# Dumbell collision, COM and rotational motion

1. Feb 14, 2016

### Sturk200

Say you've got a dumbbell-type object sitting still in space, and a mass comes in from below and strikes it. We've got a collision. If the mass strikes the dumbbell away from the center of mass, then it will cause the dumbbell to do two things (1) rotate, (2) the center of mass will move up. My question is, does the location at which the incoming mass strikes the dumbbell have any effect on the subsequent center of mass motion?

I know that striking further from the center of mass means the incoming projectile starts with rotational momentum, which is conserved, so the rotation is greater. But if I'm not mistaken, the COM motion only cares about linear momentum, not rotational, so that it wouldn't matter where on the object you strike it, the center of mass will always respond in the same way.

But then I get tripped up because this would mean that you can start with the same amount of energy in your incoming projectile, and get a more or less violent response from the collision simply from changing the geometry, which seems sort of like it would be violating energy conservation?

Does the discussion change it all for elastic, as compared to inelastic collisions?

Thanks.

2. Feb 15, 2016

### Staff: Mentor

The translational motion of the center of mass of the system (dumbbell + mass) will be unaffected by the collision.

3. Feb 15, 2016

### Buzz Bloom

Hi Doc Al:

My impression is that your answer assumed a different question to the one I think Sturk was asking. Would the velocity vector of the center of mass of the dumbbell relative to the projectile be different depending on where on the dumbbell the projectile hits it?

Regards,
Buzz

4. Feb 15, 2016

### jbriggs444

It depends on what is being held constant across the various collision possibilities. One natural assumption might be that the impact velocity is held constant and that the collision is 100% elastic in all cases.

With that assumption, the amount of momentum that is transferred will vary depending on the point of impact. The amount of energy that is transferred and the final velocity of the impacting mass will also vary depending on the point of impact.

[And I think I owe Doc Al an apology for jumping ahead of his intended exposition]

Last edited: Feb 15, 2016
5. Feb 15, 2016

### Staff: Mentor

I realize that, but I wanted to start with a statement that was easily accepted and go on from there.

In general, yes.

If you assume that the impulse transmitted to the dumbbell is held constant, then the motion of the dumbbell's center of mass would not depend on the location of the impact. But you cannot assume that, in general.

6. Feb 15, 2016

### Sturk200

Thanks for all the answers. I think I'm starting to be able to tease out the dependence on the impact location. Here is what I'm getting.

Assumptions: There is a dumbbell of length 2l, on the either end of which is a mass m. In addition, there is an incoming mass, also m, from below, with initial velocity v1i. The center of mass of the dumbbell is obviously directly in the center. Let the point of collision be a distance x from the center of mass. The collision is perfectly elastic, so both momentum and energy are conserved.

When I do energy and momentum conservation I get the following three equations:

v1i == 2 v2f - v1f (linear momentum)
v1i^2 == 2 v2f^2 + v1f^2 + 2 l^2 ω^2 (kinetic energy)
x v1i == 2 l^2 ω - x (v1i + v2f) (angular momentum),

where v1f is the velocity of the incoming (now outgoing) projectile after the collision, v2f is the velocity of the dumbbell center of mass after the collision, and ω is the angular velocity of the dumbbell after the collision. I can solve these for the unknowns and get some complicated expressions, all of which end up having some dependence on x. Is this a valid analysis of the motion?

$\left\{\left\{\text{v2f}\to \frac{2 \left(-\text{v1i} x+\frac{14 l^4 \text{v1i} x}{12 l^4+l^2 x^2}-\frac{2 l^2 \sqrt{-l^2 \text{v1i}^2 x^2 \left(-l^2+4 x^2\right)}}{12 l^4+l^2 x^2}\right)}{x},\text{v1f}\to \frac{-5 \text{v1i} x+\frac{56 l^4 \text{v1i} x}{12 l^4+l^2 x^2}-\frac{8 l^2 \sqrt{-l^2 \text{v1i}^2 x^2 \left(-l^2+4 x^2\right)}}{12 l^4+l^2 x^2}}{x},\omega \to \frac{2 \left(7 l^2 \text{v1i} x-\sqrt{l^4 \text{v1i}^2 x^2-4 l^2 \text{v1i}^2 x^4}\right)}{12 l^4+l^2 x^2}\right\},\left\{\text{v2f}\to \frac{2 \left(-\text{v1i} x+\frac{14 l^4 \text{v1i} x}{12 l^4+l^2 x^2}+\frac{2 l^2 \sqrt{-l^2 \text{v1i}^2 x^2 \left(-l^2+4 x^2\right)}}{12 l^4+l^2 x^2}\right)}{x},\text{v1f}\to \frac{-5 \text{v1i} x+\frac{56 l^4 \text{v1i} x}{12 l^4+l^2 x^2}+\frac{8 l^2 \sqrt{-l^2 \text{v1i}^2 x^2 \left(-l^2+4 x^2\right)}}{12 l^4+l^2 x^2}}{x},\omega \to \frac{2 \left(7 l^2 \text{v1i} x+\sqrt{l^4 \text{v1i}^2 x^2-4 l^2 \text{v1i}^2 x^4}\right)}{12 l^4+l^2 x^2}\right\}\right\}$

I don't really know what that says. Is it correct to say that as x gets larger, v2f gets smaller? What about v1f?

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