Physics of Rolling Ball Coasters

1
Physics of Rolling Ball Coasters

• Cross Product
• Torque
• Inclined Plane
• Inclined Ramp
• Curved Path
• Examples

2
Cross Product (1)

• The Cross Product of two three-dimensional
  vectors a lta1,a2,a3gt and b ltb1,b2,b3gt is
  defined as follows
• If q is the angle between the vectors, then

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Cross Product (2)

• Important facts about the cross product
• The cross product is always perpendicular to the
  vectors a and b.
• The direction of the cross product is given by
  the right hand rule (see diagram, where
  ).
• The cross product is greatest when
• While the dot product produces a scalar, the
  cross product produces a vector. Therefore it is
  sometimes called a vector product.

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Digression

• Earlier, the definition of angular velocity was
  given, but details on the use of the cross
  product were not explained. Now that we have the
  cross product, we define the relation between
  tangential and angular velocity in general
• For circular motion, the velocity is always
  perpendicular to the position vector and this
  reduces to v r w.
• Similarly we define the relationship between
  tangential and angular acceleration

5
Torque (1)

• Before dealing with a rolling ball, we must
  discuss how forces act on a rotating object.
• Consider opening a door. Usually you grab the
  handle, which is on the side opposite the hinge,
  and you pull it directly toward yourself (at a
  right angle to the plane of the door). This is
  easier than pulling a handle in the center of the
  door, and than pulling at any other angle. Why?
• When causing an object to rotate, it is important
  where and how the force is applied, in addition
  to the magnitude.
• Torque is a turning or twisting force, and it is
  a measure of a force's tendency to produce
  rotation about an axis.

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Torque (2)

• There are two definitions of torque. First is in
  terms of the vectors F and r, referring to the
  force and position, respectively
• Second is in terms of the moment of inertia and
  the angular acceleration. (Angular acceleration
  is the time derivative of angular velocity).
• (Note the similarity to Newtons Second Law, F
  m a. Here all the terms have an angular
  counterpart.)

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Inclined Plane

• Consider a ball rolling down an inclined plane as
  pictured. Assume that it starts at rest, and
  after rolling a distance d along the ramp, it has
  fallen a distance h in the y-direction.

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Inclined Plane (2)

• We will now consider the energy of the system.
  The system is closed, so energy must be
  conserved. Set the reference point for potential
  energy such that the ball starts at a height of
  h.
• Initially the ball is at rest, so at this instant
  it contains only potential energy. When it has
  traveled the distance d along the ramp, it has
  only kinetic energy (translational and
  rotational).
• We can also express h in terms of d.
• This gives us the square velocity after the
  particle moves the distance d.

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Inclined Plane (3)
From the previous slide

• If you know the square velocity of a particle
  after it travels a distance d, and you know that
  the acceleration is constant, then that
  acceleration is unique. This derivation shows
  why, using definitions of average velocity and
  average acceleration.

Eliminating t and vi0, these expressions give
Comparing this result to the previous slide, we
can see that
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Inclined Track (1)

• When using physics to determine values like
  acceleration, there are often two perfectly
  correct approaches one is using energy (like we
  just did), and a second is by using forces. While
  energy is often simpler computationally, it is
  not always as satisfying. For this next
  situation, the previous approach would also work,
  with the only difference being that
  However, to demonstrate the physics more
  explicitly, we will take an approach using
  forces.
• When we build a track for a rolling ball
• coaster, there will actually be two
• contact points, one on each rail. Because
• the ball will now rest inside the track, we
• need to re-set the stage. The picture shows
• a sphere on top of a 2-rail track, with the
• radius R and the height off the track b marked in.

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Inclined Track (2)

• These are all the forces acting on the ball
  friction, gravity, and a normal force.
• The black square in the center represents the
  axis of rotation, which in this case is the axis
  connecting the two points where the ball contacts
  the track.
• The yellow arrow represents friction and the blue
  arrow represents the normal force. Neither of
  these forces torque the ball because they act at
  the axis of rotation. Thus the vector r is 0.
• The green arrow represents gravity.
• Convince yourself that the total torque is given
  by

12
Inclined Track (3)

• We also have a second definition of torque
• Setting these equal and solving for acceleration
  down the track
• Notice that if b R, then this reduces to the
  previous expression for acceleration

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Curved Paths

• Until now we have considered only straight paths.
  While these are much simpler, they would make a
  very boring rollercoaster. Now we need to put
  together all the theory discussed to this point.
• Given a parameterized path r(s), define , ,
  and as the principal unit vectors in the
  tangential, normal, and binormal directions,
  respectively.
• At any instant along the path there are two
  vectors acting on the particle, gravity and a
  force exerted by the track which we will call the
  normal force.
• Note Do not confuse the normal force with the
  normal direction . While they coincide in 2D
  systems, in a 3D system the normal force may
  point in any direction along the plane defined by
  the unit normal and unit binormal vectors.
• Finding the normal force will tell us how much
  force the track must be able to withstand at a
  given point.
• Also important are the total (resultant) forces
  on the system. They will be discussed after the
  normal force.

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Curved Paths (2)

• To apply Newtons second law, consider all forces
  in the normal direction.
• Acting in the positive direction is gravity,
  and in the negative direction is the normal
  force, N.
• The sum of these forces must result in curved
  motion around the instantaneous radius R.

NN refers to the component of the normal force in
the normal direction
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Curved Paths (3)

• This formula has three parts. First is finding an
  expression for . Second is finding an
  expression for k. And third is getting an
  expression for v2. We will take these one at a
  time.
• The most convenient expression for is given
  by
• where r is the second derivative of the path,
  rN is the acceleration in the normal direction,
  rT is the acceleration in the tangential
  direction, and is the unit tangent vector.

with
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Curved Paths (4)

• Next is the curvature, which is most useful
  expressed as
• As for the v2 term, we will get this from energy.
  We assume that friction is negligible, and since
  the system is closed, energy is conserved. In
  general, the initial types of energy include
  potential as well as both kinetic energies, and
  at any position s along the track there are the
  same types.
• Using the definition of w, we can relate it to v
  by . Then

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Curved Paths (5)

• Thus the general expression for the magnitude of
  the normal force in the normal direction is
• where r(s) is the path of the center of mass, m
  is the mass of the object, I is the moment of
  inertia of the object, ry(s) is the height of the
  center of mass at position s and b is the height
  of the center of mass from the axis connecting
  the points of contact. If the track is banked
  such that there are no forces acting in the
  binormal direction (so no lateral forces), then
  the normal force is in the direction of the unit
  normal vector.

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Curved Paths (6)

• Now that the expression for the normal vector is
  found, we can focus on forces important to the
  rider. For these we consider only forces that
  push on your skin. To prove this to yourself,
  consider an astronaut in orbit around the earth.
  They are in constant free-fall, so gravity is
  acting on them. However, they feel weightless.
  What this means to us is that only the normal
  force of the track can be felt by the ball (and
  only the normal force of the seat on a bobsled
  coaster is felt by the rider). So in this case,
  the only force felt by the ball is the normal
  force.

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Example - Parabola

• Set r(s)lts,-s2,0gt. Consider a ball starting at
  rest, with b? R. Then we have the expression for
  the magnitude of the normal force.

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The graph looks like this (vertical axis in gs,
horizontal axis has arbitrary units)
21
Some may wonder why this curve has the shape it
does. The reason for this is that the resultant
(or net) force is predefined by the track and the
initial conditions. The graph of N is the
difference of the component of gravity and the
resultant force. Graphically that means the
normal force is the difference between the two
black curves, which explains the shape of
the graph for the magnitude of the normal force
(blue).
Force of gravity
N
Resultant force
22
Example - Cosine
where A is the amplitude of the curve, and l is
the wavelength.
23
Example Unit Normal Direction,
The unit normal always points toward the center
of curvature. Thus when making calculations
involving the normal direction, take special care
around points of inflection! (In this diagram
there are arrows immediately before and after the
inflection point, but at the inflection point
there is no normal direction defined)
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Normal force pushing up out of the track
A ball, starting at rest, b2R/3, amplitude is 1,
wavelength is 2p.
This shows the normal force and its direction at
different points on the curve.
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As before, here is a breakdown of the different
components of the forces on the the system.
N
Resultant force
Force of gravity
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This shows the normal force at different
initial speeds. Where are they coincident and
why?
Speeds increase 1 m/s with each line. Blue is
initially at rest, and orange is initially at 5
m/s.
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Loop-de-Loop
Unfortunately, we were unable to find the actual
parameterization used in the design of coasters.
However, we think its something like this
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This shows the local coordinate system for
certain points along the path.
29
This shows the normal vector along the track.
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One last comment on forces in the normal and
binormal directions. We have calculated the
necessary resultant force for the normal
direction, but there has been no discussion on
the binormal direction. Since there is no
curvature in this direction, the only force that
will act in the binormal direction is gravity. To
prevent the ball from having any net lateral
force, the binormal component of gravity should
be balanced by a component of the normal force.
Consider the formula and diagrams below