I want to know an expression for the Momentum of this car using the variables in the photo (all of which are known).
The lateral force for the left and right wheel are the same, but different from the front wheels. The longitudenal force for the left and rear wheels are different.
Delta = steering angle.
a = distance between front axle and Centre of Gravity, b = distance between rear axle and Centre of Gravity, c = half the width of the car.
Flong l = lateral force of the left wheel, Flong r = lateral force of the right wheel.

Can anyone help me with this? Thanks in advance!

Good morning, Toby and welcome.

 

Your photo looks much better the correct way up.

 

 

I am not sure what you are trying to achieve so please explain further.

 

Momentum is a vector that can be resolved into components parallel to the motion and at right angles to it (tangentially and radially) in this case.

 

The tangential momentum is another word for the (instantaneous) linear momentum.

 

Then there is the angular momentum.

 

So over to you?

I'm looking for the angular momentum. And thanks for turning the picture.

We seem to be short of a radius.

You know that if the motion is circular, the net force must be toward the center and will have a magnitude of v^2/r. I suppose it's possible that you might be able to determine F and the turning radius from the diagram, and if you can do that, you can determine v. After that, the angular momentum is r x p (for a circle, the magnitude is just mvr)

Another way to look at it is that force is the rate of change of momentum.

 

http://www.bbc.co.uk/schools/gcsebitesize/science/add_ocr_pre_2011/explaining_motion/forcesandmotionrev4.shtml

But that doesn't tell you what the momentum is.

You know that if the motion is circular, the net force must be toward the center and will have a magnitude of v^2/r. I suppose it's possible that you might be able to determine F and the turning radius from the diagram, and if you can do that, you can determine v. After that, the angular momentum is r x p (for a circle, the magnitude is just mvr)

 

I was just thinking that - but it does assume no understeer nor oversteer; ie perpendiculars of both sets of tyres pass through same point which is the axis of rotation

 

Also I guess as the OP gave dimensions of the car he may be thinking of rigid body angular momentum of a box - which would make a calculable but probably insignifant (in the face of the other estimations) difference