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 = and b = is defined as follows: If q is the angle between the vectors, then

3. 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.

4. 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.

6. 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 Newton’s Second Law, F = m a. Here all the terms have an angular counterpart.)

7. 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.

8. 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.

9. Inclined Plane (3) 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 v i =0, these expressions give Comparing this result to the previous slide, we can see that From the previous slide:

10. 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.

11. 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

13. 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.

14. Curved Paths (2) To apply Newton’s 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. N N refers to the component of the normal force in the normal direction

15. 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 v 2. 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, r’’ N is the acceleration in the normal direction, r’’ T is the acceleration in the tangential direction, and is the unit tangent vector. with

16. Curved Paths (4) Next is the curvature, which is most useful expressed as As for the v 2 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

17. 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, r y (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.

18. 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.

19. Example - Parabola Set r(s)=. Consider a ball starting at rest, with b=⅔ R. Then we have the expression for the magnitude of the normal force.

20. The graph looks like this: (vertical axis in g’s, 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 Resultant force N

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)

24. Normal force pushing up out of the track A ball, starting at rest, b=2R/3, amplitude is 1, wavelength is 2 p. This shows the normal force and its direction at different points on the curve.

25. Force of gravity Resultant force N As before, here is a breakdown of the different components of the forces on the the system.

26. 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.

27. 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:

28. This shows the local coordinate system for certain points along the path.

29. This shows the normal vector along the track.

30. 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: