Ok, two rigid bars of length l1, l2 and masses m1 and m2. They have linear velocities v1 and v2 as vectors in 3d space. They have angular velocities w1 and w2. My questions: 1)w should be a vector, but how is it specified so I know what axis the bar is spinning about. 2)I detect that the two bars collide. The collision point is at point x1 on bar1 and x2 on bar2. I know from basic mechanics that momentum must be conserved, and energy loss is modelled using a restitution coeeficient (e=final relative velocity/original relative velocity, right?) But what about angular considerations? At first I though angular momentum had to be conserved, but imagine a brick thrown at the end of a pole - initially there''s only linear momentum, afterwars there''s angular motion too. If the collision is at the midpoint of the bar then there will be no angular effect, but how does one calculate the impulse in this situation, and how do you apply impulse in a rotational frame? 3)If the collision is right at the end of the bar, that doesn''t mean the bar will spin but not move, but what proportion of the energy will be rotational and what proportion linear, kinetic energy. Thanks.

I''m not strong at physics, but the first question answers itself. The missing point may be that the axis of rotation passes through the center of mass. So you have a point and you need a direction. Hint, hint, nod, nod.

1) Like lilBuddy said, question 1 answers itself. The bar is spinning about the w axis.

2) Angular momentum is also a conserved quantity. In your example, the brick itself has angular momentum about the center of the pole before the collision (and probably afterwords). Some of that angular momentum is transferred to the pole during the collision.

3) The center of mass of the bar will move exactly as it would if the bar were hit in the center, but, unlike hitting in the center, a hit at the end will also cause the bar to spin.

The constraints of linear momentum conservation, angular momentum conservation, and energy conservation are sufficient to resolve any elastic collision.

-- Zeno

No my example has no angular momentum: I throw a brick in a straight line and it hits the end of a stationary rod. Afterwards the brick still is travelling in a (different) straight line, but the rod now has angular momentum.

btw, are rotational and linear KE summed or treated separately?

This sounds way too specific to _not_ be a homework question.

> 2) Angular momentum is also a conserved quantity.

Yes and no. Angular momentum is conserved about any point. E.g. you could take a fixed point in the world and look at the angular momentum of all objects about it.

But when we refer to the angular moentum of an object we are usually referring to the angular momentum about it''s own centre. As two colliding objects have different centres you cannot sum these angular momentums and get a meaningful result, and it''s certainly not conserved.

> The constraints of linear momentum conservation, angular
> momentum conservation, and energy conservation are sufficient
> to resolve any elastic collision.

No. First don''t use conservation of energy: it''s fine for pong or arkanoid but corresponds to nothing in the real universe and looks very unrealistic. It also means energy put into your system does not dissipate, whic can cause problems as objects are added unless you find another, artificial, way to remove energy.

But more generally you need more information than just the angular and linear momentum and KE. E.g. if I drop a cube onto a flat surface a dozen times and it bounces back (perfectly elasically or not) it can bounce back differently each time even if the momentum and energy are the same, whether or not they are conserved.

Just a tiny pedantic point on terminology... angular momentum is NOT a vector. Angular momentum is defined as the rate of change with respect to time of the angle of a reference line that intersects the centre of mass of the object. Since angles are not vectors, then neither are derivatives of angles. (Angular rotations are not vectors because they are noncommutative). Actually, derivatives of vectors are not vectors, but we won''t go into that!

Cheers,

Timkin

quote:

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Original post by Timkin
Just a tiny pedantic point on terminology... angular momentum is NOT a vector. Angular momentum is defined as the rate of change with respect to time of the angle of a reference line that intersects the centre of mass of the object. Since angles are not vectors, then neither are derivatives of angles. (Angular rotations are not vectors because they are noncommutative). Actually, derivatives of vectors are not vectors, but we won''t go into that!

Cheers,

Timkin

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According to my physics text book angular momentum IS a vector. You define an angular velocity by the axis about which rotation occurs and the rate at which it happens. => It MUST be in vector form to tell us what axis the thing is rotating around.

quote:

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No. First don''t use conservation of energy: it''s fine for pong or arkanoid but corresponds to nothing in the real universe
and looks very unrealistic. It also means energy put into your system does not dissipate, whic can cause problems as
objects are added unless you find another, artificial, way to remove energy.

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I wasn''t going to use energy conservation but restitution ie you basically specify to what extent the collision is elastic. e=1, elastic, e=0, totally non-elastic.


Isn''t angular momentum just the sums of momentums of many points moving in a circle?

Does it matter which part of bar1 (end/middle) hits bar2 as to how bar2 responds?
Or is the impulse the same and I just apply it to wherever on the bar the collision occurs?

How do I apply an impulse to a bar - some will go towards linear velocity and some to angular, but how?

Thought of a better argument against conservation of angular momentum:
I run towards a roundabout and jump on. I change from travelling in a straight line to travelling in a circle. Maybe it''s the sum of angular and linear momentum that''s conserved?

For such an easy problem, I''m surprised no-one''s offered a solution - considering some of you must have made proper collosions between 3d objects rather than rods...

who ever said garbage about homework, go play with yourself..

how do you do homework? is it required that you read a book and regurgitate an answer?

even if it was homework, there is no guarantee that the answer someone receives is right, thus causing further research.. THEN, more is learned because the researcher has to look at different solutions and determine which is better. (or they can copy an incorrect answer and receive a bad grade... either way the appropriate grade is earned)

Also, (to continue my rant) if someone didn''t know how to answer a homework question, they could ask the teacher, right? Well, they asked us instead.

FINALLY, if you can''t tell that the guy isn''t some moron spouting a "two trains leave" problem, you need help. If everyone who asked a question somebody else thought was elementary didn''t ask the question, NOBODY would ever learn anything.