IPhO 1991 · Theory problem 1

Spinning ball bouncing on the floor

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The figure 1.1 shows a solid, homogeneous ball radius \(R\). Before falling to the floor its center of mass is at rest, but the ball is spinning with angular velocity \(\omega_0\) about a horizontal axis through its center. The lowest point of the ball is at a height \(h\) above the floor.

Figure 1

Figure 1.1

When released, the ball falls under gravity, and rebounds to a new height such that its lowest point is now \(ah\) above the floor. The deformation of the ball and the floor on impact may be considered negligible. Ignore the presence of the air. The impact time, although, is finite.

The mass of the ball is \(m\), the acceleration due the gravity is \(g\), the dynamic coefficient of friction between the ball and the floor is \(\mu_k\), and the moment of inertia of the ball about the given axis is: \[ I=\frac{2mR^2}{5} \]

You are required to consider two situations, in the first, the ball slips during the entire impact time, and in the second the slipping stops before the end of the impact time.

Situation I: slipping throughout the impact.
Find:
a) \(\tan\theta\), where \(\theta\) is the rebound angle indicated in the diagram;
b) the horizontal distance traveled in flight between the first and second impacts;
c) the minimum value of \(\omega_0\) for this situations.

Situation II: slipping for part of the impacts.
Find, again:
a) \(\tan\theta\);
b) the horizontal distance traveled in flight between the first and second impacts.
Taking both of the above situations into account, sketch the variation of \(\tan\theta\) with \(\omega_0\).

Official English text of the problem.

Answers

Let \(D\) be the horizontal distance traveled during the rebound flight. In the usual short-impact approximation, with gravity’s impulse during contact neglected:

  • Situation I(a): \[ \boxed{\tan\theta=\mu_k\frac{1+\sqrt a}{\sqrt a}} \]

  • Situation I(b): \[ \boxed{D=4\mu_k h\sqrt a(1+\sqrt a)} \]

  • Situation I(c): \[ \boxed{\omega_{0,\min}=\omega_c =\frac{7\mu_k}{2R}\sqrt{2gh}\,(1+\sqrt a)} \] At equality, slipping stops exactly at separation. Nonzero slip at separation requires \(\omega_0>\omega_c\).

  • Situation II(a): \[ \boxed{\tan\theta=\frac{2R\omega_0}{7\sqrt{2agh}}} \]

  • Situation II(b): \[ \boxed{D=\frac{4R\omega_0}{7}\sqrt{\frac{2ah}{g}}} \]

  • Requested sketch: \[ \boxed{ \tan\theta= \begin{cases} \displaystyle\frac{2R\omega_0}{7\sqrt{2agh}}, &0\leq\omega_0<\omega_c,\\[6pt] \displaystyle\mu_k\frac{1+\sqrt a}{\sqrt a}, &\omega_0\geq\omega_c. \end{cases}} \] The graph rises linearly from the origin to a corner at \[ \left(\omega_c,\ \mu_k\frac{1+\sqrt a}{\sqrt a}\right), \] then remains horizontal. It is continuous at the corner.

Solution

Flight speeds and impact equations

Let \(u\) be the downward speed immediately before impact and \(w\) the upward speed immediately after impact. Mechanical energy is conserved during each free flight, so \[ u=\sqrt{2gh}, \qquad w=\sqrt{2gah}=\sqrt a\,u. \] The centre of mass falls by \(h\) and rises by \(ah\), because its height always exceeds that of the lowest point by \(R\).

Choose the positive horizontal direction to be the direction of the friction force on the initially spinning ball. Write \(v\) for the horizontal centre-of-mass velocity, and take positive \(\omega\) in the initial spin direction. The horizontal velocity of the lowest point is then \[ v-R\omega. \] Initially this is \(-R\omega_0\), so friction accelerates the centre of mass and reduces the spin.

Let the normal impulse be \[ P=\int_{\text{impact}}N\,dt. \] For contact duration \(\tau\), vertical impulse–momentum gives \[ P-mg\tau=m(u+w). \] In the standard short-impact approximation, \[ g\tau\ll u+w, \] and hence \[ P\simeq m(u+w). \] The impact can have a finite duration while satisfying this approximation.

Let \(J\) be the accumulated horizontal frictional impulse. Linear and angular impulse–momentum give \[ mv=J, \qquad I(\omega-\omega_0)=-RJ. \] Using \(I=2mR^2/5\), \[ v=\frac{J}{m}, \qquad \omega=\omega_0-\frac{5J}{2mR}. \] Consequently, \[ \boxed{R\omega-v=R\omega_0-\frac{7J}{2m}}. \] This quantity decreases monotonically while the ball slips. Slipping stops when it reaches zero.

Situation I(a): rebound angle

If slipping continues throughout impact, the friction force has magnitude \(\mu_kN\) throughout. Therefore \[ J=\mu_kP=\mu_km(u+w), \] and the outgoing horizontal speed is \[ v=\mu_k(u+w). \]

The angle \(\theta\) in the figure is measured from the vertical, so \[ \tan\theta=\frac{v}{w} =\mu_k\frac{u+w}{w}. \] Since \(w=\sqrt a\,u\), \[ \boxed{\tan\theta=\mu_k\frac{1+\sqrt a}{\sqrt a}}. \]

Situation I(b): horizontal distance

The ball leaves and returns to the same centre-of-mass height, \(R\). Its flight time \(T\) therefore satisfies \[ 0=wT-\frac12gT^2, \] giving \[ T=\frac{2w}{g}. \] The horizontal velocity remains constant in flight, so \[ D=vT=\frac{2\mu_k(u+w)w}{g}. \] Substituting \(u^2=2gh\) and \(w=\sqrt a\,u\), \[ \boxed{D=4\mu_kh\sqrt a(1+\sqrt a)}. \]

Situation I(c): minimum initial angular velocity

Because \(R\omega-v\) decreases monotonically during slipping, checking its value at the end of impact is sufficient. At the boundary between the two situations, it becomes zero exactly at separation: \[ R\omega_0-\frac{7J}{2m}=0. \] With \(J=\mu_km(u+w)\), this gives \[ \omega_c=\frac{7\mu_k}{2R}(u+w). \] Thus \[ \boxed{\omega_{0,\min}=\omega_c =\frac{7\mu_k}{2R}\sqrt{2gh}(1+\sqrt a)}. \]

For \(\omega_0>\omega_c\), slipping persists even at separation. For \(\omega_0=\omega_c\), it persists until the final instant of contact.

Situation II(a): rebound angle

When \(0<\omega_0<\omega_c\), slipping stops before the normal impulse is complete. Let \(J_*\) be the frictional impulse accumulated at that instant. The condition \(v=R\omega\) gives \[ R\omega_0-\frac{7J_*}{2m}=0, \] so \[ J_*=\frac{2mR\omega_0}{7}. \] Therefore, at the instant slipping stops, \[ v_*=\frac{2R\omega_0}{7}, \qquad \omega_*=\frac{2\omega_0}{7}. \]

These tangential velocities remain unchanged during the rest of contact. To see this, suppose a static friction force \(f\) acts while \(v=R\omega\). Then \[ m\dot v=f, \qquad I\dot\omega=-Rf, \qquad \dot v=R\dot\omega. \] Together these imply \[ f\left(\frac1m+\frac{R^2}{I}\right)=0, \] and hence \(f=0\). No further tangential impulse is required.

The outgoing horizontal speed is therefore \[ v=\frac{2R\omega_0}{7}, \] and \[ \boxed{\tan\theta=\frac{v}{w} =\frac{2R\omega_0}{7\sqrt{2agh}}}. \]

The result also includes the limiting case \(\omega_0=0\), when there is no slipping and the rebound is vertical.

Situation II(b): horizontal distance

Using the same flight time \(T=2w/g\), \[ D=\frac{2R\omega_0}{7}\frac{2w}{g}. \] Thus \[ \boxed{D=\frac{4R\omega_0}{7g}\sqrt{2agh} =\frac{4R\omega_0}{7}\sqrt{\frac{2ah}{g}}}. \]

Requested sketch

For fixed \(h,R,a,\mu_k\), the graph has two portions:

  • From \(\omega_0=0\) to \(\omega_c\), it is a straight line through the origin with slope \[ \frac{2R}{7\sqrt{2agh}}. \]
  • At \(\omega_c\), it joins a horizontal line of height \[ \mu_k\frac{1+\sqrt a}{\sqrt a}. \]

The joining values agree because \[ \frac{2R\omega_c}{7\sqrt{2agh}} =\mu_k\frac{1+\sqrt a}{\sqrt a}. \] At larger initial spins, friction already acts throughout contact, so increasing the spin cannot increase the available tangential impulse.

The solution was checked against the official solution.