IPhO 1982 · Theory problem 2

Oscillating coat hanger

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A (suitably made) wire coat hanger can perform small amplitude oscillations in the plane of the figure around the equilibrium positions shown. In positions a) and b) the long side is horizontal. The other two sides have equal length. The period of oscillation is the same in all cases.

What is the location of the center of mass, and how long is the period?

Figure 1

The figure does not contain any information beyond the dimensions given. Nothing is known, e.g., concerning the detailed distribution of mass.

Official English text of the problem.

Answers

  • Center of mass: Halfway along the triangle’s altitude, on its symmetry axis: \(5.0\,\mathrm{cm}\) from the long side and \(5.0\,\mathrm{cm}\) from the opposite vertex.
  • Common period: \[ \boxed{T=2\pi\sqrt{\frac{0.050+\sqrt{0.210^2+0.050^2}}{g}} \simeq 1.03\,\mathrm{s}}, \] where the lengths are in metres and \(g=9.81\,\mathrm{m\,s^{-2}}\).

Solution

Treat the hanger as a rigid physical pendulum, with a fixed suspension point in each configuration and negligible friction. Let its mass be \(M\), and let \(I_G=Mk^2\) be its moment of inertia about the axis through its center of mass perpendicular to its plane.

If the center of mass lies a distance \(d\) below the suspension point, the parallel-axis theorem and gravitational torque give \[ M(k^2+d^2)\ddot{\theta}=-Mgd\sin\theta. \] For small oscillations, \(\sin\theta\simeq\theta\), so \[ T=2\pi\sqrt{\frac{k^2+d^2}{gd}}. \] Thus a common period means that \[ \ell\equiv\frac{gT^2}{4\pi^2} =d+\frac{k^2}{d} \] has the same value for all three suspension points.

Location of the center of mass.

Denote the vertex opposite the long side by \(A\), the midpoint of the long side by \(B\), and the corner used for suspension in c) by \(C\). The given dimensions are \[ AB=h=0.100\,\mathrm{m}, \qquad BC=a=0.210\,\mathrm{m}. \]

In position a), equilibrium requires the center of mass \(G\) to lie vertically below \(A\). Since the triangle is isosceles and its long side is horizontal, \(G\) lies on the line \(AB\). In position b), the suspension point must likewise lie vertically above \(G\), so it is \(B\).

Both equilibria must be stable. Consequently, \(G\) lies between \(A\) and \(B\). Writing \(BG=x\), we have \[ 0<x<h, \qquad d_a=h-x,\qquad d_b=x,\qquad d_c=\sqrt{a^2+x^2}. \]

Equality of the periods in a) and b) gives \[ h-x+\frac{k^2}{h-x} =x+\frac{k^2}{x}, \] or \[ (h-2x)\bigl[x(h-x)-k^2\bigr]=0. \] Hence either \[ x=\frac h2 \quad\text{or}\quad k^2=x(h-x). \]

The second possibility would imply \[ \ell=x+\frac{x(h-x)}{x}=h. \] But in configuration c), \[ \ell=d_c+\frac{k^2}{d_c}>d_c\ge a =0.210\,\mathrm{m}>h. \] This contradicts the common period. Therefore \[ \boxed{x=\frac h2=0.050\,\mathrm{m}}. \]

The center of mass is thus halfway between the long side’s midpoint and the opposite vertex. This conclusion does not require a uniform mass distribution.

Period of oscillation.

With this center of mass, the suspension distances are \[ d_a=d_b=d=0.050\,\mathrm{m}, \qquad d_c=r=\sqrt{0.210^2+0.050^2} =0.215870\ldots\,\mathrm{m}. \]

Equating the periods for distances \(d\) and \(r\), \[ d+\frac{k^2}{d}=r+\frac{k^2}{r}. \] Since \(d\ne r\), this yields \[ k^2=dr. \] The common equivalent pendulum length is therefore \[ \ell=d+\frac{dr}{d}=d+r =0.265870\ldots\,\mathrm{m}. \] Finally, \[ \boxed{ T=2\pi\sqrt{\frac{d+r}{g}} =2\pi\sqrt{\frac{0.265870\ldots}{9.81}} \simeq1.03\,\mathrm{s}. } \]

The solution was checked against the official solution.