# Two Point Collision Response?

by Thales
Tags: collision, point
 P: 13 Hello, I am attempting to determine a two point collision response for a rigid body. One point collision I can calculate, but I can't seem to extend it to two points. So, I thought, maybe I could get some insight here. :) Set up (It'd be great to have the ability to draw this one out!): I know the mass, velocity, angular velocity and moment of intertia of a rigid body, which then simultaneously collides at two contact points. For this case, I'm just going to assume a perfectly elastic collision. The mass it is colliding with will be much, much bigger than the rigid body, so I can assume it infinite. This would be akin to an airplane landing on a run way, for instance, where the two rear wheels make contact simultaneously. I then am looking for the angular and linear reponse velocities. Well, the response velocities should be equal but opposite to the surface of collision (for the perfectly elastic collision). However, determining which portion of the response goes to rotational velocity , and which to linear velocity I'm not sure of. I did try to set this up using impulse equations for both torque and force, but to no avail. Thanks for any feed back.
 P: 524 You are speaking in very general terms, but from what I can decipher, you are actually talking about accelerations, rather than velocities. You need to give some more specifics regaurding the situation you are modeling before anyone can give you a definite answer.
 PF Patron HW Helper Sci Advisor Thanks P: 11,935 I will sketch a technique of dealing with impacts that easily incorporates possible changes in angular velocities and multiple simultaneous collisions. By Newton's 3.law, the impulses acting upon the two objects are of course action/reaction pairs (so I usually just write "impulse" rather than "impulses" when focusing upon thedynamics at a contact point) With an "eccentric impulse" I mean an impulse whose direction is not collinear with the direction vector from an object's C.M. to the point at which the impulse acts. Clearly, only eccentric impulses can change a rigid body's angular velocity! With a "direct impact" between two objects, I mean that the associated impulse acts along the common vector normal at the point of contact. This means that only the magnitude of the impulse is unknown. The general impact case in which the impulse vector's direction is also unknown, is called an "oblique impact". I will in the following limit myself to consider a multi-collision direct-impacts model with eccentric impulses in 3-D. 1. and 2.phase of the impact period: We divide the impact period in 2 phases. At the end the 1st, which lasts physically (ideally speaking) to the moment of maximal deformation of the two objects, we require that the normal velocities of the two objects at all contact points must be equal. Together with the 12 impulse equations for the two objects (for linear and angular momentum changes), these conditions form a closed system of equations. We can solve for intermediate quantities like the two C.M.-velocities at the end of the first phase, the two angular velocity vectors at the end of the the first phase, and the (magnitude(s) of) the primary impulse(s) that has acted upon the objects during the 1.phase. Clearly, associated to each contact point there will correspond an impulse In the 2.phase, a secondary impulse(s) is produced; it is associated with the reversal of elastic deformations. In order to simplify the 2.phase, it is usual to assume it is proportional to the primary impulse, so that we have the relations: $$\vec{I}_{sec}=k\vec{I}_{prim},\vec{I}_{tot}=(1+k)\vec{I}_{prim}$$ where the total impulse that has acted upon an object (at a given point of contact) is $$\vec{I}_{tot}$$ The scalar k is called the restitution coefficient, and in the 1-collision case between two objects, it is easy to show that the choice k=1 conserves the energy of the system (elastic collision) Evidently, then k=0 is the inelastic collision case (which means there is no second impact phase) If you need anything further, let me know..
P: 524

## Two Point Collision Response?

 You need to give some more specifics regaurding the situation you are modeling before anyone can give you a definite answer.

I stand corrected
Good work Arildno!
 P: 13 Gza, I'm using the conservation of momentum, because I want the final velocities after collision. I realize it's general, but I want to find the general solution, presently. arildno, Good feedback! Right now I'm trying to follow my own line of thinking on this matter. I have the following possible solution, but it seems too good to be true, so I doubt it's right. Here's what I have. For the sake of simplicity, imagine a thin rod, and one force, F1, is applied at one side of the C.M. and another, F2, at the other side. The forces will be at right angles to the radius of rotation, and both are going in the same direction, upward. Also, I'm sticking with 2D, to keep the problem to its essence. The rod will have mass M, and moment I. I tried using the latex, but I'm having trouble with it. So, I'll stick with ascii for now. I"ll use the slash, /, for integral. /F1 dt + /F2 dt = / M dv Where dv is change in velocity of c.m. of rod. /F1 r1 dt + /F2 r2 dt = / I dw Where dw is change in angular velocity of rod. If I were to measure the total velocity at F1, call it v1... It'd be 1> v1 = dv + dw*r1 The velocity at F2, call it v2... 2> v2 = dv + dw*r2 So, if I'm solving for two point collisiion, the two parameters I'd know at the start are v1 and v2 From equations 1 and 2 above I have two equations and two unknowns. Thus, I can find the linear and angular components of the rod after collision.
 PF Patron HW Helper Sci Advisor Thanks P: 11,935 I'm sorry, I don't understand what you are doing here: Do you know F1 and F2??(you don't) Each object will gain new translational velocities and angular velocities. At the very least, this gives you 6 unknowns...
 P: 13 It wasn't well presented. What I'm really trying to do is find the resultant linear and angular velocity components after impact. I know, at the outset, the velocities of each collision point before impact, as well as their points of impact. So, with a perfectly elastic collision, they must be the same after impact. F1 and F2 I'm only putting there to show that they in fact sum up to the final linear and angular velocities. I don't need to find their values. Also, dv is the change in linear velocity, and dw is the change in angular velocity.
 PF Patron HW Helper Sci Advisor Thanks P: 11,935 Hi Thales, First a definition: With "normal collision velocity", I mean the normal component of the velocity difference between the contact points on each object. (That is, how fast they crash into each other at a given point of contact). For a fully elastic collision, you are right in asserting that the "final normal response velocity" at a given contact point will be the negative of the normal collision velocity at that point (Newton's condition) (with "final normal response velocity" I mean the normal component of the velocity difference after impact, i.e, how fast the objects part from each other) This follows quite easily from direct impact theory. In your 2-D case, you assumed that one of the objects (the ground) will have zero linear and angular velocity. In addition, you have assumed that the direction of the final linear velocity of the other object is known(the plane). Hence, you really have only two unknowns (the magnitude of the linear velocity and the induced angular velocity), and you should therefore be able to cope with only two equations (for example, the contact points' demands) Before commenting at this, however, I would like to say that in the general 2-D case, you will have 6 fundamental unknowns (the 4 linear velocities and the two angular velocities after impact). Nothing can reduce this number, except highly specific situations/geometries where some of these fundamental unknowns are known at the outset (as in your case). To mention one tricky feature of fully elastic direct impacts, the contact point tangential velocity may change from its original value if rotations are induced! If you require no sliding velocity between the two objects during a fully elastic impact (i.e, an oblique impact), the tangential response velocity will become the negative of the "tangential collision velocity" I will comment on your plane scenario somewhat later.
P: 13
 Quote by arildno Hi Thales, First a definition: With "normal collision velocity", I mean the normal component of the velocity difference between the contact points on each object.
Understood.

 In your 2-D case, you assumed that one of the objects (the ground) will have zero linear and angular velocity. In addition, you have assumed that the direction of the final linear velocity of the other object is known(the plane). Hence, you really have only two unknowns (the magnitude of the linear velocity and the induced angular velocity), and you should therefore be able to cope with only two equations (for example, the contact points' demands)
This is what I gathered. :)

 Before commenting at this, however, I would like to say that in the general 2-D case, you will have 6 fundamental unknowns (the 4 linear velocities and the two angular velocities after impact).
Right, this is true.

 To mention one tricky feature of fully elastic direct impacts, the contact point tangential velocity may change from its original value if rotations are induced!
I was thinking that if I were to apply a frictional force, say at the wheels of an airplane, as it lands, this would be at right angles to the collision force.

And, given that

$$\mu_k = \frac{f}{F}$$

$$\int F\mu_k dt = \int f dt = \int M dv_x$$

Since

$$\int (F_1 + F_2) dt = M dv_y$$

I can write

$$\frac{ \mu_k \int (F_1 + F_2) dt}{M} = \mu_k \int dv_y = \int dv_x$$

Okay, enough latex

You'd do the same thing for the torque part.

 I will comment on your plane scenario somewhat later.
I'll definitely be interested.

 P: 13 I've been thinking about this, and I have another question, but this time with only a one point collision. I have a bit of a conundrum. If I have a thin rod with Mass M, and moment of inertia I, and the rod is hit by an impulse force at some distance r from its c.m., will the I get the following resultant angular, W, and linear, V, velocities? The rod is initially not rotating, and bangs into a huge obstacle with a perfectly elastic collision. Impulse equations: 1> For Forces in y direction (direction of force) F*dt = m *dV 2> For torques F*r*dt = I * dW Then, if I know the force and the short duration dt... I can substitute F*dt of eq 1 into eq 2, thus: 3> m*dV*r = I * dW This tells me that the angular and linear components of velocity are related by equation 3. My conundrum is that the linear velocity remains the same regardless of where along the rod the force is applied, while the angular component is dependent upon how far from the c.m. the force is applied. Does this make sense?
 PF Patron HW Helper Sci Advisor Thanks P: 11,935 Okay, I was a bit tired yesterday, so I misread your problem. Sorry about that, I'll check upon your actual question.
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P: 11,935
 Quote by Thales If I have a thin rod with Mass M, and moment of inertia I, and the rod is hit by an impulse force at some distance r from its c.m., will the I get the following resultant angular, W, and linear, V, velocities? The rod is initially not rotating, and bangs into a huge obstacle with a perfectly elastic collision.
Is the following interpretation correct?
The rod has an inital linear velocity, and bangs elastically into a extremely massive object at distance r from C.M?
 PF Patron HW Helper Sci Advisor Thanks P: 11,935 I will assume that I finally read your problem correctly! I'll use impact theory for this. Let $$v_{0}\vec{i}$$ be the rod's initial (linear) velocity, and m its mass, $$\mathcal{I}$$ its moment of inertia with respect to C.M., and r distance to the impact point. Hence, we have in the 1. impactperiod: $$-I_{1}=m(v_{1}-v_{0})$$ $$r\vec{j}\times(-I_{1}\vec{i})=\mathcal{I}\omega_{1}\vec{k}$$ $$v_{1}-r\omega_{1}=0$$ (the last equation is the requirement that the normal velocity at the contact point must be zero during the first phase of impact) Solving for the primary impulse $$I_{1}$$ we get: $$I_{1}=\frac{m\mathcal{I}v_{0}}{\mathcal{I}+mr^{2}}$$ 2.Impact period: In an elastic collision, the secondary impulse is set equal to the primary impulse, and hence, we may solve for the final values: $$v_{f}=\frac{(mr^{2}-\mathcal{I})v_{0}}{\mathcal{I}+mr^{2}}$$ $$\omega_{f}=\frac{2mrv_{0}}{\mathcal{I}+mr^{2}}$$
 P: 13 arildno, Many thanks Sorry for the late response. I've been having Internet troubles on this end, and haven't been able to respond. You have the problem exactly right. However, I have a question, which I assume is simple. How do you manipulate the cross product in that equation? I understand cross products, but I'm not really adept at manipulating them. I looked at the identities for them, but still I'm not sure. What I'm asking is how you used those first three equations to get the fourth, where you find $$I_1$$ . The cross product throws a wrench in the works for me, so that I'm not readily able to solve it. Btw, I note that you broke the problem down exactly as you mentioned in your first response to me in this thread. I'll have to remember that. It's a good conceptual way of picturing such problems.
 PF Patron HW Helper Sci Advisor Thanks P: 11,935 Ok: I have used the convention: $$\vec{i}\times\vec{j}=\vec{k},\vec{j}\times\vec{k}=\vec{i},\vec{k}\times \vec{i}=\vec{j}$$ Since the cross-product is anticommutative, we have, for example $$\vec{j}\times\vec{i}=-\vec{k}$$ Hence, the 2.equation becomes:$$I_{1}r\vec{k}=\mathcal{I}\omega_{1}\vec{k}$$ or: $$\omega_{1}=\frac{{I}_{1}r}{\mathcal{I}}$$ I'd like to say that even if you in many cases may skip the explicit calculation of the impulse, I find that retaining it makes the solving of problems much easier..
 P: 13 Thanks arildno, I very much appreciate your deeper insight here. I'm looking again at my equations, then are your equations. If I take your equations, and do the following $$-I_{1}=m\Delta v$$ $$r\vec{j}\times(-I_{1}\vec{i})=\mathcal{I}\Delta\omega\vec{k}$$ And, if I know the impulse value at the start... Then, wouldn't the following relationship be true: $$\frac{\Delta v}{\Delta w} = \frac{-\mathcal{I}}{mr}$$ And wouldn't it further be true that the change in velocity $$\Delta v$$ doesn't change regardless of the value of r?
 P: 13 I decided to run an experiment. I tied a mechanical pencil to a piece of yarn and hung it from a floor lamp. The yarn is about 27 inches long. The pencil was balanced at CM. I then have a device that delivers a pretty constant force all of the time, a stapler Anyway, I applied this same force to various points along the pencil. Right at the CM, as expected, no rotation results, and its linear displacement is pretty good. When I applied the force away from the CM, as expected, a rotation is observed, and the futher from the CM the faster the rotation. However, the further from the CM the less the linear displacement or swing. I ran this experiment several times to confirm the results. I eyeballed the movement, but the magnitude of the quantities was such that it was easy to make comparisons that way. Anyway, the long and the short of what I'm getting at is that the linear displacement changes as you move further from the CM, yet I can't make sense of how the equations are correctly describing this. Any ideas?
 P: 13 Okay, I think I have figured this out. I apologize for thinking aloud on the board, but I'm always looking for feedback. Here is where I erred in my assessment of that experiment I posted above. Even though the spring force has the same velocity when let go, it doesn't supply the same force to the rod (mechanical pencil) at all points along the radius. When the force is applied to the CM all of the displacement, during acceleration, goes to the pencil's linear velocity. However, as you move further out, more of it goes to angular velocity. This means that the force has to travel further in the same time as you move further out from the CM. So, the vital component is the work done. The impulse force doesn't change, but the work done must change. The work is Force*distance. The impulse is Force*time. The impulse force has to cover more distance in the same time as you move further out along the radius, in order for the impulse to be the same. If this happens, then the amount of swing should be the same if the impulse is the same, regardless of where along the rod the force is applied. It's amazing how something seemingly so simple can be such a bear. But, I think I have this one right, for now. Any feedback would be appreciated!

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