IPhO 1995 · Theory problem 3

Cylindrical Buoy

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(a) (3 marks)

A buoy consists of a solid cylinder, radius \(a\), length \(l\), made of lightweight material of uniform density \(d\) with a uniform rigid rod protruding directly outwards from the bottom halfway along the length. The mass of the rod is equal to that of the cylinder, its length is the same as the diameter of the cylinder and the density of the rod is greater than that of seawater. This buoy is floating in sea-water of density \(\rho\).

In equilibrium derive an expression relating the floating angle \(\alpha\), as drawn, to \(d/\rho\). Neglect the volume of the rod.

Figure 1

(b) (4 marks)

If the buoy, due to some perturbation, is depressed vertically by a small amount \(z\), it will experience a nett force, which will cause it to begin oscillating vertically about the equilibrium floating position. Determine the frequency of this vertical mode of vibration in terms of \(\alpha\), \(g\) and \(a\), where \(g\) is the acceleration due to gravity. Assume the influence of water motion on the dynamics of the buoy is such as to increase the effective mass of the buoy by a factor of one third. You may assume that \(\alpha\) is not small.

Figure 2

(c) (8 marks)

In the approximation that the cylinder swings about its horizontal central axis, determine the frequency of swing again in terms of \(g\) and \(a\). Neglect the dynamics and viscosity of the water in this case. The angle of swing is assumed to be small.

Figure 3

(d) (5 marks)

The buoy contains sensitive accelerometers which can measure the vertical and swinging motions and can relay this information by radio to shore. In relatively calm waters it is recorded that the vertical oscillation period is about 1 second and the swinging oscillation period is about 1.5 seconds. From this information, show that the floating angle \(\alpha\) is about \(90^\circ\) and thereby estimate the radius of the buoy and its total mass, given that the cylinder length \(l\) equals \(a\).

[You may take it that \(\rho\simeq1000\ \mathrm{kg\,m^{-3}}\) and \(g\simeq9.8\ \mathrm{m\,s^{-2}}\).]

Official English text of the problem.

Answers

  • (a) With \(\alpha\) in radians, \[ \boxed{\frac d\rho=\frac{\alpha-\sin\alpha\cos\alpha}{2\pi}}. \]

  • (b) The vertical oscillation frequency is \[ \boxed{f_z=\frac{1}{2\pi} \sqrt{\frac{3g\sin\alpha}{2a(\alpha-\sin\alpha\cos\alpha)}}}. \]

  • (c) The swinging frequency is \[ \boxed{f_\theta=\frac{1}{2\pi}\sqrt{\frac{12g}{29a}}}. \]

  • (d) \[ \boxed{\alpha\simeq90^\circ,\qquad a\simeq0.23\ \mathrm m,\qquad M\simeq20\ \mathrm{kg}}. \] Using the nominal periods exactly gives \(\alpha\simeq91.1^\circ\).

Solution

(a)

Let the mass of the cylinder be \(m\). The rod has the same mass, so \[ m=d\pi a^2l,\qquad M=2m. \]

The submerged cross-section is a circular segment subtending an angle \(2\alpha\) at the cylinder’s centre. Its area is the sector area minus the triangle area: \[ A_{\rm sub} =\frac12a^2(2\alpha)-\frac12a^2\sin2\alpha =a^2(\alpha-\sin\alpha\cos\alpha). \] Here and below, angles in these expressions are in radians.

Define \[ S(\alpha)=\alpha-\sin\alpha\cos\alpha. \] Neglecting the rod’s volume, the displaced volume is \(a^2lS(\alpha)\). By Archimedes’ principle, equilibrium requires \[ \rho g a^2lS(\alpha)=2d\pi a^2lg. \] Consequently, \[ \boxed{\frac d\rho=\frac{S(\alpha)}{2\pi} =\frac{2\alpha-\sin2\alpha}{4\pi}}. \]

(b)

At equilibrium, the width of the cylinder’s cross-section at the water surface is \[ w=2a\sin\alpha. \] Thus the horizontal area cut by the water surface is \(wl=2al\sin\alpha\).

Take the displacement \(z\) as positive downwards. To first order in \(z\), the increase in displaced volume is \[ \Delta V=2al\sin\alpha\,z. \] The equilibrium buoyancy already balances the weight. The additional buoyancy therefore gives the restoring force \[ F_z=-\rho g\Delta V=-2\rho gal\sin\alpha\,z. \]

The stated one-third increase in effective mass gives \[ M_{\rm eff}=M+\frac13M=\frac43M. \] From part (a), the actual mass satisfies \[ M=\rho a^2lS(\alpha). \] Hence the equation of motion is \[ \frac43\rho a^2lS(\alpha)\,\ddot z +2\rho gal\sin\alpha\,z=0. \]

This is simple harmonic motion, with \[ \omega_z^2=\frac{3g\sin\alpha}{2aS(\alpha)}. \] The frequency in cycles per second is therefore \[ \boxed{f_z=\frac{1}{2\pi} \sqrt{\frac{3g\sin\alpha} {2a(\alpha-\sin\alpha\cos\alpha)}}}. \]

Only the vertical displacement has been linearized; no small-\(\alpha\) approximation has been made.

(c)

Use the prescribed approximation that the cylinder rotates about its fixed horizontal central axis.

Because the cylinder has a circular cross-section, rotation about this axis leaves the submerged geometry unchanged. The buoyancy force acts vertically through the central axis and produces no torque about it. The cylinder’s own weight also acts through this axis. The restoring torque therefore comes entirely from the rod’s weight.

The rod extends from a distance \(a\) to a distance \(3a\) from the axis. Its centre of mass is consequently at distance \(2a\). For a displacement \(\theta\) from the downward vertical, \[ \tau=-2mga\sin\theta\simeq-2mga\,\theta, \] where the last approximation uses \(|\theta|\ll1\).

The cylinder’s moment of inertia about its central axis is \[ I_{\rm cyl}=\frac12ma^2. \] For the thin uniform rod, the parallel-axis theorem gives \[ I_{\rm rod} =\frac{1}{12}m(2a)^2+m(2a)^2 =\frac{13}{3}ma^2. \] Thus \[ I=I_{\rm cyl}+I_{\rm rod} =\frac{29}{6}ma^2. \]

Neglecting water dynamics and viscosity as instructed, \[ I\ddot\theta=-2mga\,\theta, \] or \[ \ddot\theta+\frac{12g}{29a}\theta=0. \] Therefore, \[ \boxed{\omega_\theta=\sqrt{\frac{12g}{29a}}, \qquad f_\theta=\frac{1}{2\pi}\sqrt{\frac{12g}{29a}}}. \]

(d)

Let the vertical and swinging periods be \(T_z\) and \(T_\theta\). Taking the ratio of the squared frequencies eliminates \(a\) and \(g\): \[ \left(\frac{T_\theta}{T_z}\right)^2 =\frac{\omega_z^2}{\omega_\theta^2} =\frac{29\sin\alpha}{8S(\alpha)}. \]

Using \(T_z\simeq1\ \mathrm s\) and \(T_\theta\simeq1.5\ \mathrm s\), \[ \frac{S(\alpha)}{\sin\alpha} \simeq\frac{29}{8(1.5)^2} =\frac{29}{18} \simeq1.611. \] At \(\alpha=90^\circ=\pi/2\), \[ \frac{S(\alpha)}{\sin\alpha}=\frac{\pi}{2}\simeq1.571, \] and the predicted period ratio is \[ \frac{T_\theta}{T_z} =\sqrt{\frac{29}{4\pi}} \simeq1.519, \] consistent with the approximate measured ratio \(1.5\). Solving with the nominal periods gives \(\alpha\simeq91.15^\circ\). Thus \[ \boxed{\alpha\simeq90^\circ}, \] so approximately half the cylinder is submerged.

The swinging period gives the radius directly: \[ T_\theta=2\pi\sqrt{\frac{29a}{12g}}, \] hence \[ a=\frac{3gT_\theta^2}{29\pi^2} =\frac{3(9.8)(1.5)^2}{29\pi^2}\ \mathrm m \simeq0.231\ \mathrm m. \] Therefore, \[ \boxed{a\simeq0.23\ \mathrm m}. \]

Since \(l=a\), equilibrium gives the total mass as \[ M=\rho a^3S(\alpha). \] For \(\alpha\simeq\pi/2\), \[ M\simeq\frac{\pi}{2}\rho a^3 \simeq\frac{\pi}{2}(1000)(0.231)^3\ \mathrm{kg} \simeq19.4\ \mathrm{kg}. \] Using the nominal-period value \(\alpha\simeq91.15^\circ\) gives \(M\simeq19.9\ \mathrm{kg}\). At the accuracy warranted by the stated periods, \[ \boxed{M\simeq20\ \mathrm{kg}}. \]

The solution was checked against the official solution.