
#1
Dec610, 06:16 PM

P: 26

1. The problem statement, all variables and given/known data
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. 2. Relevant 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 3. 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? 



#2
Dec610, 07:59 PM

P: 23

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. 



#3
Dec610, 08:02 PM

P: 26

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




#4
Dec610, 08:08 PM

P: 23

Two Car Eccentric Impact
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. 



#5
Dec610, 08:13 PM

P: 23

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




#6
Dec610, 08:19 PM

P: 26

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




#7
Dec610, 08:20 PM

P: 23

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) 



#8
Dec610, 08:21 PM

P: 23

r = distance from c.m. to orthogonal velocity




#9
Dec610, 08:22 PM

P: 26

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 nonzero velocity.




#10
Dec610, 08:23 PM

P: 26

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 nonzero 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.




#11
Dec610, 08:25 PM

P: 23

what do you have so far?




#12
Dec610, 08:30 PM

P: 26

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




#13
Dec610, 08:38 PM

P: 26

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 precollision, 1, and post collision, 2. e is the coefficient of restitution. 



#14
Dec610, 08:39 PM

P: 26

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.




#15
Dec610, 08:42 PM

P: 26

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). 



#16
Dec610, 09:17 PM

P: 23

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... 



#17
Dec610, 09:20 PM

P: 23

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.




#18
Dec610, 09:39 PM

P: 26

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).



Register to reply 
Related Discussions  
Creating a vert. drop impact that is equivalent to a faster, but lighter hor. impact  General Physics  5  
Eccentric orbits  General Astronomy  0  
Eccentric test  General Discussion  3  
Why eccentric orbits?  General Astronomy  2 