Solving for Angular and Linear Velocities in a Two Car Collision

[trogtothedor]

Homework Statement

From the given picture, we know:
VA before collision
VB before collision
the radius to the center of mass of each car from the collision point P
e
the masses of both
the mass moment of inertia of both

We must find the velocities of the centers of mass of each car and their respective angular velocity.

Homework Equations

e = Vbp(y)2 - Vba(y)2 / Vap(y)1 - vbp(y)1
equations relating Vap to Va and Vbp to Va
(Hp)1 = (Hp)2

The Attempt at a Solution

I have worked out all of the relations and equations until I have a correct (Hp)1 = (Hp)2 formula. I have it in terms of omega(A) and Vay after impact.

The equation I have is correct since it works with the answers given in the problem. However, I cannot for the life of me figure out how to relate omega(A) to Vay after impact. I tried finding the instantaneous center of zero velocity, but that introduced more unknowns. Maybe there is a different relationship I missed between other variables which can be converted into terms depending on omega(A) and Vay?

 

[Yafa]

One approach you may consider:

omega(A) is related to Vay through the acceleration vector.

acceleration = a(tangential) + a(normal)
a(tangential) = r X alpha
a(normal) = omega X (omega X r) = omega X V(tangential)

Make sure you take all the rotations about the center of mass.

Find the net acceleration of your center of mass, use relative accelerations to get to it. Then take your cross products symbolically. Plug into the equation in component form. Solve for omega.

 

[trogtothedor]

We don't have a time period for the collision so I don't see how I could go about using accelerations.

 

[Yafa]

If you have rotation, you always have a normal component of acceleration (centripital acceleration). If there is no moment or other external forces applied to the system, then the net acceleration of any point in the system other than the centroid is just the cross product of omega and the perpindicular component of velocity.

In simpler terms, you could just take cross product of the component of velocity orthogonal to the line connecting the center of mass to the point of impact just after collision. That will give you the angular velocity.

 

[Yafa]

make sure all your velocities and lines are in vector component form when you cross them.

 

[trogtothedor]

I still don't understand. Wouldn't the cross product of Velocity x Distance result in units of meters squared over seconds?

 

[Yafa]

you're right, it's v/r. my mistake. And for that one, you use the magnitudes...

I had them mixed up. omega cross r = velocity(orthogonal)

 

[Yafa]

r = distance from c.m. to orthogonal velocity

 

[trogtothedor]

Right, but that seems like you're taking the point of impact as the instantaneous center of zero velocity. However, point P for both cars has a non-zero velocity.

 

[trogtothedor]

Or in your other post, it seems as if the IC is taken to be the center of mass. Both the center of mass and the impact points have non-zero velocities. If I knew the point about which they were rotating, or possibly a component of the velocity at point P, this would be easy. However, I know neither.

 

[Yafa]

what do you have so far?

 

[trogtothedor]

I will upload a picture of my equation since it will be too confusing to type.

 

[trogtothedor]

As stated, the only unknowns are omega(A) and V(A2Y), which is the velocity of the center of mass of A in the y direction after impact.

In the equation, a variable followed by an x or y means that it is the x or y component of that vector.

Also, a 1 or 2 denotes pre-collision, 1, and post collision, 2. e is the coefficient of restitution.

 

[trogtothedor]

Also the masses of both cars and the moments of inertia are similar, so m is the mass of either car and I is either cars mass moment of inertia.

 

[trogtothedor]

I can scan the equations that led up to that one if that is helpful. However, I know that they are correct.

I used the velocities in the y direction at point P, along with the coefficient of restitution, to arrive at an equation for omega(B) which used omega(A) and V(A2Y). I used conservation of inertia in the y direction to arrive at :
V(B2Y) = V(A1Y) - V(A2Y).
Also, the velocities of the centers of mass in the x direction remain unchanged, so V(A1X) = V(A2X) and V(B1X) = V(B2X).

 

[Yafa]

right. Have you tried using the relative velocity equations?

V(A2) = V(P2) + V(P/A2)

V(P/A2) = r (P/A) X omega (P/A)

since they are rigid bodies, then omega (P/A) should be the same omega for the entire body...

 

[Yafa]

you could also try to get V(A) by using the parallel axis theorem and working backwards to get it. Find I about P, then work backwards from there.

 

[trogtothedor]

Yes I did that. I had velocities at P in the equation relating to the coefficient of restitution, then i substituted values of the velocity at A in order to remove those unknowns, which left me with V(A2Y).

 

[trogtothedor]

and omega(A)

 

[Yafa]

OK, I'm thinking that the I.C.R. must be perpindicular to both velocity vectors for the body correct? If you look at the direction of the velocity of P on A just after the collision, there is only one way to draw a perpindicular line to that vector so that you could draw another perpindicular vector with its tail at A. If you tool around with some trig, I bet you could find V(A) using I.C.R.

 

[trogtothedor]

I was trying to do that but kept ending up with nonsense or arriving at the same equations i already had.

 

[Yafa]

do you have V(by)?

 

[Yafa]

If you have V(bpy), that should be equal to and opposite the impulse applied to point p on A. Use the principle of impulse and momentum to find omega(A)

 

[trogtothedor]

Do you mean after impact? I solved for V(B2Y) in terms of the initial and post-collision velocities of A. I'm not exactly sure how to set up the equation that you're taking about

 

[Yafa]

mv + integral of F dt from t=0 to t=t is equal to mv final. Since V(bp1y) is zero and V(bp2y) is known, V(bp2y) must equal the vertical component of impulse (integral of Fdt).

Then use H(A1) + integral of Mdt from t=0 to t equals H(A2). H(A1) = 0 since it isn't rotating, so integral Mdt = H(A2) = I(A)omega, where M = impulse in the y direction times the horizontal distance from A to p.

 

[Yafa]

More generally, the integral of the sum of all the forces acting on the body over time is equal to the net change in momentum. This can be broken down into component vectors.

integral of sigma Fdt = delta G

Likewise, the integral of all the moments applied over time is equal to the net change in angular momentum.

integral of sigma Mdt = delta H(about G)

 

[Yafa]

I*omega is also equal to r X mv.

 

[trogtothedor]

Thank you for you're assistance, but now I am stuck on one final element. I am unsure whether or not I have the correct angular impulse equation about point p. Should it be : m(ra X Va1) + m(rb X Vb1) = I*omega(a) + I*omega(b) + m(ra X Va2) + m(rb X Vb2) ?

 

[trogtothedor]

Sorry for being so lost. Our book has nothing near as complex as this problem and the most difficult example we solved in class was a simple rod with a fixes point of rotation.

 

[trogtothedor]

When I plug the values given to us as the answers into that equation, it does not work. I don't know if the values could be wrong, which they have been in the past, or if I did not set up my angular impulse equation correctly.

 

[Yafa]

I think you have to treat each body separately. you have a net loss of energy since your coefficient of restitution is less than 1. You should break up the collision into 3 parts. Find the energy before, the change during, and the energy after the collision.

Check this out, it should be of some assistance. http://www.myphysicslab.com/collision.html

Scroll all the way down near the bottom of the page.

 

[trogtothedor]

I understand that there is energy loss. However, momentum is conserved I thought... which is why conservation of momentum is applied. If that's the case then I don't understand why you can't apply conservation of angular momentum to point P and treat the cars as an entire system.