IPhO 1998 · Theory problem 1

Rolling of a hexagonal prism

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Consider a long, solid, rigid, regular hexagonal prism like a common type of pencil (Figure 1.1). The mass of the prism is \(M\) and it is uniformly distributed. The length of each side of the cross-sectional hexagon is \(a\). The moment of inertia \(I\) of the hexagonal prism about its central axis is

\[ I=\frac{5}{12}Ma^2 \tag{1.1} \]

Figure 1

Figure 1.1: A solid prism with the cross section of a regular hexagon.

The moment of inertia \(I'\) about an edge of the prism is

\[ I'=\frac{17}{12}Ma^2 \tag{1.2} \]

a) (3.5 points) The prism is initially at rest with its axis horizontal on an inclined plane which makes a small angle \(\theta\) with the horizontal (Figure 1.2). Assume that the surfaces of the prism are slightly concave so that the prism only touches the plane at its edges. The effect of this concavity on the moment of inertia can be ignored. The prism is now displaced from rest and starts an uneven rolling down the plane. Assume that friction prevents any sliding and that the prism does not lose contact with the plane. The angular velocity just before a given edge hits the plane is \(\omega_i\) while \(\omega_f\) is the angular velocity immediately after the impact.

Show that we may write

\[ \omega_f=s\omega_i \tag{1.3} \]

and write the value of the coefficient \(s\) on the answer sheet.

¹Authors: Leó Kristjánsson and Thorsteinn Vilhjálmsson

Figure 2

Figure 1.2: A hexagonal prism lying on an inclined plane.

b) (1 point) The kinetic energy of the prism just before and after impact is similarly \(K_i\) and \(K_f\).

Show that we may write

\[ K_f=rK_i \tag{1.4} \]

and write the value of the coefficient \(r\) on the answer sheet.

c) (1.5 points) For the next impact to occur \(K_i\) must exceed a minimum value \(K_{i,min}\) which may be written in the form

\[ K_{i,min}=\delta Mga \tag{1.5} \]

where \(g=9.81\ \mathrm{m/s^2}\) is the acceleration of gravity.

Find the coefficient \(\delta\) in terms of the slope angle \(\theta\) and the coefficient \(r\). Write your answer on the answer sheet. (Use the algebraic symbol \(r\), not its value.)

d) (2 points) If the condition of part (c) is satisfied, the kinetic energy \(K_i\) will approach a fixed value \(K_{i,0}\) as the prism rolls down the incline.

Given that the limit exists, show that \(K_{i,0}\) may be written as:

\[ K_{i,0}=\kappa Mga \tag{1.6} \]

and write the coefficient \(\kappa\) in terms of \(\theta\) and \(r\) on the answer sheet.

e) (2 points) Calculate, to within \(0.1^\circ\), the minimum slope angle \(\theta_0\) for which the uneven rolling, once started, will continue indefinitely. Write your numerical answer on the answer sheet.

Official English text of the problem.

Answers

  • a) \(\displaystyle s=\frac{11}{17}\approx0.6471.\)
  • b) \(\displaystyle r=\frac{121}{289}\approx0.4187.\)
  • c) \(\displaystyle \delta=\frac{1-\cos(\pi/6-\theta)}{r}.\)
  • d) \(\displaystyle \kappa=\frac{\sin\theta}{1-r}.\)
  • e) \(\displaystyle \theta_0=6.5724\ldots^\circ\approx\boxed{6.6^\circ}.\)

The coefficients \(s,r,\delta,\kappa\) are dimensionless.

Solution

We treat each impact as instantaneous, neglecting gravity’s impulse during it. Between impacts the prism rotates about its stationary contact edge; the contact force does no work, so mechanical energy is conserved. We retain the exact trigonometric functions throughout.

a) Change of angular velocity at impact

In a cross section, let \(A\) denote the old contact edge, \(B\) the next edge to strike the plane, and \(C\) the centre of mass. For a regular hexagon,

\[ AC=BC=AB=a, \qquad \angle ACB=60^\circ. \]

Immediately before impact, the prism rotates about \(A\), so

\[ \mathbf v_C=\boldsymbol\omega_i\times\overrightarrow{AC}. \]

At impact the old edge releases and the new edge becomes the pivot. The impulsive contact force acts through \(B\), so angular momentum about the fixed line where \(B\) strikes is conserved.

Before impact, the angular momentum about \(B\) comprises rotation about the centre of mass and the centre of mass’s orbital angular momentum:

\[ \begin{aligned} \mathbf L_{B,i} &=I\boldsymbol\omega_i+ \overrightarrow{BC}\times M\mathbf v_C\\ &=\left(I+M\overrightarrow{BC}\cdot\overrightarrow{AC}\right) \boldsymbol\omega_i\\ &=\left(I+\frac12 Ma^2\right)\boldsymbol\omega_i. \end{aligned} \]

After impact, the prism rotates about \(B\), giving

\[ \mathbf L_{B,f}=I'\boldsymbol\omega_f. \]

Consequently,

\[ \left(\frac{5}{12}+\frac12\right)Ma^2\omega_i =\frac{17}{12}Ma^2\omega_f, \]

and therefore

\[ \boxed{\omega_f=\frac{11}{17}\omega_i}, \qquad \boxed{s=\frac{11}{17}}. \]

b) Change of kinetic energy at impact

Immediately before and after impact, the motion is pure rotation about a stationary edge. Both edges have the same moment of inertia \(I'\), so

\[ K_i=\frac12 I'\omega_i^2, \qquad K_f=\frac12 I'\omega_f^2. \]

It follows that

\[ \frac{K_f}{K_i} =\left(\frac{\omega_f}{\omega_i}\right)^2=s^2. \]

Thus

\[ \boxed{K_f=\frac{121}{289}K_i}, \qquad \boxed{r=\frac{121}{289}}. \]

c) Minimum energy needed for the next impact

After impact at \(B\), the centre of mass must rise until it is vertically above \(B\), then descend toward the next impact.

At the instant of impact, \(C\) lies a perpendicular distance \(\sqrt3a/2\) above the plane and a distance \(a/2\) uphill from \(B\). Its vertical height relative to \(B\) is therefore

\[ h_{\mathrm{start}} =\frac{\sqrt3a}{2}\cos\theta+\frac a2\sin\theta =a\cos\left(\frac{\pi}{6}-\theta\right). \]

For the small slopes under consideration, \(0\leq\theta<\pi/6\), the centre of mass subsequently reaches the maximum height

\[ h_{\mathrm{top}}=a. \]

The gravitational potential-energy barrier is

\[ \Delta U =Mga\left[1-\cos\left(\frac{\pi}{6}-\theta\right)\right]. \]

The available kinetic energy just after impact is \(rK_i\). To pass the highest point and reach the next edge, it must satisfy

\[ rK_i>\Delta U. \]

Hence the threshold pre-impact energy is

\[ K_{i,\min} =\frac{Mga}{r} \left[1-\cos\left(\frac{\pi}{6}-\theta\right)\right], \]

so

\[ \boxed{\delta= \frac{1-\cos(\pi/6-\theta)}{r}}. \]

The inequality is strict: at exactly the threshold energy, the prism approaches the unstable balance position without crossing it.

d) Limiting pre-impact kinetic energy

Successive impact configurations are identical apart from a translation by one side length \(a\) down the plane. Thus the centre of mass loses a vertical height \(a\sin\theta\) between impacts.

Let \(E_n\) be the kinetic energy immediately before impact \(n\). Impact reduces it to \(rE_n\), and gravity supplies \(Mga\sin\theta\) before the following impact. Therefore, for successive completed steps,

\[ E_{n+1}=rE_n+Mga\sin\theta. \]

Taking the stated limit \(E_n\to K_{i,0}\),

\[ K_{i,0}=rK_{i,0}+Mga\sin\theta. \]

Thus

\[ \boxed{K_{i,0}=\frac{Mga\sin\theta}{1-r}}, \qquad \boxed{\kappa=\frac{\sin\theta}{1-r}}. \]

Indeed, the recurrence gives

\[ E_n=K_{i,0}+r^n(E_0-K_{i,0}), \]

which converges because \(0<r<1\), provided the prism continues to cross each energy barrier.

e) Minimum slope for indefinite rolling

If \(K_{i,0}<K_{i,\min}\), any finite initial energy eventually decreases enough that the prism cannot cross the next barrier. If \(K_{i,0}\geq K_{i,\min}\) and the initial energy satisfies \(E_0>K_{i,\min}\), the recurrence keeps every finite subsequent \(E_n\) above the threshold.

The critical angle is therefore determined by

\[ K_{i,0}=K_{i,\min}, \]

or

\[ \frac{\sin\theta_0}{1-r} =\frac{1-\cos(\pi/6-\theta_0)}{r}. \]

Substituting \(r=121/289\),

\[ 121\sin\theta_0 =168\left[1-\cos\left(\frac{\pi}{6}-\theta_0\right)\right]. \]

Expanding the cosine yields

\[ 205\sin\theta_0+84\sqrt3\cos\theta_0=168. \]

Using \(t=\tan(\theta_0/2)\), this becomes

\[ (168+84\sqrt3)t^2-410t+(168-84\sqrt3)=0. \]

The root in the physical interval \(0<\theta_0<30^\circ\) is

\[ t=\frac{3(2-\sqrt3)}{14}. \]

Consequently,

\[ \theta_0 =2\arctan\left[\frac{3(2-\sqrt3)}{14}\right] =6.5723657\ldots^\circ, \]

and to the requested precision,

\[ \boxed{\theta_0=6.6^\circ}. \]

At the critical slope, an initial energy strictly above the threshold permits every subsequent step, although the prism spends progressively longer near each unstable balance position.

The solution was checked against the official solution.