and we assume the wheels are turning whatever RPM it takes to do 65MPH, what will slamming on the brakes while in mid air do? Vehicle is 6000lbs Wheels are alloy 17" Tires are LT315/70R-17. Are the wheels and tires heavy enough or moving fast enough have an impact on the angle of the flight? Some assumptions will need to be made about mass distribution... Reason for the question: http://www.youtube.com/watch?v=JKQdlXvbWSU

Once upon a time I had the rear end of a rear wheel drive card up in the air. I forget what I was working on. I think I was just rotating the two rear tires. This was back when I was doing one wheel peels all over the place, and was worried that the tires were wearing unevenly. lol. Anyway I put it in gear and stepped on the gas and I remember laughing as the speedo pegged. Then I hit the brakes. The force exerted is such that it drops the front end and lifts the rear end, ie a counterclockwise motion if viewed from the driver's side window. But it is a very insignificant amount of force. I doubt it would do much to any hypothetical vehicle traveling midair at 65 mph. To understand the physics of it, you just need to keep in mind that the wheels are spinning in a counterclockwise direction (as viewed from the driver side). Applying force to the brake pads is absorbing that counterclockwise force. Whatever is holding the pads (ie the rest of the vehicle) is going to move in the same direction at the same vector as the tires. Its the same with anything. If you grab hold of a rope moving at 3 mph, you will be pulled in the direction the rope is moving. Even if the rope stops when you grab it, you still get pulled in that direction, even if just a little bit.

The energy gets converted to heat, but angular momentum must still be conserved. The angular momentum of the spinning wheels will be converted to rotation of the truck about it's rotation point. Using some very rough numbers: Moments of inertia: I_wheel = 18.4 N.m.s^2 I_truck = 23,263 N.m.s^2 Angular velocity of a wheel: w_wheel = 65 [mi/h] * 1600 [m/mi] * 3600 [s/h] = 66.9 rad/s Angular momentum of the wheels: L_wheels = 4 * 18.4 [N.m.s^2] * 66.9 [rad/s] = 4,924 [N.m.s] Conserving angular momentum: w_truck = 4,924 [N.m.s] / 23,263 [N.m.s^2] = 0.212 rad/s So the truck will start rotating at about 2rpm in the same direction as the wheels were rotating.

I count roughly 2 seconds of hang time so if the guy slammed on his brakes right away it could have pitched him nose down about 12 degrees.

What would have happened, if he hit the gas though? I expect he'd get the nose up, but also induce a slight rotation around the longitudinal axis, due to the drive shafts momentum change. It should be possible to counter-steer against that though, using the front wheel offset to cancel out the shaft-induced angular momentum.

Its a little more complex with a car. Motocross you have one tire you are accelerating or decelerating. In a car if you just jumped and braked you have all 4 wheels moving, and more importantly, in different spots is relation to the CG. The front wheel will want move the CG up, the rear will want to force it down. Seems like the front and rear wheels would cancel if they the same mass/distance from the CG. Edcarman's example seems like it assumes the wheels are at the CG.

The rear will force the CG down from the rear which results in a nose down force while the front will force the CG up from the front which is also a nose down force. They add up not cancel out.

to me it makes sense they add up (the center of gravity doesn't change its trajectory but the truck receives angular momentum) but I don't think it's possible to do a simple calculation because the distances are big, and this fact makes it much less effective imho (it's just a sensation as I can't do those calculations). Also a truck is flexible, maybe this makes midwayman's idea more meaningful, as the front of the truck gains a momentum, while the back of the truck gains another, and in the center of the truck these contrast each other (if we assume the truck is a rubber band or something). In reality the truck is a bit rigid so it doesn't happen completely I guess? edcarman's calculation are for a 100% rigid item outside of any gravitational field or attrition force, with the axis of all 4 wheels being exactly centered in the axis of rotation of the truck. So in reality it's going to be less than that.

I lol at the 12 degrees. In a van you also have a drive shaft that is turning perpendicular to the wheels, and you have a complex rear differential which is partially contributing to the inertia of the wheels, including the absorption of an unknown amount of the torque transferred from hitting the brakes. It would be very difficult and ultimately very inaccurate to try to calculate the actual angular velocity changes of the vehicle if brakes or gas was applied. We really couldnt even say for sure whether the vehicle would tilt up/down (due to the wheels) or to the left/right (due to the driveshaft/transmission.). And dont even mention the crankshaft's mass and momentum!

It doesn't make this assumption. When you've got bodies that are both spinning about their own axes, as well as rotating about some other axis, the toal angular momentum is the sum of the angular momentum due to the spin, as well as that due to motion about the other axis. In this example, before braking, the system has no angular motion about it's central axis. The only angular motion is the spin of the wheels about their own axes. These rotation vectors are all pretty much parallel (think of the right hand rule), so they do not cancel out. Once the wheels are stopped, there's no rotation about the wheel axes, so there must be rotation about some other axis to conserve angular momentum. The direction of this axis must be the same as the wheel axes.

That's a cop-out. I agree, given the simplifying assumptions that have been made here, it's not possible. But then those calculations can be done on the back of an envelope, using generic data from the internet, in less than 30min. It is possible to accurately calculate or measure the moments of inertia of all rotating components, and it is likewise possible to accurately calculate the relative rotating speed of all of these components. Overall, it's a fairly simple system, since, aside from rotation, most of the massive components do not move relative to each other. The only exception being the suspension components. Internal torques will not affect the final angular momentum of the system - only external torques can do this. The only external force in this case are gravity, which acts through the centre of mass and so will no produce a net external torque, and air resistance, which may have some external effect. Finally, unless your vehicle is made of spaghetti, it a very reasonable to assume that it behaves in a rigid manner. Internal deformation of the structure would not have a major effect on the final angular momentum state. Given good input data (e.g. from drawings, or measurement) it should be possible calculate an accurate answer to this rigid dynamics problem, though it may take a few hours.

Would seem so, but try it. Grab something long and semi-flexible. Torque it in the same direction on both ends. It doesn't want to rotate, it wants to bend into a S-shape. I think the fact that the angular momentum is being applied into two different places relative to the CG is very important. What we are really talking about is 3 axis. 1- CG truck 2- Front wheel axle 3- Rear wheel axle Stop the front wheels and the truck will want to conserve momentum on that axis. That means the CG truck will start to rotate clockwise slowly about the front axle. The same happens in the rear- The CG truck rotating clockwise about the rear axle. Those are pretty simple to visualize. However what happens as a whole depends on where the CG truck is in respect the the front and rear axles as well as their respective mass. For simplicity lets say that CG truck lies equidistant between the front and rear axle and that the wheel mass is equal. That means the clockwise force from the front is pushing CG truck up at the same time and force that the rear is pushing down. This is pretty well bore out by simple experimentation.

This is an incorrect thought experiment since it confuses internal and external forces. Although you are are applying a torque to twist the rod, it's also apply an equal and opposite torque on you. The overall system, consisting of you and the rod experiences no net change. If you want a very good illustration of the the actual anuglar momentum effect, consider a four wheel drive radio control car going off a jump at half throttle. If you stop the throttle (stopping all four wheels), the car pitches forward. If you increase to full throttle (accelerating all four wheels), the car pitches back. By feathering the throttle somewhere between these two points, you can actually control the attitude of the car in flight. You can see this described in a number of RC car guides, like this one: http://www.rccaraction.com/blog/2010/12/08/fix-that-nose-dive-off-jumps-2/ If you want to see the mathematical derivation for the angular momentum of a body rotating about both its own COM and about some arbitrary point, you can read these two pages: http://www.kwon3d.com/theory/moi/am.html http://emweb.unl.edu/NEGAHBAN/EM373/note20/note20.htm

Actually it does want to rotate. If you do it one hand at a time it will spin correct? The same way each time correct? By grabbing the semi-flexible doohickey with both hands you are restricting the motion to only the directions your arms and hands are capable of moving. If you grab the doohickey by the "front" with your right only and turn it cw the "back" raises up and the "front" goes down relatively speaking. If you grab the doohickey by the "back" with your left hand only and turn it cw the "front" is going to go down. When you grab it with both hands you have pinned the doohickey artificially. Your arms will counteract the twisting your hands do.