(adsbygoogle = window.adsbygoogle || []).push({}); 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.

A simple graphic I made using Autocad: (attached)

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.

[tex]\sum[/tex]M[tex]_{G}[/tex]=I[tex]\alpha[/tex]

[tex]\sum[/tex]F=ma

U[tex]_{1\rightarrow2}[/tex]=[tex]\int[/tex]F[tex]\cdot[/tex]dr + [tex]\int[/tex]M[tex]\cdot[/tex]d[tex]\theta[/tex]

U[tex]_{1\rightarrow2}[/tex]+T[tex]_{1}[/tex]+V[tex]_{1}[/tex]=T[tex]_{2}[/tex]+V[tex]_{2}[/tex]

M=F*r (r=distance from axis of rotation to arm of force)

a_{G}= a_{o}+ a_{G/O}(G relative to O)

a_{G/O}= [tex]\alpha[/tex] X r + [tex]\omega[/tex] X ([tex]\omega[/tex] X r)

T = 1/2 mv^{2}

V = mgh

adx = vdv

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

Next, I went after Md[tex]\theta[/tex]. 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 [tex]\theta[/tex] since d changed with [tex]\theta[/tex]. Using trig, I came up with d_{friction}= 2.78sin[tex]\theta[/tex], and d_{normal}= -2.78cos[tex]\theta[/tex]. So, evaluating my integral for Md[tex]\theta[/tex] gave me the work performed to rotate the system from [tex]\theta[/tex]_{i}(34 degrees) through 90 degrees (balance point). This yielded 665 ft.-lbs.

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

T_{2}= 1/2 mv^{2}. But integrating adx = vdv to find v final tells me that v final is 2a_{G}*x, so T_{2}= ma_{G}*x where a_{G}is the acceleration of G.

V_{2}= 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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# Dynamics: Finding distance to reach balance point of a wheelie.

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