}\sqrt{c^2-a^2}\) \(QX = OX - OQ = \frac{c^2-a^2}{c}-(c-a) = \frac{ac-a^2}{c}\) Therefore the surface area is: \begin{align*} S &= 4 \pi a^2 - 2\pi \cdot a \cdot QX+ \pi PX \cdot OP \\ &= 4 \pi a^2 - 2\pi a \cdot \frac{ac-a^2}{c}+\pi \frac{a}{c}\sqrt{c^2-a^2}\cdot \sqrt{c^2-a^2} \\ &= 4\pi a^2 -2\pi \frac{a^2c-a^3}{c}+\pi \frac{ac^2-a^3}{c} \\ &= \pi a \frac{(a+c)^2}{c} \end{align*}

A thin uniform elastic band of mass \(m,\) length \(l\) and modulus of elasticity \(\lambda\) is pushed on to a smooth circular cone of vertex angle \(2\alpha,\) in such a way that all elements of the band are the same distance from the vertex. It is then released from rest. Let \(x(t)\) be the length of the band at time \(t\) after release, and let \(t_{0}\) be the time at which the band becomes slack. Assuming that a small element of the band which subtends an angle \(\delta\theta\) at the axis of the cone experiences a force, due to the tension \(T\) in the band, of magnitude \(T\delta\theta\) directed towards the axis, and ignoring the effects of gravity, show that \[ \frac{\mathrm{d}^{2}x}{\mathrm{d}t^{2}}+\frac{4\pi^{2}\lambda}{ml}(x-l)\sin^{2}\alpha=0,\qquad(0< t< t_{0}). \] Find the value of \(t_{0}.\)

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Solution:

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\begin{align*} \text{N2}(\nwarrow): && T\delta \theta \sin \alpha &= -m\frac{\delta \theta}{2\pi} \ddot{d} \end{align*} Notice that \(r = d \sin \alpha\) and \(x = 2 \pi r\), so \(x = 2\pi d \sin \alpha\) and \(\ddot{x} = 2\pi \sin \alpha \ddot{d} \Rightarrow \ddot{d} = \ddot{x} \frac{1}{2 \pi \sin \alpha}\) Notice also that \(T = \frac{\lambda}{l}(x-l)\) so. \begin{align*} && \frac{m}{4 \pi^2 \sin\alpha} \ddot{x} &= -\frac{\lambda}{l}(x-l) \sin\alpha \\ \Rightarrow && \frac{\mathrm{d}^{2}x}{\mathrm{d}t^{2}}+\frac{4\pi^{2}\lambda}{ml}(x-l)\sin^{2}\alpha&=0 \end{align*} The solution to the differential equation we have is: \begin{align*} && x(t) &= A \sin \left (\sqrt{\frac{4 \pi^2 \lambda}{ml}\sin^2 \alpha} \cdot t \right) + B \sin \left (\sqrt{\frac{4 \pi^2 \lambda}{ml}\sin^2 \alpha} \cdot t \right) + l \\ &&&= A \sin \left (2 \pi \sin \alpha\sqrt{\frac{ \lambda}{ml}} \cdot t \right) +B \sin \left (2 \pi \sin \alpha\sqrt{\frac{ \lambda}{ml}} \cdot t \right) + l\\ && \dot{x}(0) = 0 \\ \Rightarrow && B &= 0 \\ && x(t) &= (x(0)-l) \sin \left (2 \pi \sin \alpha\sqrt{\frac{ \lambda}{ml}} \cdot t \right) + l \\ && x(t_0) &= l \\ \Rightarrow && t_0 &= \frac{1}{4\sin \alpha} \sqrt{\frac{ml}{\lambda}} \end{align*}