Problems

An octopus propels itself horizontally from rest by jet propulsion: while at rest it sucks a volume \(V\) of water into an internal cavity. It then propels itself by squirting this water out at a constant rate of \(Q\) units of volume per unit time, through a nozzle of cross- sectional area \(A\). Let the mass of the octopus plus the water contained in the cavity at time \(t\) after an ejection begins, be \(m(t)\), let its speed be \(u(t)\) and let the drag force exerted on the octopus by the surrounding water be \(D(t)\). Show that, during ejection, \begin{align} m\frac{du}{dt} = \frac{\rho Q^2}{A} - D \end{align} where \(\rho\) is the density of water. Given that \(D = ku^2\) (\(k\) constant), show that the speed attained by the octopus at the end of ejection is \begin{align} u_1 = Q\left(\frac{\rho}{kA}\right)^{\frac{1}{2}}\frac{\alpha-1}{\alpha+1} \end{align} where \begin{align} \alpha = \left(1+\frac{\rho V}{m_0}\right)^{2\frac{k}{(\rho A)^{\frac{1}{2}}}} \end{align} and \(m_0\) is the value of \(m\) before intake of the volume \(v\). State a condition to be satisfied by \(\alpha\) for the drag to be negligible during water ejection. Find the time after the end of ejection at which \(u = u_1/10\).

A uniform block of ice of mass \(m\) has the form of a circular cylinder of radius \(a\) and moment of inertia \(\frac{1}{2}ma^2\) about its axis. It is contained in a close-fitting, thin-walled cylindrical drum with the same radius, mass \(M\) and moment of inertia \(\frac{1}{2}Ma^2\) about its axis. The surroundings are sufficiently warm that a very thin layer of water forms between the ice and the drum, so that slipping can occur between them, resisted by a couple equal to \(k\) times the relative angular velocity. Starting from a state in which both drum and ice are at rest, the drum rolls without slipping down the line of greatest slope of an inclined plane which makes an angle \(\alpha\) with the horizontal. Show that the difference, \(\omega\), between the angular velocities of the ice and the drum tends to the constant value \(\omega_0\), where \begin{align} \omega_0 = \frac{mga\sin\alpha (m + M)}{k(3m + 4M)} \end{align} and that \(\omega = (1 - e^{-t})\omega_0\) after a time \begin{align} \frac{ma^2(m + 2M)}{k(3m + 4M)} \end{align}

A heavy uniform circular cylinder of radius \(r\) rests on a rough horizontal plane. A heavy uniform rod of length \(l\) lies across it, touching the plane at its end \(A\) and touching the cylinder tangentially at a point \(B\). The rod lies in a vertical plane perpendicular to the axis of the cylinder, and its centre of gravity lies between \(A\) and \(B\). The coefficient of friction at both points of contact on the rod is \(\mu\) with \(0 < \mu < 1\). Friction is limiting at both \(A\) and \(B\), and the cylinder does not slip or roll on the plane. Show by a geometrical method or otherwise that \[l\mu^3 + r\mu^2 - l\mu + r = 0.\]

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A bead is free to slide on a curved rough wire which lies in a vertical plane. The coefficient of friction between the bead and the wire is \(\mu\). When the bead is projected down the wire with speed \(v\), it continues to move at a constant speed. Use a coordinate system \((x,y)\) as shown, in which the \(x\)-axis is inclined at an angle \(\alpha\) above the horizontal with \(\tan\alpha = \mu\), to show that the wire satisfies \[\frac{dx}{ds} = \tanh(c-as), \quad \frac{dy}{ds} = \text{sech}(c-as),\] where \(a = g/(v^2\sin\alpha)\), \(c\) is constant and \(s\) is the arc length along the wire. With the aid of the substitution \[\sin u = \tanh(c-as), \quad \cos u = \text{sech}(c-as),\] or otherwise, show that when the origin is suitably chosen, the wire must form part of the curve with equation \[ax = \ln\cos ay.\]