Quick physics question (linear momentum and angular momentum)

[uniqueuponhim]

I just had a question about angular momentum and linear momentum, which is probably best illustrated with an example.

Imagine a universe with a perfectly spherical asteroid of mass M, on which is firmly attached a massless cannon, pointed in a direction tangential to the surface of the asteroid. In the cannon is a cannonball of mass m. If the cannon fires the cannonball with a velocity v, giving it momentum mv, how exactly is the motion of the asteroid described? I can only assume it is given both linear and angular momentum, but how exactly do you determine how much of each? Can angular momentum ever take the place of linear in the law of conservation of momentum?

[GOURDHEAD]

Only angular momentum would be imparted to the asteroid in your example because to impart linear momentum, some portion of the force from firing the cannon would have to be directed through the center of mass of the asteroid which is not the case as you described the problem. If none of the force were directed tangentially, then the transfer of momentum would be totally linear. Most real world cases fall in between these two.

[crosscountry]

yep, bring up angular momentum then you have to add in torque.

you're cannon (by releasing the cannon ball) torqued the sperical asteroid. The force of the shot times the radius of the asteroid is the torque. this gives it angular momentum.

as Gourdhead said, if the cannon was not fired entirely tangential, the there would be a radial component of the imparted momentum, thus would change the linear momentum of the asteroid.

[swansont]

yep, bring up angular momentum then you have to add in torque.

But, viewed as a system, the torque is internal. You can apply conservation of angular momentum and describe the behavior - you just have to make sure you are measuring the angular momentum of both the ball and asteroid about the same axis.

[uniqueuponhim]

Ok, but if you had a second cannon on the opposite side of the planet and they both fired simultaneously, then you would get only linear momentum, and no angular momentum added to the asteroid. And if You fired one first, then waited exactly one rotation and fired the other one, you would achieve the same effect.
Also, would it be correct to state that only the total momentum is conserved, but neither linear nor angular momentum alone have to be?

[snarkophilus]

Ok, but if you had a second cannon on the opposite side of the planet and they both fired simultaneously, then you would get only linear momentum, and no angular momentum added to the asteroid. And if You fired one first, then waited exactly one rotation and fired the other one, you would achieve the same effect.
Also, would it be correct to state that only the total momentum is conserved, but neither linear nor angular momentum alone have to be?

It would be correct to say that in this case, but there are cases where angular momentum (and therefore linear momentum) are conserved -- orbiting bodies, for instance. If the cannonball was bound to the planet in some way (say, through gravity, where the cannonball is launched at less than escape velocity), then angular momentum would have to be conserved (but you'd need to consider the angular momentum of the cannonball as well).

The cases are different because it doesn't make sense to talk about angular momentum when dealing with objects that aren't bound. Notice that when you talk about a spinning planet, you're really talking about a whole bunch of bodies orbiting an axis. If you introduce an object not in this orbital system, then you can exchange linear and angular momenta.

This leads to some neat results in quantum mechanics, coincidentally.

[crosscountry]

But, viewed as a system, the torque is internal. You can apply conservation of angular momentum and describe the behavior - you just have to make sure you are measuring the angular momentum of both the ball and asteroid about the same axis.

genius!!!! you sir have correctly stated the issue!!!!

[adiffer]

ummm...

The firing of the first cannon (post #1) would impart linear and angular momentum to the larger mass. If the cannonball leaves with linear momentum 'mv' then the asteroid must have linear momentum '-mv'. This is a must because there are no external forces to change the linear momentum of the whole system.

The angular momentum of the departing cannonball is 'mvR' if you place an axis at the center of M. Again, there are no external torques, so the asteroid must have angular momentum '-mvR'.

Look at these from outside. Find a frame that makes all forces internal and life gets simpler.

[adiffer]

oops. The angular momentum values should use the rotational internia of the asteroid and won't wind up as 'mvR'. duh. 8)

[hhEb09'1]

The cases are different because it doesn't make sense to talk about angular momentum when dealing with objects that aren't bound.

Are you sure about that? I don't see why not. What if you were able to suddenly bind them--you wouldn't "create" angular momentum out of nothing.

[snarkophilus]

Are you sure about that? I don't see why not. What if you were able to suddenly bind them--you wouldn't "create" angular momentum out of nothing.

Well, if you don't have objects moving about a point (or an axis or whatever), then sort of by definition you can't talk about angular momentum, right?

Suppose you were to bind two objects that were moving parallel to each other in opposite directions just as they passed each other. They would gain angular momentum about a point between them, but at the expense of their individual linear momenta (relative to that point). So you're not creating it out of nothing. Total linear momentum is conserved (and energy is conserved), but the acceleration toward each other generates angular momentum.

To illustrate why it doesn't make sense to talk about angular momentum with unbound objects, imagine a plane with x and y axes. There are two unbound objects, A and B. Object A is at (1,1) moving left at some speed, and object B is at (-1,-1), moving right at the same speed. Where do you define the point about which the angular momentum exists (assuming it exists)?

Suppose you use (0,0) as that point (it really doesn't matter which point you pick -- the argument still holds -- but (0,0) is a nice, symmetric choice). Pick three points along the path of A (or B -- it's the same) so that the middle point is halfway between the other two (so that the time taken to travel between these points is equal). I'll take (1,1), (1,1/2), and (1,0). Now, what are the angles generated by connecting these points to (0,0)?

From (1,1) to (0,0) to the x axis, it's 45 degrees. From (1,1/2) to (0,0) to the x axis it's about 63 degrees. From (1,0) to (0,0) to the x axis, it's 90 degrees.

This means that the angular velocity in the first part of the trip, from (1,1) to (1/2,1) is (63-45)/t = 18/t, while the angular velocity in the second part is (90-63)/t = 27/t. Angular momentum has not been conserved!

This is compounded when you consider that (0,0) was an arbitrary choice of axis. Using different points will give you different amounts of non-conservation.

[hhEb09'1]

Well, if you don't have objects moving about a point (or an axis or whatever), then sort of by definition you can't talk about angular momentum, right?

No, you can. You can choose any center of rotation that you like.

This means that the angular velocity in the first part of the trip, from (1,1) to (1/2,1) is (63-45)/t = 18/t, while the angular velocity in the second part is (90-63)/t = 27/t. Angular momentum has not been conserved!

Angular momentum is always conserved. Go ahead and compute it.

[snarkophilus]

No, you can. You can choose any center of rotation that you like.
Angular momentum is always conserved. Go ahead and compute it.

Forgot about the distance component. That is what I get for staying up too late.

I still stand by the statement that this is only true in the absence of external forces, however. (Although one could argue that, in the universal sense, there are no universal forces. Then I'd have nothing. )

[crosscountry]

Forgot about the distance component. That is what I get for staying up too late.

I still stand by the statement that this is only true in the absence of external forces, however. (Although one could argue that, in the universal sense, there are no universal forces. Then I'd have nothing. )

angular momentum is always conserved. if you torque a system then it's angular momentum changes, but the torque (external force) makes up for it.

[crosscountry]

ummm...

The firing of the first cannon (post #1) would impart linear and angular momentum to the larger mass.

not if (as in Post #1) it was fired 100% tangentially. if you hold a bicycle tire and spin it tangentially, then you add no linear momentum.

[adiffer]

Try it. Strike it tangentially on one side and you'll find you have to hold back the axle.

Any external force leads to a change in linear momentum for a system. F=dP/dt

[crosscountry]

Try it. Strike it tangentially on one side and you'll find you have to hold back the axle.

Any external force leads to a change in linear momentum for a system. F=dP/dt

only forces directed through the center of mass. this can be partial forces if you are not 100% tangential. just like torque cannot be applied through the CM, linear momentum cannot change when the force is not through it.



all in a perfect world. in our world the bicycle tire may move a little, but that is due to other properties.

[adiffer]

No. Sorry.

Linear momentum and angular momentum are not things that can be traded like kinetic and potential energy. All forces change momentum. All torques change angular momentum.

The two dimensional version of the asteroid and cannon is an air hockey table, the puck, and one player. Strike the puck off center and it spins AND takes off in the direction the player struck it.

You can also see this on a microscopic level. The particle actually struck can't know where the center of mass of the rigid body is. It knows it is locked in a lattice, so it transfers forces along those lines. However, a force on the particle IS through its center of mass, so it must pick up linear momentum. That means the rigid body also picks up linear momentum.

One last way to see it is to look at the asteroid and cannonball as a system. The whole system experiences no external forces, so the net momentum and angular momentum cannot change when the cannon is fired. However the cannonball departs with some linear momentum, so the asteroid must have an oppositely directed momentum of the same magnitude to ensure no net change is made to the system. The cannonball also departs with angular momentum because its linear momentum is off-centered. The same argument shows then that the asteroid must be spinning the other way after the cannon is fired to ensure no net change to angular momentum occured.

People often think linear and angular momentum are tradeable. It is unfortunate that we used 'momentum' in both names because that helps to create the confusion. They are similar ideas, but they are very different structurally. The units are different and the geometric ranks are different. The causal force is the only thing that connects them, but all it can do is create change for both separately.

[alainprice]

I'm still trying to decide who I agree with.

Even though in a perfect world, a rocket attached to a tire tangentially would indeed simply accelerate the tire's spin. However, in the cannonball example, the ball itself goes flying in a straight line. Therefore, I don't see why the projecting body does not travel in the opposite direction with equal momentum.

This one is not natural.

[crosscountry]

I'm still trying to decide who I agree with.

Even though in a perfect world, a rocket attached to a tire tangentially would indeed simply accelerate the tire's spin. However, in the cannonball example, the ball itself goes flying in a straight line. Therefore, I don't see why the projecting body does not travel in the opposite direction with equal momentum.

This one is not natural.

the cannon ball wouldn't actually travel in a straight line.



I'm starting to see what adiffer is saying however. when the cannon ball is part of the asteroid/cannon system there is a momentum associated with it.


anytime two objects, previously connected, seperate momentum is conserved.

in the case of the bicycle tire, I am an outside force acting on it with outside torque. the bicycle tire experiences no change in linear momentum.



BUT, with the cannon ball, the torque is internal. After the two objects seperate the initial linera momentum must equal the sum of the final momentums.





so....


asteroid and cannon ball ...... both linear and angular momentum



bicycle tire and my arm, .... just angular momentum.

[swansont]

I'm still trying to decide who I agree with.

Even though in a perfect world, a rocket attached to a tire tangentially would indeed simply accelerate the tire's spin. However, in the cannonball example, the ball itself goes flying in a straight line. Therefore, I don't see why the projecting body does not travel in the opposite direction with equal momentum.

This one is not natural.

The forces and torques are all internal. Both linear momentum and angular momentum will be conserved.

[hhEb09'1]

Forgot about the distance component. That is what I get for staying up too late.

Angular momentum, angular velocity--they both have the same first name.

the cannon ball wouldn't actually travel in a straight line.

Why not? If it is not affected by any other force, it should, shouldn't it?

[alainprice]

Yeah, what happened to Newton's laws of motion? Especially the first law: When the net forces acting on a particle are 0, it will remain at rest(if originally at rest), or will move with constant speed in a straight line(if originally in motion).

Let's add some numbers to the problem.

I will assume the rotating body is a circular disk, of mass m1(10 kg), initally at rest with respect to both rotation and translation. I will also assume the cannonball is of mass m2(1 kg), and begins with a speed vo=0 m/s, and ends with a speed of v1=10 m/s.

Ok, cannonball has 10 kg*m/s, therefore the disk will also have this linear momentum.
The disk(having m1=10 kg) will travel in the opposite direction with a speed of 1 m/s.

Okay, now for the angular momentum. We will assume our disk has a radius of 100 m. Angular velocity is = V / 100 m. Angular momentum is the vector product of the moment of inertia and angular velocity. Moment of inertia is given with respect to the axis that is perpendicular to the disk's surface and passes through its center. Since we are dealing with a disk of even mass distribution, its moment of inertia is given by the formula: I = 1/2 * m * r^2 = 50 000 kg * m^2

Knock yourself out...

[crosscountry]

Angular momentum, angular velocity--they both have the same first name. Why not? If it is not affected by any other force, it should, shouldn't it?

yea it is. the central force of the asteroid. the cannon ball experiences it from start to finish. that central force pulls it from a straight path.


think of curved space.

[adiffer]

quite true. The cannonball is essentially in orbit, though the orbit may be open if the cannon imparts enough energy.

We know from Kepler and Newton that the angular momentum in the orbit is conserved along the path if no other torques are present, so the spin of the asteroid won't be changing during the cannonball's flight whether the path is open or closed.

[crosscountry]

ah, but the spin changes during the impulse of the shot.

[crosscountry]

and you forgot, if the asteroid has large bodies of water, tidal slowing can take place

[adiffer]

all true. A closed orbit would have two impulses. Tidal friction would transfer angular momentum. Reality can be quite a mess. 8)

The core point, though, is that forces change linear momentum and torques change angular momentum. F=dP/dt and N=dL/dt. There is no way for the cannon shot in post #1 not to affect both P and L for the asteroid. There is also no way to trade P and L.

Roughly speaking, the cannon provides and impulse dP(=m dv) for the cannonball. That means the asteroid must also receive an impulse -dP(=M dV). We also know the cannonball departs with some angular momentum dL(=m R dv using the asteroid's center as an axis) so the asteroid picks up an amount -dL. The balance for the two is required because the force is internal.

[Kaptain K]

and you forgot, if the asteroid has large bodies of water, tidal slowing can take place

Name an asteroid with "large bodies of water"!

[crosscountry]

Earth