Transcript Physics of Rolling Ball Coasters

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
m gh  m v  I
2
2
2
2
2  v 
2m g(d sin  )  m v   m R  
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 | mbgsin 

We also have a second definition of torque:

 2
a
|  | I |  |  m R2  m b2  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