Physics of Rolling Ball Coasters
Cross Product
Torque
Inclined Plane
Inclined Ramp
Curved Path
Examples
Cross Product (1)
• The Cross Product of two threedimensional vectors a = <a1,a2,a3> and b =
<b1,b2,b3> is defined as follows:
a  b  a2b3  a3b2 , a3b1  a1b3 , a1b2  a2b1
• If q is the angle between the vectors, then
| a  b || a || b | sin 
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 a  b  c ).
• The cross product is greatest
when a  b
• While the dot product produces
a scalar, the cross product
produces a vector. Therefore it
is sometimes called a vector
product.
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:
v r
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:
a  r
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.
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:
 
  r F

Second is in terms of the moment of inertia and the angular acceleration.
(Angular acceleration is the time derivative of angular velocity).


 I
(Note the similarity to Newton’s Second Law, F = m a. Here all the terms
have an angular counterpart.)
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.
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.
1 2 1 2
mgh  mv  I
2
2
2
 v 
2mg (d sin  )  mv 2   mR 2  
5
 R 
7 2
2 gd sin   v
5
5

v 2  2d  g sin  
7

2
Inclined Plane (3)
5
7


From the previous slide: v 2  2d  g sin  
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.
vaverage 
a
vi  v f
v f  vi
t
v 2f  2d a 
2

d
t
Eliminating t and vi=0, these expressions give
5
Comparing this result to the previous slide, we can see that a  g sin 
7
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   v b . 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.
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:

 

|  || b  mg | m | b || g | sin 

Inclined Track (3)


|  || b  mg | mbg sin 

We also have a second definition of torque:

 2
a
|  | I |  |  mR 2  mb2  CM
5
 b
Setting these equal and solving for acceleration down the
track:
aCM
g sin 

 2R 2 
 2  1
 5b

Notice that if b = R, then this reduces to the previous
expression for acceleration
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 Tˆ , N̂ , and B̂ 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 N̂. 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.
Curved Paths (2)
To apply Newton’s second law,
consider all forces in the normal
direction.
Acting in the positive N̂ direction
is gravity, and in the negative N̂
direction is the normal force, N.
The sum of these forces must result
in curved motion around the
instantaneous radius R.
2
mv
m g  Nˆ  N N 
R
N  m g  Nˆ  mv 2
N
NN refers to the component of
the normal force in the normal
direction
Curved Paths (3)
This formula has three parts. First is finding an expression
for N̂. 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 N̂ is given by
r ' '( s) | rT ' ' ( s) | Tˆ
ˆ
N ( s) 
rN ' ' ( s)
with
r '( s )
ˆ
T ( s) 
| r '( s ) |
where r’’ is the second derivative of the path, r’’N is the
acceleration in the normal direction, r’’T is the
acceleration in the tangential direction, and Tˆ is the unit
tangent vector.
Curved Paths (4)
•
•
Next is the curvature, which is most useful expressed as
| r '( s)  r ' '( s) |

| r '( s) |3
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.
1
1
1
1
mgry (0)  mv02  I 02  mgry ( s )  mv( s ) 2  I ( s ) 2
2
2
2
2
•
v
b
Using the definition of w, we can relate it to v by   . Then
v( s ) 2  v02 
2mg
(ry (0)  ry ( s ))
(m  I 2 )
b
Curved Paths (5)
Thus the general expression for the magnitude of the normal force in
the normal direction is
 r ' '( s )  r '( s ) 
r '( s )
r ' '( s )  


 | r '( s ) |2 
2mg
 2
 | r '( s )  r ' '( s ) | 


| N N ( s ) | 0,0,mg 
 m vi 
(ry (0)  ry ( s )) 
 r ' '( s )  r '( s ) 
(m  I 2 )

 | r '( s ) |3 
r '( s )
b


r ' '( s )  
 | r '( s ) |2 


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.
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.
Example - Parabola
1
Set r(s)=<s,-s2,0>.
Consider a ball
starting at rest,
with b=⅔ R. Then
we have the
expression for the
magnitude of the
normal force.
| N ( s) | mg
-5
-10
-15
19  36s 
-20
2
19(1  4s )
2
3
2
-25
2
3
4
5
The graph looks like this:
(vertical axis in g’s, horizontal axis has arbitrary units)
1
0.8
| N ( s) | mg
0.6
19  36s 
2
19(1  4s )
2
3
2
0.4
0.2
1
2
3
4
5
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
1
black curves, which explains the shape of the
graph for the magnitude of the normal force (blue).
0.8
Force of gravity
0.6
N
0.4
0.2
Resultant force
0.5
1
1.5
2
Example - Cosine
r ( s)  s, A cos
2s

,0
where A is the amplitude of the curve, and l is the wavelength.
Example – Unit Normal Direction, N̂ .
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)
Normal force pushing up out of the track
3
2.5
2
A ball, starting at rest,
b=2R/3, amplitude is 1,
wavelength is 2p.
1.5
1
1
2
2
3
4
5
6
1.5
1
0.5
1
-0.5
-1
2
3
4
5
6
This shows the normal
force and its direction at
different points on the
curve.
As before, here is a breakdown of the different components of
the forces on the the system.
3
2
N
1
Resultant force
1
2
3
4
5
6
Force of gravity
-1
5
4
This shows the normal
force at different initial
speeds. Where are they
coincident and why?
3
2
1
1
-1
2
3
4
5
Speeds increase 1 m/s with each line. Blue is
initially at rest, and orange is initially at 5 m/s.
6
Loop-’de-Loop
Unfortunately, we were unable to find the actual parameterization
used in the design of coasters. However, we think it’s something
like this:
s
 s

r ( s )   5 cos( s  , ,5 sin( s  )
3
2 10
2
0.6 -2
0.4
0.2
0
0
2
4
6
5
2.5
0
-2.5
-5
0.5
0
-0.5
-1
-2
0
2
4
6
5
This shows the local
coordinate system for
certain points along the
path.
2.5
0
-2.5
-5
0.75
0.5
-2
0.25
0
-0.25
0
2
4
6
This shows the
normal vector
along the track.
5
2.5
0
-2.5
-5
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:
ˆ
ˆ ˆ
N ( s)  ( N N ) N  (m g.B) B