You knew this was coming. When I talked about the car on the Wall of Death, I mentioned that the motorcycle was cooler. It has to lean to go around the Wall of Death - of course by Wall of Death, I mean a giant cylinder with the motorcycle riding on the inside vertical wall. Like this:
Since I already posted a short explanation of how a car can do this, I want to look at the angle this bike has to lean in order to ride the wall. Yes, the bike has to lean. Why? Torque. Before I talk about torque, I need to remind you about fake forces. The problem is that I want to look at the bike in the equilibrium position as it is relative to the wall. But the bike is not in equilibrium. Instead it is accelerating since it is moving in a circle. If I want to look at the reference frame where the bike is stationary (and not accelerating), I have to add in a fake force.
If you wanted to give this fake force a name, that name would be the centrifugal force (not the centripetal force - here is the difference between the two). In general, a fake force is a force you need to add to an object so that in an accelerating reference frame it behaves like it is in a non-accelerating frame. For most systems (but not all), you can just say the fake force is the following:
For a first approximation, you could take the motorcycle as the reference frame. This frame is accelerating towards the center of the circle that it is moving in (centripetal acceleration). If you want to treat the frame as though it is not accelerating, you would need to add a fake force in the opposite direction to the acceleration - so radially outward. This would be the centrifugal force. It would have a value of:
There is a small problem, but let me forget about that for right now. Suppose the motorcycle was driving such that the vehicle was completely perpendicular to the wall. Here is what the force diagram for that would look like in the accelerating frame.
Here, the problem is that the torque about the point where the wheel touches the wall is not zero. If I consider the gravitational force to act at the center of mass, the motorcycle would "fall over" in this frame. The only way for the torque to be zero would for the bike to be tilted, like this:
You can see that in this case, the torque from the fake force can be the opposite of the torque from gravity (torques about the point of contact). In the previous case, the fake force produced no torque since the force was directed straight towards the point of contact. At this point, you would think I could just find the angle that the torques from these two forces add up to zero. Well, I can do that if I assume that the length of the bike is small compared to the radius of the circular path. In that case, the following should be true:
Note that I am actually showing the z-components of the torques. Anyway, that shows the relationship between the angle and the radius and angular speed. If the bike goes faster, the angle θ gets smaller (more perpendicular to the wall). That makes sense.
Ok, then why is this wrong? The problem is that I have calculated the torque by the fake force as though it were a single force acting at the same location as the gravitational force (center of mass). This isn't really true. The part of the bike (maybe we could include the person on the bike as part of the bike) that is closer to the center of the circular motion will have a smaller fake force on it since it has a smaller acceleration. The part of the bike closer to the wall will have a greater fake force.
Notice that you don't have to do this with the gravitational force. It is uniform at all locations. This means that the sum of the gravitational force on all pieces of the bike is the same as a single gravitational force acting at the center of mass on the bike. I should probably explicitly show this some time, but you know how things go. But what about the total torque due to this changing-with-location fake force? Of course, we need to start with a picture. Here is the force on a just a small piece of the motorcycle.
Here I am calling the length of the motorcycle (and maybe person) L. The radius of the cylinder is R and the radius of the circular motion of the piece is r. Oh, I guess I left off the mass of this small piece. It has a mass of dm. One more thing. I am going to write this fake force as just dF - without the vector sign. All of these forces are in the same direction, so I don't need to treat them as vectors.
Let me go ahead and also call the location of this small piece of the motorcycle a distance of s from the contact point so that the length of this small motorcycle piece (which would technically be called a mini bike - you know like those Honda 50's?) ds. I probably should have drawn these on the diagram, but I didn't. Ok, here is the value of r for that typical piece.
Here, R is the distance all the way to the wall so I need to subtract the horizontal component of the location of the motorcycle piece. As a check, if s = 0, then r = R - that makes sense. Now, here is an expression for the little fake force (dF) on that little piece of the motorcycle (dm):
Now, I can use this to find the total torque from this fake force. The torque from this one little piece will be:
Before I simplify this any more, let me change the dm variable to something more useful. Clearly, I would like to integrate along the s axis - so it would be nice to get a integration variable of ds. If I assume uniform density (which probably wouldn't be true for a person on a motorcycle), I can write:
Here, M is the mass of the whole thing. This means that the torque from that small piece would be:
This looks crazy, but really it isn't. Remember that the only thing that changes as I move along the motorcycle would be the value of s. The angle (θ) is the same for all of these little pieces of the motorcycle. To find the total torque, I can just integrate from s = 0 to s = L.
Ok, just a quick check. Both terms in this result have the correct units for torque (N*m). Also, if θ is 0, the torque would also be zero - which is why it has to tilt. But what now? I said the magnitude of this torque must be the same as the torque from the gravitational force. This means that:
How about a nice plot? Let me use a radius for the cylinder of 5 meters. In this case, I can plot the lean angle as a function of angular speed. This is for a uniform-density motorcycle person with a height of 1.5 meters.
The mass of the motorcycle doesn't matter (for a uniform distribution) as you would probably expect. As the motorcycle goes faster, it is closer to being perpendicular to the wall. Of course, this assumes that the frictional force is always enough to hold the bike up, but at some low limit of speed, it would slip down. But what about the height of the motorcycle? Here is the same thing except for three motorcycles, 1, 1.5, and 3 meters tall.
The red line is for the taller motorcycle and the green is the shortest one. So, this say for the same speed, a taller bike would have to lean more. Why? Well, since more of this bike is closer to the center of the cylinder it will have less torque from the fake force (centrifugal force).
What about the car? Why doesn't this happen with the car as it goes around the wall of death? Here is a force diagram for a car doing the same thing. Note that I am just going to guess at the location for the centrifugal force on the car (it should be lower than the center of mass though).
For this car, it doesn't have to lean. Why? Because there is an interaction between the two (four) tires and the wall. These sets of tires are not in the same line as the center of mass and the "center of centrifugal force". Remember, the problem with the motorcycle was that there was nothing to provide a torque in the opposite direction to the torque from the gravitational force.
So, the car is simpler but also more boring.
Back to the Wall of Death
Why did I start this whole thing? Really, I sometimes forget. I think I was trying to get an estimate for the speed of the motorcycle. First, what about the angle? I think this is my best shot:
Looking at the background, I estimate the lean for this guy to be about 20°. Now, I will need something else. I can either get a value for the angular speed, or the radius of the cylinder. Since this is just a quick shot, let me go with the radius of the cylinder. I don't have to guess about the size. Nik Berg (@dadorak) posted that it has a diameter of 31 feet. While I am speaking of Nik, you really should check out his "the making of video" for this Wall of Death stunt. Actually, I suppose I need to also estimate the height of this rider. Here you can see I am already cheating. The rider plus the bike is clearly not of a uniform density. But things like this never stopped me before. Why should they stop me now? I will just go with a height of 1.5 meters. Really, I don't think this is too bad of a problem.
Using these values along with the calculation gives an angular speed of 2.77 radians per second. If the radius is 4.72 meters, this would be a speed of 13 m/s (about 29 mph). That seems reasonable. I am happy.
What about the g-force?
The video Nik posted describing how the thing was made claims that the rider experiences an acceleration of about 3 g's. This isn't so straight forward to calculate because different parts of the body have different accelerations. So, let me go back to the small force on the small piece of the motorcycle - which I called dF. If I just add up all these forces and divide by the mass, I will get an estimate for the "apparent weight" of the rider.
Using the angular speed for the above example, this would be an effective acceleration of 30.46 m/s2 or about 3.1 g's. Sometimes, I am surprised when stuff works out like I thought it would. Of course, like I said this is just the "net apparent acceleration". Just image if your bike was 4.72 meters tall. In this case your head would be at the center of the circle and not really accelerating.
