# Dynamics: Finding distance to reach balance point of a wheelie.

1. Dec 6, 2010

### Yafa

1. The problem statement, all variables and given/known data
Background:

This is a problem I made up as part of an extra credit assignment for my first dynamics (engineering) class. I know the system will of course roll over, and he will end up looking like a fool, but that is besides the point. Keep in mind also that I can modify the problem any way I want so long as it is still a dynamics problem.

The problem:

A student is trying to impress a group of sorority girls at school. Fortunately, he rode his Suzuki GSX-R1000 to school today. The bike’s specifications are:
• Weight: 444 lb.
• Wheelbase: 55.3”
• Seat height: 32”
• Rear wheel diameter: 17”
• Tire profile: 55mm
• Torque at full throttle: 70 ft-lbs. (at the rear wheel)
• Radius of gyration about the centroid of the system (with rider): 1.44’
He decides that it will be easier for him to track distance covered than time in seconds to reach the balance point of a wheelie (the center of mass is directly above the contact point of the rear wheel). If the student weighs 170 lbs, and the distance from the seat to the top of his helmet is 2 feet, help him impress the girls by finding the total distance he should apply full throttle to reach the balance point. You may neglect air friction. Assume the tire does not slip, he starts from rest, and the system is perfectly rigid. You may model the system as a triangle with bottom corners at the contact points of the wheels to the ground, and the top corner at the top of the student's head. Assume the centroid of the triangle coincides with the center of mass of the system.

2. Relevant equations

Here is my very MODEST first attempt at using the "Latex Reference." I couldn't figure out how to get the subscripts to come out right, so some of them appear as superscripts for some reason.

$$\sum$$M$$_{G}$$=I$$\alpha$$
$$\sum$$F=ma
U$$_{1\rightarrow2}$$=$$\int$$F$$\cdot$$dr + $$\int$$M$$\cdot$$d$$\theta$$
U$$_{1\rightarrow2}$$+T$$_{1}$$+V$$_{1}$$=T$$_{2}$$+V$$_{2}$$
M=F*r (r=distance from axis of rotation to arm of force)
aG = ao + aG/O (G relative to O)
aG/O = $$\alpha$$ X r + $$\omega$$ X ($$\omega$$ X r)
T = 1/2 mv2
V = mgh
aG(x) is the acceleration of G in the direction of x.

Let G be the location of the center of mass of the system and O be the contact point between the tire and the road.

3. The attempt at a solution

I had some problems getting my full attempt up in one post, so I'm going to try to break it up over several posts.

OK, so I figured that since I only need to equate forces, moments and distance that the correct approach should be to use energy methods.

Using U$$_{1\rightarrow2}$$=$$\int$$F$$\cdot$$dr + $$\int$$M$$\cdot$$d$$\theta$$ , I started going after each component one at a time. I started with F dot dr. That was straight forward, yielding simply Fx*x = maG(x)*x, where x is what I'm after (the linear distance covered by G), and aG(x) is the acceleration of G in the direction of x.

Next, I went after Md$$\theta$$. M=F*d where d is the distance from the center of gravity (G) to the arm of the force. There were 2 forces in this case: the friction of the road on the tire and the normal force. I found the friction by dividing the torque at the center of the rear wheel by the distance "r" to the edge of the tire. r = wheel radius + tire profile = .708' + .18' = .89'. F = T/r = 70 lb.-ft./.89 ft. = 78.65 lbs. The normal force must be equal and opposite to weight, so that gives us 625 lbs.

Since the system is rotating about point O, the moment about G needed to be expressed as a function of $$\theta$$ since d changed with $$\theta$$. Using trig, I came up with dfriction = 2.78sin$$\theta$$, and dnormal = -2.78cos$$\theta$$. So, evaluating my integral for Md$$\theta$$ gave me the work performed to rotate the system from $$\theta$$i (34 degrees) through 90 degrees (balance point). This yielded 665 ft.-lbs.

Since it starts from rest, T1 and V1 both must be zero. This tells us that U(1->2) = /int (F dot dr) + /int (M dot d(theta)) = T2 and V2.

T2 = 1/2 mv2. But integrating adx = vdv to find v final tells me that v final is 2aG*x, so T2 = maG*x where aG is the acceleration of G.

V2 = mgh, where h is the change in height of G as it rotates about O until it gets to the balance point. Again, using trig, I get 2.78-1.56 = 1.22.

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Last edited: Dec 6, 2010
2. Dec 6, 2010

### Yafa

So plugging everything back into my energy equation, I get the following:

maG(x)*x + 181.3 = maG*x + maG*x + 1.22mg

Next, I need to find the acceleration components of G. Using aG = ao + aG/O, I can easily see that ao is just the friction force divided by the mass. This yields 4.05 m/s2.

Going after aG/O proves to be a bit painful, but after a few cross products, I end up with (sorry, the latex was spitting out some really unusual stuff for some reason) [-311alpha(sub o)*cos(theta) - 2.78alpha(sub o)*sin(theta)] i(hat) + [-311alpha(sub o)*sin(theta) + 2.78alpha(sub o)*cos(theta)] j(hat), where alpha(sub o) is the angular acceleration of the system about O.

At this point, I still have 2 unknowns (theta and alpha(sub o)). Since the only part I'm interested in is the x component of acceleration of G, I can ignore the j(hat) components. ao is just the acceleration caused by the force of friction at that point, the net acceleration of point o is 4.05 ft/sec2 i(hat). So, aG = ao + aG/O yields aG = [4.05 - 311alpha(sub 0)*cos(theta) - 2.78alpha(sub o)*sin(theta)] i(hat).

Summing the moments about G, setting them equal to I*alpha yields alpha(sub G) = [5.43sin(theta) - 43.19cos(theta)] k(hat).

Plugging all that back into my energy equation, I get x = 1562alpha(sub o)sin(theta) - 13.96alpha(sub o)cos(theta).

In summary, I have the acceleration of the center of mass as a function of theta and the angular acceleration about o. I also have the angular acceleration of the center of mass as a function of theta.

So far, this is all I can get. I have used up all my equations of motion. I have applied the equation of moments and energy. I don't know where to go from here.

Last edited: Dec 6, 2010
3. Dec 6, 2010

### Yafa

I went back through the problem, and found a few careless mistakes. I now know which way to go, but I cant figure out how to get around this differential equation:

(alpha) = (omega)' = (theta)'' = 5.43sin(theta) - 43.19cos(theta)

If someone could help me solve that for theta, then I think I can use the initial conditions to find time, and then use that to find x.

Maybe I'll just post up that diff eq in the math section.

4. Dec 8, 2010

### Yafa

for anyone who looked at this and is curious, here is an update.

I thought this would be a straightforward dynamics problem given all the assumptions; however, it turns out to be something beyond the scope of what is solvable in a beginning dynamics course. The reason is the variable moment about G. Even though the friction force is constant, the moment increases as the motorcycle rotates about the contact point since the distance from the friction force to G increases with theta.

The best I could come up with was that expression for alpha above. I ran it by my professor, and he too was surprised at how quickly it became messy. He advised me not to attempt a solution since the expression for alpha is nonlinear differential equation that is not immediately solvable, even though I have the initial conditions.

Since the class is already over, and he accepted my project based on what I had done already, I'm going to lay this one to rest. If anyone decides to take a poke at this one in the future, I'd be curious to see how it turns out.