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Homework Help: What actually causes a motorcycle to do a wheelie?

  1. Apr 3, 2004 #1
    I'm not wanting an answer but I'm looking to be pointed int the right direction.
    First of all what actually causes a motorcycle to do a wheelie??
    Does it have something to do with torque??
  2. jcsd
  3. Apr 3, 2004 #2
    Yes, Torque and static friction.
  4. Apr 3, 2004 #3
    I don't think that you include static friction thoug.
    how do you relate the torque to center of mass and the distance between the two wheels?
  5. Apr 3, 2004 #4
    No sorry this problem has stumped me.
    I thought I might know how to relate the torque and that but I just can't see where to start.
    I mean you're given that the distance between the two wheels is 1.55 m, and that the centre of mass of the bike (including the rider) is .88m above the ground, and half way between the distance of the two wheels.
    THen it says to assume that the mass of each wheel is small in comparison to the body of the motorcycle. the engine only drives the rear wheel.
    I mean how can you find the horizontal acceleration from this info??
    I think my main problem is that I don't exactly know what exactly goes on when the motorcycle does a wheely.
    Any point in the right direction would be great :)
    My test is tomorrow and i'm wasting too much time wanting to know how to approach questions like this.
  6. Apr 3, 2004 #5
    The torque required to do a wheelie will be the lever arm length from the rear wheel axil to the center of mass. The center of mass can be found a couple of ways. I usually use this: [tex]x_{cm}=\frac{m_1x_1+m_2x_2+...+m_nx_n}{\sum m}[/tex]

    which can be rewritten as:

    [tex]x_{cm}=\frac{\sum^n_{i=1} m_ix_i}{M}[/tex]
    [tex]y_{cm}=\frac{\sum^n_{i=1} m_iy_i}{M}[/tex]
    [tex]z_{cm}=\frac{\sum^n_{i=1} m_iz_i}{ M}[/tex]

    Set the origin at the rear wheel axil, and assume the motorcycle is 2D thus elliminating the z.

    The x and y component can now be used to find the moment about the rear axil [tex]\tau=(Mg)r[/tex] where Mg is the weight of the bike at the center of mass (total wieght) and g is the gravitational acceleration and r is the radius from the moment axis to the point of force.

    where [tex] r= \sqrt{x_{cm}^2+y_{cm}^2}[/tex]

    You now have a radius from the rear axil to the center of mass and the center of mass is found by knowing the weight distribution of the bike (distance between the wheels, engine placement, rider placement/orientation, rider weight, weight of wheels).

    In your problem, the y component would be the height (.88m) and the x component would be the distance along the x (1/2 of 1.55m). All you need is the force (weight of the bike and rider). Think about what is going on. The engine via a set of sprockets and a chain/belt must produce enough torque to cause the center of mass to rotate about the rear axil. Its just like using a torque wrench to tighten a bolt. You have a lever arm and a force.

    This doesn't set the torque at the rear axil but torque is a free vector (able to move) so it will work.

    EDIT: Sorry, I was typing a longer more complex solution but I noticed you had replied when I previewed so I changed my respons a little to suit your question.
    Last edited: Apr 3, 2004
  7. Apr 3, 2004 #6
    This is a cool question. I never thought to analyze the physics behind a wheely on a motorcycle. I used to be quite an avid rider. It was my only mode of transportation for 3+ years, and I continued to ride high mileage thereafter. Recently, there is no money or time for a bike.

    Anyway...has anyone here actually performed wheelies before? If you have, then you noticed that weird feeling of finding a "balance point." This question has caused me to think about this. Is the balance point that you feel acheived when the CM is over the rear wheel? Or is there something else at work here?
  8. Apr 4, 2004 #7
    I have a Yamaha R6 and a YZ250, had a CBR F3 and various other dirt bikes so I've done more than my share. Yes, the balance point is where the CM is directly over the rear wheel axil. At that point you can let off the throttle and ride the wheelie. On either side of that point you have to accelerate the bike to keep the wheel up or decelerate (accelerate in the opposite direction) to keep from tipping over.

    A better way to say the above I guess would be the point where the radius vector is parallel to the force vector

  9. Apr 4, 2004 #8
    so i thought that the radius would have been from the center of mass to the rear axle. And wouldn't that just be 1.55/2????
  10. Apr 4, 2004 #9
    And you calculate the the inertia as well don't you?
    would the bike kinda be like a long tube with the center of mass in the middle.
    That is, I=(1/12)mD^2 + mD^2.
    Then after you do that you solve for angular acceleration using T=Ia.
    Then angular acceleration is related to the horizontal acceleration by the equation.
    A=r*a , where a is the angular acceleration.
    and you have your radius value.
    Is that right???
    Sorry I'm so stupid :)
    And thanks a lot for your help
  11. Apr 4, 2004 #10
    Hahahah can't believe what i was thinking.
    Yep nevermind about that radius question hahaha.
    for some reason i thought that the CM was along the line of the distance between the two axles hahahaha.
    but yep you find it using pythag's theorem.
    Got ya ;)
    But i still just need to know if i can class the bike as that sort of system with regards to the inertia???
  12. Apr 4, 2004 #11


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    The easiest way to do this is to calculate the torque about the point of contact of the rear wheel with the road. There are two components; when they cancel, there is no weight on the front wheel. What are the two components? the mass, acting vertically as if from the centre of mass, and the reaction force due to acceleration. This latter force also acts from the centre of mass, but is horizontal. Let us say the centre of mass is d ahead of the rear axle, and h above the road. Then the first torque is mgd, and the second is -mah. They sum to zero when gd=ah. IOW, when a is d/h g's, the front wheel has no weight on it. If a is any larger, you are wheelying.

    In cars, d is much larger than h. Street tires are only good for a=1g at the most, so this explains why ordinary street cars don't wheely.
  13. Apr 4, 2004 #12
    well i get the d value to be 1.17.
    I use that r=sqrt(xcm^2+ycm^2) = 1.17
    so that is the distance from the wheel to the COM.
    The height is 0.88m.
    But one of the tutors at our uni has apparently told someone that the answer is about 11.9m/s^2 and that mgh=mah.
    I'm lost now
    Please help me as soon as possible if you can, because i really want to go into my test knowing how to do this question.
    Thank you in advance to anyone who can help me.
  14. Apr 4, 2004 #13


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    I think you misunderstand. d is not r. d is the horizontal component of the vector from the rear wheel centre (or contact patch) to the COM. Let's put it another way: Define the point of contact of the rear wheel with the road to be the origin. Then the location in cartesian coordinates of the COM is (d,h). For this problem you do not need to calculate r=sqrt(d^2+h^2), because the 2 forces are resp. horiz. and vertical, and to get the torque, you only need the distance component at right angles to the force.
  15. Apr 4, 2004 #14
    Thank you all soooo much
    I understnad it now.
    I was wondering why the torques weren't in opposite directions when I was using r haha.
    THanks soo much.
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