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How to use Moments of Inertia to find Max Acceleration (sportbike pre-wheelie)

  1. Mar 18, 2016 #1
    So in this problem there's a lot of leeway as far as parameters go. The numbers I give for weights, dimensions, etc, may not be super realistic, but that's not too important to me because I can plug in realistic numbers later. I more of need help understanding the concept of moments of inertia.

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

    Determine the maximum acceleration that can be achieved by a motorcycle without causing the front wheel to lift off the ground. Select a real-life motorcycle and research its geometry and inertia properties including the inertia properties of the wheels. Don’t forget the rider! Also, determine how the maximum acceleration depends on the horizontal and vertical positions of the center of mass with respect to the points of contact between the ground and the wheels.

    Vehicle starts from rest.
    Assume no slipping.

    2. Relevant equations
    sum of moments around point of contact of rear wheel with ground = 0
    moment of inertia wheel = mr^2
    moment of inertia motorcycle body (solid cuboid) = (m/12)(w^2+h^2)
    moment of inertia rider (solid cuboid) = (m/12)(w^2+h^2)
    M=I*alpha
    parallel axis theorem = I+md^2

    3. The attempt at a solution
    I'm not super interested in the numbers quite yet, just the concept. Essentially what I have done is I have figured out the moments of inertia for the two wheels, the motorcycle body, and the rider, all about the point of contact of the rear wheel with the ground, which I am calling point O (I've included a rough diagram below). I'm treating the body and the rider as a single rigid body, though I am keeping their center of masses separate.

    From there I figured that I can't just sum up the moments of the weight forces of the wheels, body, and rider around point O, because the instant that the bike starts accelerating, the moments of inertia of the bodies around point O add an extra torque around point O. Therefore, I need to sum the moments of the weight forces around point O, as well as using the parallel axis theorem around point O on the moments of inertia of each body. That will leave me with an equation that looks like this (all torques around point O):

    (weight torque of front wheel)+(weight torque of back wheel)+(weight torque of bike)+(weight torque of rider)-[(sum of moment of inertias around point O)(angular acceleration around point O)]=0

    I can then solve for the angular acceleration and take it from there. The problem is, I'm not sure if I fully understand the concept here. The other thing I'm not sure about is to me it would seem that the instant the bike starts to accelerate, each mass would have a horizontal forward facing force as well as its downward facing weight force. If this is the case, then I have to account for more moments than I was thinking, but I don't know if that's flawed thinking or not.

    Thanks guys!
     

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    Last edited: Mar 18, 2016
  2. jcsd
  3. Mar 18, 2016 #2

    jack action

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    First, find the center of mass of your bike, wheels and rider combined. This is the point where gravity (weight) and longitudinal acceleration act.

    You have 3 forces acting on your bike: 1 vertical at each wheel and 1 horizontal at your rear wheel (point O). Don't forget your bike is accelerating, so ma has to react at the CG.

    What do you know from the problem statement? «without causing the front wheel to lift off the ground». If the wheel lifts off the ground, that means that the vertical force at the front wheel is zero; then this is your limit and this is a known force.

    Knowing that, redo your summation of moments about point O.
     
  4. Mar 19, 2016 #3
    Thank you for your response jack action. So when I sum up the moments, do I include the torque of F=ma at the CG? And should I also include the moment of inertias of each body since they will oppose the angular rotation of the bike which would add another torque? For example, this is the equation for the summation of the moments about point O according to your advice:

    (entire weight torque)+(ma at CG torque)+[(sum of moment of inertias around point O)(angular acceleration around point O)]=0

    Is that correct?
     
  5. Mar 19, 2016 #4

    jack action

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    I'm not sure I fully understand what you are saying - but I think you got it - so I will repeat with a more standard wording.

    This is a very simple problem of dynamics. You need to do a free body diagram where:
    [tex]\sum F_x = ma_x[/tex]
    [tex]\sum F_y = ma_y[/tex]
    [tex]\sum M_{@O} = I_{@O} \alpha[/tex]
    So, there's the weight ([itex]a_y = g[/itex]), no rotation motion ([itex]\alpha = 0[/itex]), [itex]\sum F_x[/itex] is the traction force of the tire and [itex]\sum F_y[/itex] is composed of the 2 reaction forces at the tires (and we already determined the front one should be zero). [itex]\sum M_{@O}[/itex] is the summation of the product of each force times its perpendicular distance to point O (which could be anywhere, but it will ease calculations if you choose the rear tire-road contact point).

    You can still choose to sum up the different components [itex]i[/itex], which will only change the previous equations this way:
    [tex]\sum F_x = \sum(m_i) a_x[/tex]
    [tex]\sum F_y = \sum(m_i) a_y[/tex]
    [tex]\sum M_{@O} = \sum(I_{@Oi}) \alpha[/tex]
    All components are under the same acceleration in any direction (including angular). It's the moment arm of each force that will determine the effective CG position of the assembly.

    Do the free body diagram. Everything should become clear and easy.
     
  6. Mar 19, 2016 #5
    Thank you for all you've written. That makes sense. I feel like it would be more simple if I wasn't so thrown off about the inertia properties of the wheels. The part I'm still trying to figure out is the part in the prompt where it says "Determine the maximum acceleration...including the inertia properties of the wheels." For instance, when a dirt-bike rider goes off a jump, while he's in the air he can pitch the bike by giving the bike gas (causing the back wheel to rotate creating a moment that pitches the bike) or by applying the brake.

    My understanding is that a similar situation is happening with the bike in this problem. The instant the bike accelerates, the wheels begin to rotate. The rotational acceleration of each wheel will create a moment at that instant ([itex]M=I\alpha[/itex]), further preventing the bike from doing a wheelie, adding to the sum of the moments around point O. I feel if we were to neglect the inertia properties of the wheels, your last post would be perfect for solving the question. But maybe I just don't understand inertial properties haha.

    Does what I'm thinking make sense?

    Here's an image of a problem that is very similar to the one that I am asking, but says to ignore the rotational inertia of the wheel. It prescribes a very similar method to what you gave, but in the problem I am working on, I need to include the rotational inertia of the wheel.
    xzNg9LG.png
     
  7. Mar 20, 2016 #6

    jack action

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    I have to say, you put a doubt in my mind, but I would say that rotational inertia is irrelevant to this problem. But others are welcome to chime in.

    Here's how rotational inertia affects your bike. You have your torque input [itex]T[/itex] (from the engine) that fights your wheel inertia [itex]I\alpha[/itex]. Whatever's left, will go to the tire-road contact patch in the form [itex]F_Ar[/itex], where [itex]r[/itex] is the tire radius. So:
    [tex]T = F_Ar + I\alpha[/tex]
    [tex]\frac{T}{r} = F_A + I\frac{\alpha}{r}[/tex]
    But we know that [itex]F_A=ma[/itex] and [itex]a = \alpha r[/itex], such that:
    [tex]\frac{T}{r} = ma + I\frac{a}{r^2}[/tex]
    [tex]\frac{T}{r} = \left(m + \frac{I}{r^2} \right) a = \left(1 + \frac{I}{mr^2} \right) ma[/tex]
    Where [itex]m + \frac{I}{r^2}[/itex] is known as the equivalent mass. There is another way to find it on this page, based on conservation of energy.

    In the end, [itex]F_A[/itex] still equals [itex]ma[/itex] on your free body diagram. So the calculations in example 7.1 would still hold. But to create the force [itex]F_A[/itex], you will need a larger torque than initially thought, since you have to also fight rotational inertia.

    As for your dirt bike example, this is true because there is nothing restricting the bike motion. The torque created by the engine (or the brakes) will fight the inertia of the wheel, but the reaction will be to try to rotate the bike frame around the wheel as much as trying to rotate the wheel with respect to the frame.
     
  8. Mar 22, 2016 #7
    Thanks again for sticking with me. I was able to speak with my professor today about this problem and in the next day or so I'm going use the method he prescribed but on the sample problem (in my last post above). I'll see if the answer turns out any different (if I can figure it out) and I'll post the results here. Thanks again jack action! I'll be back in a day or two.
     
  9. Aug 7, 2016 #8
    Sorry for not getting back sooner everyone. I got my work back from my professor a few months ago and everything turned out correct, I just haven't posted anything here yet.

    When including the moment of inertia of the rear wheel, the max acceleration the bike can achieve is less than what can be achieved when disregarding the moment of inertia of the wheel (an example of disregarding the moment of inertia of the wheel can be found in the picture of the example in post #5). The difference in accelerations when regarding/disregarding the moment of inertia isn't too large, although as far as accuracy is concerned, including the moment of inertia of the wheel gives a more accurate acceleration.

    A way to think about it is that the moment of inertia is a measurement of something's resistance to rotation, therefore the rear wheel is going to resist rotation to a certain extent, which has an effect on the rest of the bike. The rear wheel will exert a moment on the bike, and the bike will exert a moment on the rear wheel.

    This won't be a perfect example, but imagine if the moment of inertia of the rear wheel was so large that it was as if the rear wheel was bolted to the ground (like I said, this isn't perfect, since if it was actually bolted to the ground, the center of mass would change as would other things). If you pulled the throttle with your bike's rear wheel "bolted" to the ground, you would instantly do a wheelie and would probably rotate all the way around until you smashed the top of your head on the ground. Essentially, the greater the moment of inertia of the rear wheel, the slower you have to accelerate to avoid popping a wheelie. This idea applies to the problem.

    You have to start with the rear wheel and draw a free body diagram for it alone, as well as a kinetic diagram for it.

    FullSizeRender.png
    The FBD and KD above illustrate the equation ##\sum M_P = \sum (M_k)_P##, or in other words, the sum of the moments around point P (FBD) is equal to the sum of the "kinetic moments" around point P (KD). Below is a key to the diagram above:

    ##W_T =## total weight of bike;
    ##F_f =## friction force
    ##N_T =## normal force of entire bike (front wheel has a normal force of zero)
    ##F_3 =## force that the rest of the bike exerts on the rear wheel the instant acceleration begins
    ##M =## moment that causes rear wheel to rotate
    ##R =## radius of wheel
    ##I =## moment of inertia of rear wheel
    ##\alpha =## angular acceleration of rear wheel
    ##m_r =## mass of rear wheel
    ##a =## linear acceleration of rear wheel

    From the FBD and KD above you can get three necessary equations. The next step is then to draw a FBD and KD for the rest of the bike without the rear wheel, and get two necessary equations from that. After that use a system of equations to solve for the variables in the five equations and you'll be able to find the max acceleration the bike can achieve without popping a wheelie.

    Hopefully that helps anyone who's facing this same problem.
     
  10. Aug 8, 2016 #9

    jack action

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    You cannot compare the two situations as a motorcycle with a bolted wheel wouldn't linearly accelerate and therefore there wouldn't be any weight transfer associated with this acceleration (there would only be one due the angular acceleration). It is a completely different problem.
    Absolutely not true. If the rear wheel inertia is greater, you need a greater wheel torque to produce the required acceleration to pop a wheelie. But the needed acceleration is still the same.

    I'll start by repeating stuff from post #6. Everything start with the wheel torque, right? So, let's study how it transforms by doing the sum of moments about the center of the wheel, considering only the wheel ##w## itself:
    [tex]T_w = F_A r + I_w\alpha_w[/tex]
    We know that ##a = \alpha_w r## at the perimeter of the wheel. But we also know that at the wheel-ground contact point, there is no relative motion; therefore it must be the center of the wheel that linearly accelerate with an acceleration ##a##. Since the wheel center pushes on the rest of the bike, then the whole bike accelerate with acceleration ##a##.

    Now, let's look at the bike as a whole. The diagram in post #5 applies.

    From the sum of forces in the x-direction, we can establish that ##F_A = ma##.

    From all of this, we know (see post #6 for reference):
    [tex]ma = \frac{T_w}{r\left(1 + \frac{I_w}{mr^2}\right)}[/tex]
    If we didn't consider the inertia of the wheel (##I_w = 0##), then the equation would reduce to ##ma = \frac{T_w}{r}##.

    What do we learn from this? The larger the wheel inertia, the larger must be the wheel torque to maintain a desired level of acceleration.

    Going back to the «whole bike» situation. We can do a sum of moments about points A and B. From this we will find that:
    [tex]N_A = mg\frac{d}{w} + ma\frac{h}{w}[/tex]
    [tex]N_B = mg\frac{w-d}{w} - ma\frac{h}{w}[/tex]
    What do we learn from this?
    1. ##N_A + N_B## always equals ##mg##.
    2. If ##a=0## then ##N_A = mg\frac{d}{w}## and ##N_B = mg\frac{w-d}{w}##. These are the static front & rear weight distribution.
    3. The weight transfer due to the acceleration is ##ma\frac{h}{w}##. It transfers from the front to the rear.
    4. If ##mg\frac{w-d}{w} \lt ma\frac{h}{w}## then the front reaction force is less than zero and the rear reaction force is greater than the weight of the bike, which is impossible. So we have to consider that the bike frame will start rotating under the extra moment.
    Case where ##mg\frac{w-d}{w} \lt ma\frac{h}{w}##:

    The bike frame ##f## is turning around the wheel center. Let's do a sum of moment about the rear wheel center:
    [tex]F_A r + ma (h-r) = mg (w-d) + I_f\alpha_f[/tex]
    But we already know that ##F_A = ma##, so:
    [tex]mar + ma (h-r) = mg (w-d) + I_f\alpha_f[/tex]
    Or:
    [tex]mah - mg (w-d) = I_f\alpha_f[/tex]
    What do we learn from this?
    1. If, and only if, ##mah \gt mg (w-d) ##, then the bike frame will begin rotating with angular acceleration ##\alpha_f## (which has nothing to do with ##\alpha_w##).
    2. As the bike frame start moving up, ##h## will increase and ##d## will decrease, so by keeping a constant acceleration ##a## then ##\alpha_f## will increase as the frame goes up.
    If you want to relate this to the wheel torque:
    [tex]\frac{T_w h}{r\left(1 + \frac{I_w}{mr^2}\right)} - mg (w-d) = I_f\alpha_f[/tex]
    So once the bike frame is up, the driver can play with the wheel torque to set ##\alpha_f## positive or negative to move the front wheel up or down. It even allows him to drive with the front wheel continuously up in the air, like if it was floating in that position.
     
  11. Aug 8, 2016 #10
    I can see how my assumptions were incorrect. I just redid all my math with a greater moment of inertia for the rear wheel and can see where I went wrong in assuming that you have to accelerate slower to avoid popping a wheelie. I had changed some values without changing others that were critically linked to those changed. Thanks for pointing that out.

    As far as I'm aware though, the method for solving the problem with the FBD and KD in post #8 still yields the correct results. I don't say that as some physics wiz, but because my professor said my work was correct haha.
     
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