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Ok, so here's the setup of the problem. A motorcyclist tries to jump over some cars by using a ramp. Right after he leaves the take-off ramp, he notes that his motorcycle is angled slightly upward and has zero angular velocity (meaning that if he did nothing and stayed in the same position, the bike would remain in the same orientation with respect to the ground). If this tilt is maintained, then a problem will arise when he hits the landing ramp as he should be tilted downwards to ensure a smooth landing. The question is what should the biker do in order to make his bike tilt forward? Also, ignore any energy loss/air resistance.

According to my professor, the biker should hit the brakes. The reasoning is as follows.

The wheels are spinning quickly in the forward direction at takeoff, so they have substantial angular momentum. If the brakes are applied, then some of this angular momentum is transferred to the main body of the bike, because the total angular momentum of the system is conserved. The bike will therefore rotate forward somewhat, as desired.

I understand that the motorcycle-biker system must have its angular momentum conserved (only gravity acts on the system, but since it acts on the center-of-mass, it provides no torque) with respect to its center-of-mass. What does the statement "the wheels...have substantial angular momentum" really mean? With respect to what? Intuitively, it feels as there should be angular momentum, as the object is spinning. But, doesn't angular momentum have to be measured with respect to a fixed point/axis or the center-of-mass of the motorcycle-biker system? If so, what is the fixed point/axis in this situation?

I want to understand the solution, but I cannot understand what point the angular momentum is measured about. If someone could either explain the reasoning behind the solution or tell me what axis/point the wheels would have angular momentum, I would greatly appreciate it.

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    $\begingroup$ I find it fascinating that the bike will behave similarly whether it's on the ground or in the air. Everyone intuitively understands the situation on the ground, that the bike will pitch forward when you hit the brakes, and tilt back when you hit the gas. Kinda neat to see that the ground isn't entirely responsible for that effect, and that it happens when the wheels aren't even touching anything. $\endgroup$ May 7 '20 at 14:00
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    $\begingroup$ Yea, when I was thinking of this problem I intuitively knew that hitting the brakes would make me go forward, but I had no idea why that was the case. $\endgroup$
    – Imajinary
    May 7 '20 at 16:14
  • $\begingroup$ I answered the question for myself in a flash (didn't write because there was a good answer already), but now that I am thinking about it, I understand why motorcyclists steer left and/or right when jumping. I was under impression that it is just a kind of dance. It is not, it is a method of canceling or inducing a roll. Does anyone know how a motorcycle can yaw? $\endgroup$
    – fraxinus
    May 7 '20 at 16:33
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    $\begingroup$ Also note that on a motorcycle (or bicycle) you can brake the wheels individually, causing the bike to rotate around different axes. You don't want to brake strongly with just the front wheel. $\endgroup$
    – jamesqf
    May 7 '20 at 16:38
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    $\begingroup$ Cool stuff, I had never thought about it! youtube.com/watch?v=cv345HMqwFI for a real-life example $\endgroup$ May 7 '20 at 22:41
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As you already noted, the center of mass should be the best point for analyzing the motion, because you don't have any external torque when you view the system from the center of mass. The general expression of angular momentum of any rigid body about any point is

$$\mathbf L=I_{\text{COM}}\boldsymbol{\omega}+m\mathbf r \times \mathbf v_{\text{COM}}$$

where $I_{\text{COM}}$ and $\mathbf v_{\text{COM}}$ are the moment of inertia of the rigid body (about the center of mass) and the velocity of the center of mass of the rigid body (when viewed from the point of reference) respectively.

In your case, the above formula can be applied to the wheel. Since the wheel is stationary (stationary in the sense of translatory motion) in the center of mass frame, the second term reduces to zero, and the total angular momentum of the wheel (about the center of mass) would then be

$$\mathbf L_{\text{wheel}}=I\boldsymbol{\omega}$$

Thus since the $\omega$ would be naturally high, the wheels do have substantial angular momentum about the center of mass.

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Your professor is right. The angular momentum of the front wheel is measured relative to its spin axis. By touching the brakes, a little of that angular momentum is transferred via friction forces to the motorcycle and the rider. If the bike is in a nose-down state, a twist of throttle will bring the nose up because of the counter-torque that the engine applies to the motorcycle frame. But please don't try this at home!

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    $\begingroup$ "a little of that angular momentum is transferred"? If you stop the wheel (in respect to the bike's frame) surely all of it gets transferred? Well some remains on the wheel since it is now rotating at the same speed as the bike. $\endgroup$
    – NikoNyrh
    May 8 '20 at 9:35
  • $\begingroup$ Not all gets transferred if you are using mechanical brakes- there are friction losses there. $\endgroup$ May 8 '20 at 16:21
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    $\begingroup$ Friction cannot dissipate angular momentum; it is a means of transferring angular momentum. Same as for linear momentum. Momentum is a macroscopic mechanical quantity and cannot be hidden in thermal motion like energy can. $\endgroup$
    – nanoman
    May 8 '20 at 16:57
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As you have noted, angular momentum is always measured with respect to some point, but that can be any point. While there are some nice properties associated with the centre of mass, such as the fact that the principal axes of rotation pass through that point (i.e. objects rotate about their centre of mass), conservation of angular momentum still applies regardless of the point about which it is calculated (as long as you use the same point throughout). Therefore, you are free to choose whatever point makes calculations convenient, which in this case is probably going to be the centre of mass.

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Most dirt bikes are light. The rider should be standing and leaning well forwards (center of mass in front of foot pegs) on the take off, so that by moving backwards relative to the bike (pushing forwards on the handlebars) in mid-air will pitch the bike downwards. The throttle or tapping of the brakes can be used for some adjustment, but the standing rider's internal torques applied to the bike accounts for most of the pitch correction.

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