I. INTRODUCTION
When a superball is thrown so that it bounces from the
floor to the underside of a surface, such as a table, it will
bounce back to the person who threw it instead of continuing
to bounce between the table and the floor @see Fig. 1~a!#.
Strobel1 and Garwin2 independently analyzed this sequence
of collisions and their analysis has been revisited many times
in undergraduate laboratories, physics demonstrations, and
textbooks. This paper explores a variation on this demonstration:
instead of bouncing a superball from the underside of a
table, what happens when the superball is thrown down into
a vertical channel composed of two parallel walls? If this
demonstration were performed with a steel bearing, the bearing
would ricochet back and forth as it falls down the channel
under the force of gravity. However, the superball seems
to defy gravity and bounces back out and returns to the person
who threw it, as in Fig. 1~b!.
A comparison of the ball’s path in both the horizontal and
vertical channels suggests that these two sets of collisions
may be related. I explore this similarity for the more general
case of a superball bouncing between the walls of an infi-
nitely long vertical channel. The matrix description of the
superball motion is the most useful and straightforward approach
to this problem and the equation of motion which is
developed in Sec. III captures the dynamics of the ball in
both the vertical and horizontal channels. When gravity acts
along the length of the channel ~the vertical channel!, the ball
does not continuously fall, but instead is trapped in a length
of the channel, repeatedly rising and falling with a fixed
frequency. This frequency does not depend on gravity but on
the ball’s moment of inertia. Hence the motion in both the
vertical and horizontal channels is essentially the same, with
gravity affecting the amplitude and the displacement of the
oscillation.
You will probably find that even if the ball is thrown so
that it enters the channel, it sometimes does not quite make it
out. Even in this failed demonstration, however, the trajectory
of the superball is unexpected and instead of immediately
falling down the channel, it undergoes a decaying oscillation
as it descends. This motion is a consequence of the
inelasticity of the collision, and hence during the collision
there is not a perfect exchange between the linear motion and
the spin. As a result, the ball may not be able to climb out of
the channel, but it will certainly resist the descent.
To accurately describe the ball’s motion, it is necessary to
take into account the energy losses in the tangential bounce
of a realistic superball. This analysis was done by Cross, who
presented a modification of Garwin’s superball model to take
into account the tangential coefficient of restitution.3 Section
II examines this modified model and its implications for the
dynamics of the superball. Because of the simplicity of the
collision matrix with tangential and normal coefficients of
restitution, the equations of motion for the fully elastic superball,
which are developed in Sec. III, can be easily applied,
with only minor modifications, to the behavior of a
realistic superball. The discussion in Sec. IV examines these
general equations for the special case of a ball with tangential
loss, but perfect restitution normal to the surface. For this
case, the motion of the ball has the properties of both a
damped harmonic oscillator and a body falling through a
resistive medium. The ball falls down the channel at a terminal
velocity that depends on the tangential coefficient of restitution
and the moment of inertia of the ball. Before reaching
this terminal velocity, however, the ball undergoes either
a decaying oscillation or an exponential decay, once again
depending on the restitution and the moment of inertia. This
dependence is similar to the underdamped and overdamped
motion of a harmonic oscillator.
II. KINEMATICS OF A PARTIALLY INELASTIC
SUPERBALL
The simple model of an ultraelastic rough ball or superball
begins with two assumptions about the ball during its collision
with a flat surface:1,2 ~a! the kinetic energy is conserved
during the collision, and ~b! there is no slip at the point of
contact. Because the collision is assumed to occur at a point
on the ball’s surface, all of the forces act through this point
and the torque is zero. As a result, the angular momentum
about the point of contact is conserved during the collision.
From these two conservation laws, the spin and velocity
parallel to the surface after the collision are expressed in
terms of the initial spin and velocity of the ball:
vx52 ~a21!
~a11!
vx02 2a
~a11!
c0 , ~1!
and
c52 2
~a11!
vx01
~a21!
~a11!
c0 ,