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uniqueuponhim
2005-Sep-24, 01:15 PM
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
2005-Sep-24, 01:55 PM
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
2005-Sep-24, 02:25 PM
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
2005-Sep-24, 02:58 PM
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
2005-Sep-24, 03:05 PM
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
2005-Sep-25, 03:37 AM
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
2005-Sep-25, 05:42 AM
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
2005-Sep-25, 06:05 AM
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
2005-Sep-25, 06:07 AM
oops. The angular momentum values should use the rotational internia of the asteroid and won't wind up as 'mvR'. duh. 8)

hhEb09'1
2005-Sep-25, 10:55 AM
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
2005-Sep-25, 11:29 AM
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
2005-Sep-27, 07:42 PM
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 (http://scienceworld.wolfram.com/physics/AngularMomentum.html) is always conserved. Go ahead and compute it. :)

snarkophilus
2005-Sep-27, 09:12 PM
No, you can. You can choose any center of rotation that you like.
Angular momentum (http://scienceworld.wolfram.com/physics/AngularMomentum.html) 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
2005-Sep-28, 01:46 AM
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
2005-Sep-28, 01:47 AM
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
2005-Sep-28, 06:44 AM
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
2005-Sep-28, 01:32 PM
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
2005-Sep-28, 08:34 PM
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
2005-Sep-28, 10:34 PM
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
2005-Sep-28, 11:45 PM
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
2005-Sep-29, 12:05 AM
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
2005-Sep-29, 06:02 PM
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
2005-Sep-30, 01:52 AM
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
2005-Sep-30, 02:57 AM
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
2005-Sep-30, 03:29 AM
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
2005-Sep-30, 03:42 AM
ah, but the spin changes during the impulse of the shot.

crosscountry
2005-Sep-30, 03:43 AM
and you forgot, if the asteroid has large bodies of water, tidal slowing can take place

adiffer
2005-Sep-30, 05:37 AM
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
2005-Sep-30, 11:27 AM
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
2005-Sep-30, 02:34 PM
Earth

Kaptain K
2005-Sep-30, 04:18 PM
Quick! Somebody call Brian Marsden! I have it on good authority that Earth is an asteroid!!!

hhEb09'1
2005-Sep-30, 07:54 PM
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.Ah, OK. I just assumed you were ignoring the gravitational effect, because of your previous statement
The force of the shot times the radius of the asteroid is the torque.
and you forgot, if the asteroid has large bodies of water, tidal slowing can take placeTidal slowing does not require water :)

crosscountry
2005-Oct-01, 12:40 AM
ok, tidal slowing does not require water.




for the record, the difference between a planet and an asteriod was imposed by man. (arbitrary)

EGGHEAD
2009-Mar-22, 05:03 PM
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?

I'm sure the asteroid will be given no angular momentum at all only linear
momentum as there will be no torque applied to the asteroid at all!
Imagine this; The shape of the asteroid is irrelevent ,it could be any shape as the shape won't affect the type of momentum it receives.Let's say it's shape is cube and the cannonball is fired from the edge [remember that it's centre of mass remains unchanged] now imagine an identical but miiror image cube with a similar cannonball fired from it's edge, now imagine these two cubes are bound together across their adjacent surfaces, how will you descibe the motion of this larger mass? Obviously it will have only have linear momentum.Now, imagine[a lot of imagining is required!] this larger mass breaks up into two cubes after the firing of the cannonball [our original two cubes] How will you describe the motion of these two seperate cubes? They will continue to move on the straight path they were travelling before the seperation with only linear momentum.There will be NO tension between the two surfaces to bring about a change in motion,and because there is no tension implies neither cube was constrained to move in a linear motion at the time the cannonball was fired.If the two cubes were not secured together at the time the cannonballs were fired both cubes would still move together and possess only linear momentum.
You can't cause any object to spin by the application of a single non-torque force,a wheel spins because it has an axle [Newton's 3'rd law]an asteroid doesn't.

crosscountry
2009-Mar-22, 09:49 PM
no, even in your scenario angular momentum is applied. The only difference is that since you're not using a sphere the force is no longer tangential, and you get both angular and linear acceleration.

If you aim the cannon perpendicular to a line from the CM to the cannonball you'll be back to only angular momentum.




I'll repeat this:

Torque cannot act through a center of mass. Any force acting through the Center of Mass will change linear momentum.

Since our cannon ball is now fired perpendicular to the CM line only angular momentum is affected, and lm doesn't change.

And I want to go back and unsay what I said earlier about the asteroid moving. It will only spin. The ball receives linear acceleration and the equal and opposite force is applied through a lever arm the radius of the asteroid as a torque on the asteroid.

Hornblower
2009-Mar-23, 02:33 AM
In my opinion both EGGHEAD and crosscountry are mistaken. If the cannon is tangent to the asteroid, the recoil will cause the asteroid to move off in the opposite direction to that of the cannonball, and at the same time will set it spinning. That is the only way to have conservation of both linear and angular momentum.

Let's imagine the cannon on the bottom of the asteroid, aimed to the left. The system is the combination of the asteroid, the cannon and the cannonball. Before firing, everything is stationary, and the barycenter of the system is below the center of the asteroid and above the ball. This barycenter is stationary and remains stationary after firing the cannon, thus satisfying the condition of conservation of momentum. Since the ball is moving to the left after the firing, the asteroid must be moving to the right. This has to be the case regardless of whether or not the asteroid is spinning after the firing.

Now let's look at the angular momentum, which also must be conserved. The line of position through the ball, the system barycenter and the center of the asteroid is rotating clockwise after the firing. Thus these components of the system motion give us a non-zero component of angular momentum with respect to the barycenter. Since the overall resultant must remain zero, we must have a counterclockwise spin of the asteroid.

Clear as mud? I realize this can be hard to follow from words alone. Walking you step by step through some sketches in the classroom is much better.

crosscountry
2009-Mar-23, 04:39 AM
ah, but you're forgetting something.


Here's the deal, the cannon ball and asteroid+cannon are a system. You have to look at the whole picture.

The cannon ball after being fired now has angular momentum, just like a comet does so around the sun. The angular momentum is centered around the previous CG, and the cannon ball makes an arc (it does not travel in a straight line).

And since the ball does not travel in a straight line we don't measure that as linear momentum.

The total system has 0 net angular and 0 net linear momentum. After the cannon fires the cannon takes a positive(+) angular momentum and the asteroid takes a negative (-) angular momentum.

take heed: some of the force would have to go through the center of mass to have the asteroid move other than spin.






stick your arm straight out with open fist. Now, I'll punch your hand. Your body rotates about the center of mass - it does not move translationally. Just like the bicycle wheel doesn't move, it only spins.



Or, have a kid on a swing. You push the kid, and he/she moves in a pivot around the bar above. The bar doesn't move, and yet you pushed the kid tangentially. Why doesn't the bar move?

Ken G
2009-Mar-23, 06:10 AM
adiffer and Hornblower are right. This problem is a fairly straightforward application of the conservation of linear and angular momenta. We may imagine the asteroid starts off with zero of both, and so the asteroid + cannon ball will always, together, have zero of both (neglecting any other bodies, as none were mentioned in the OP, and also neglecting the momentum of any gases used to fire the cannon ball, so a catapult would be a better example). That means that whatever is the linear momentum of the cannon ball, at any later time (curved path or otherwise), that will always be equal and opposite to the linear momentum of the asteroid. It is simply not true that forces have to have a "component through the center of mass" to impart linear momentum-- indeed, forces are vectors, and so is linear momentum, the latter being given by the time integral of the former (with no regard to the location where the former is applied).

Location does come in when analyzing angular momentum and torque, which are not formally vectors, but again it is much easier here to simply note that the total angular momentum will be zero. The angular momentum of the cannon ball is defined with regard to a point, and any point may be used to do that, and the total angular momentum, around any stationary or inertially moving point, will always be zero. One natural choice is to take for our point the initial center of the asteroid (or more accurately, Hornblower's barycenter), and then it remains only to inspect the motion of the cannon ball at any later moment of our choosing to determine its angular momentum about that point, and whatever that is, will be equal and opposite to the angular momentum of the planet. However, the planet will have both spin and orbital angular momentum, the latter coming from the gravity of the cannon ball, so if we just want to know the spin angular momentum of the planet, it will just be what it started with after the cannon ball was first fired. Later changes will all have to be in the orbital angular momentum-- the spin won't change, because tidal influences from the cannon ball will be so tiny as to be utterly negligible.

Indeed, it seems to me the basic question from the OP can be best answered in the limit where the cannon ball is shot at a speed much faster than the escape velocity, and we simply neglect all gravitational effects altogether. They don't add anything fundamental to the answer. Again the asteroid will acquire linear momentum, and also angular momentum around the barycenter in the form of spin.

Note also that if we are neglecting gravity, another interesting point to choose is the point of the initial location of the cannon ball. Around that point, the angular momentum of the cannon ball is always zero (if we say it is shot with no spin of its own), so the angular momentum of the planet will also be zero about that point. That allows us to connect the spin of the planet to its linear momentum, as the latter will show up as orbital angular momentum around that point (whether the system is bound or otherwise, and indeed here I'm neglecting gravity so it is unbound).

Hornblower
2009-Mar-23, 01:44 PM
ah, but you're forgetting something.


Here's the deal, the cannon ball and asteroid+cannon are a system. You have to look at the whole picture.

The cannon ball after being fired now has angular momentum, just like a comet does so around the sun. The angular momentum is centered around the previous CG, and the cannon ball makes an arc (it does not travel in a straight line).

And since the ball does not travel in a straight line we don't measure that as linear momentum.
That simply is not true. At any time just after the launch the ball has velocity that it did not have before launch, and thus it has momentum that it did not have before launch. The fact that the asteroid's gravitational field will cause the ball's trajectory to deviate from a straight line does not invalidate this analysis. I recommend that you find an introductory physics textbook and review the definition of momentum.


The total system has 0 net angular and 0 net linear momentum. After the cannon fires the cannon takes a positive(+) angular momentum and the asteroid takes a negative (-) angular momentum.

take heed: some of the force would have to go through the center of mass to have the asteroid move other than spin.






stick your arm straight out with open fist. Now, I'll punch your hand. Your body rotates about the center of mass - it does not move translationally. Just like the bicycle wheel doesn't move, it only spins.



Or, have a kid on a swing. You push the kid, and he/she moves in a pivot around the bar above. The bar doesn't move, and yet you pushed the kid tangentially. Why doesn't the bar move?The bar does not move because it is attached to the virtually immovable ground. Our thought-experiment asteroid-ball system is in free fall.

m74z00219
2009-Mar-23, 09:59 PM
Well, there is an interesting "hole" in the question that no one seems to have noticed.

If you ignore gravity, then the asteroid would NOT be affected by the cannonball departing at its new non-zero relative velocity. It was clearly stated that the cannon was massless. Therefore, the cannonball cannot impart a reaction force (Newton's third law) onto the cannon. Hence, the cannon cannot impart a force on the asteroid.

Since, in reality, there is gravity, this wouldn't be the case. After chemical energy is converted into the KE of the cannonball, we'll have the total mechanical energy of the system (a conserved quantity ignoring gravitational radiation, so we can assumer the cannonball/asteroid is a closed system).
Regardless of what happens, this total energy has to be conserved in the system (rotational kinetic energy, linear KE, gravitational potential energy).

The system is given angular momentum (you can very conveniently choose the center of mass as the center of rotation, essentially the center of the perfectly spherical asteroid).
tau = r X F , r essentially being the radius of the asteroid. F being the force imparted by the expanding gas in the massless cannon. Though the cannonball is not attached to the asteroid, it is connected gravitationally.

However, there is also a linear momentum component because as the cannonball moves away, the two objects will continue to attract each other (they will accelerate back together, to the common center of mass).

Regardless of the details, after the initial change in mechanical energy imparted by the cannon, the new total mechanical energy must be conserved. As well, the new total momentum must be conserved.
P_total = P_linear + P_orbital_angular_momentum + P_spin_angular_momentum


m74

Hornblower
2009-Mar-23, 11:34 PM
Well, there is an interesting "hole" in the question that no one seems to have noticed.

If you ignore gravity, then the asteroid would NOT be affected by the cannonball departing at its new non-zero relative velocity.
Total non sequitur. The presence or absence of significant gravity does not affect the phenomenon of recoil when a gun is fired.

It was clearly stated that the cannon was massless. Therefore, the cannonball cannot impart a reaction force (Newton's third law) onto the cannon. Hence, the cannon cannot impart a force on the asteroid.
In this sort of thought experiment, the cannon is treated as part of the asteroid, which thus will recoil with a finite velocity. Newton's law of action and reaction is alive and well.


Since, in reality, there is gravity, this wouldn't be the case. After chemical energy is converted into the KE of the cannonball, we'll have the total mechanical energy of the system (a conserved quantity ignoring gravitational radiation, so we can assumer the cannonball/asteroid is a closed system).
Regardless of what happens, this total energy has to be conserved in the system (rotational kinetic energy, linear KE, gravitational potential energy).

The system is given angular momentum (you can very conveniently choose the center of mass as the center of rotation, essentially the center of the perfectly spherical asteroid).
tau = r X F , r essentially being the radius of the asteroid. F being the force imparted by the expanding gas in the massless cannon. Though the cannonball is not attached to the asteroid, it is connected gravitationally.

However, there is also a linear momentum component because as the cannonball moves away, the two objects will continue to attract each other (they will accelerate back together, to the common center of mass).
More non sequitur. There is a linear momentum component because of the motion imparted by the propellant, not because of subsequent gravitational action.


Regardless of the details, after the initial change in mechanical energy imparted by the cannon, the new total mechanical energy must be conserved. As well, the new total momentum must be conserved.
P_total = P_linear + P_orbital_angular_momentum + P_spin_angular_momentum


m74
That equation is invalid because linear and angular momentum terms have different units.

Overall, this post is so incoherent that I cannot tell whether you think Ken G and I are right, wrong, all of the above, or none of the above. I recommend that you find a good Physics 101 book and brush up on the fundamentals.

m74z00219
2009-Mar-24, 05:49 AM
Total non sequitur. The presence or absence of significant gravity does not affect the phenomenon of recoil when a gun is fired. In this sort of thought experiment, the cannon is treated as part of the asteroid, which thus will recoil with a finite velocity. Newton's law of action and reaction is alive and well.

It's not a nonsequitor. The problem was setup as two bodies in closed system (the only things in a universe). If the cannon is massless, it can NOT impart any force onto the asteroid. F=ma. If m is zero, force is zero.



More non sequitur. There is a linear momentum component because of the motion imparted by the propellant, not because of subsequent gravitational action.

Yes there is linear momentum (once again, not non-sequitor), but you are analyzing the problem incorrectly. You have to begin analysis after the input of energy into the system. So, AFTER the propellant imparts energy unto the cannonball, you know the total mechanical energy and have a basis for your analysis (you also "immediately" know the total angular and linear momentum). Most definitely will the mutual gravitational attraction affect the ball's linear momentum. The cannonball will slow because it's being attracted, and hence the asteroid will gain linear momentum (leaving the total linear momentum constant). Then vice versa when the cannonball is pulled back into the body of the asteroid.



That equation is invalid because linear and angular momentum terms have different units.
Overall, this post is so incoherent that I cannot tell whether you think Ken G and I are right, wrong, all of the above, or none of the above.


I made ONE error. I meant that the total angular momentum and total linear momentum are independently conserved.

Plinear = Plinear(cannonball) + Plinear(asteroid)
Pang = Pang_orbit(cannball) + Pang_spin(cannonball) + Pang_orbit(asteroid) + Pang_spin(asteroid)


My main points where that there is linear and angular momenta, the system (as presented), begins with a non-zero angular momentum, a non-zero linear momentum, and a non-zero mechanical energy when you analyze properly. Properly being after the launch because the cannon is massless and so for all intents and purposes the cannonball has a non-zero KE at time zero with an initial velocity tangent to the asteroid and an initial position at about the surface of the asteroid.



I recommend that you find a good Physics 101 book and brush up on the fundamentals.

Honestly, is that sort of comment necessary. I admittedly did make a (single) error, but you really have no right to make such criticism when your analysis is flawed.

Hornblower
2009-Mar-24, 12:26 PM
It's not a nonsequitor. The problem was setup as two bodies in closed system (the only things in a universe). If the cannon is massless, it can NOT impart any force onto the asteroid. F=ma. If m is zero, force is zero.
If our massless cannon was not attached to something else, any given impulse from the propellant would send it flying off to kingdom come, and leave the cannonball stationary.

That is not what we have here. The OP clearly states that the cannon is firmly attached to the asteroid. The sole reason for making it massless is to simplify the description of the system. That is a common trick in thought-experiment analysis, and is consistent with Newton's development of his general theory. We can make the cannon massless, or as nearly so as is good limiting-case mathematical practice, and still give it enough structural strength to transmit the recoil from the propellant. The asteroid and the cannon become a single body for the purpose of this exercise, and the lack of mass in the chamber and the barrel do not prevent the transmission of the propellant's force to the asteroid.


Yes there is linear momentum (once again, not non-sequitor), but you are analyzing the problem incorrectly. You have to begin analysis after the input of energy into the system. So, AFTER the propellant imparts energy unto the cannonball, you know the total mechanical energy and have a basis for your analysis (you also "immediately" know the total angular and linear momentum). Most definitely will the mutual gravitational attraction affect the ball's linear momentum. The cannonball will slow because it's being attracted, and hence the asteroid will gain linear momentum (leaving the total linear momentum constant). Then vice versa when the cannonball is pulled back into the body of the asteroid.

My bold for reference. The asteroid, like the cannonball, will lose linear momentum as a result of gravitational action immediately after firing.

You appear to be overlooking the vector nature of momentum. Before firing the system was stationary, hence no linear momentum. After firing, the ball has a momentum vector of nonzero magnitude in one direction, and the asteroid has a momentum vector of equal magnitude in exactly the opposite direction. The vector sum is of zero magnitude, and the system barycenter remains stationary. Gravity causes both bodies to slow down and change direction in unison. Thus the momentum vectors now have reduced but still equal magnitude and are pointing in changed but still directly opposite directions. The barycenter remains stationary. This is the basic property we call conservation of momentum.

If the initial velocity is slow enough, the two bodies will come back together as a result of the gravitational attraction. If they stick together instead of bouncing off, the combined system will once again be stationary.


I made ONE error. I meant that the total angular momentum and total linear momentum are independently conserved.

Plinear = Plinear(cannonball) + Plinear(asteroid)
Pang = Pang_orbit(cannball) + Pang_spin(cannonball) + Pang_orbit(asteroid) + Pang_spin(asteroid)


My main points where that there is linear and angular momenta, the system (as presented), begins with a non-zero angular momentum, a non-zero linear momentum, and a non-zero mechanical energy when you analyze properly. Properly being after the launch because the cannon is massless and so for all intents and purposes the cannonball has a non-zero KE at time zero with an initial velocity tangent to the asteroid and an initial position at about the surface of the asteroid.

Once again you appear to be overlooking the vector nature of these momentum attributes. The overall resultant system magnitudes of both types of momentum are zero at all times.



Honestly, is that sort of comment necessary. I admittedly did make a (single) error, but you really have no right to make such criticism when your analysis is flawed. My confidence in the basic correctness of my reasoning is unshaken. In fact it is strengthened by the exercise of writing technically sound responses to your arguments. This is in spite of being away from physics at the professional or college level for nearly 40 years. Ken G says I have it right, and he adds additional words of wisdom. My challenge to you: Can you convince him and other physicists in this forum that I am wrong?

Ken G
2009-Mar-24, 02:38 PM
Yup, you have it completely right.

swansont
2009-Mar-24, 05:56 PM
I concur as well, Hornblower.




It's incorrect to treat the cannon as some separate entity in analyzing the problem. The cannonball has mass and momentum, as does the asteroid that's what is important. The massless cannon is "physics license" to simplify the problem, nothing more.

EGGHEAD
2009-Mar-24, 09:28 PM
I believe are correct Hornblower. In my wordy analysis I simplified the system by completely ignoring the effect of gravity,however, I still stand by my analysis if the effect of gravity is ignored.

m74z00219
2009-Mar-24, 10:00 PM
Hornblower, I accept that challenge. I will promptly change my day-job of teaching when you or another can persuade me to believe that a massless object can exert a force.

Though it may seem surprising to you, I do understand your reasoning. However, I standby the fact that it does not work if the cannon is truly massless. I would say that the initial thought experiment is broken in this respect unless the problem is analyzed my way. In which case, there is no need for the cannon, just the cannon ball with some initial tangential velocity and the asteroid with zero initial velocity.

Once again, I understand the basic laws of physics and math. I know what a vector is, thank you very much. Also, I am quite familiar with the conservation laws.

I see that if the cannon were massive then it would recoil along with the asteroid, but because it is massless, the initial motion of the asteroid will be toward the cannonball. Or are you going to continue to assert that a massless object can exert a force? You cannot deny that the asteroid would move towards the cannonball (or, more accurately, towards the barycenter) immediately if there is no reason to believe the asteroid was recoiled initially.

I am aware that massless and all manner of idealized objects are used in physics thought experiments (no need to inform me), but I find it to be damaging to the student overall. Better to call the cannon massive and analyze the way Hornblower and Ken G say, then to call it massless and erroneously analyze it that way you two say.

Ken G
2009-Mar-24, 11:00 PM
I believe are correct Hornblower. In my wordy analysis I simplified the system by completely ignoring the effect of gravity,however, I still stand by my analysis if the effect of gravity is ignored.No, your analysis was not correct. The asteroid will certainly spin. You can imagine attaching it to a second asteroid, but your claim that there will be no forces between the asteriods is incorrect-- if your two asteroids exert no forces on each other, they'll simply spin in opposite directions.

Ken G
2009-Mar-24, 11:04 PM
Hornblower, I accept that challenge. I will promptly change my day-job of teaching when you or another can persuade me to believe that a massless object can exert a force.As Hornblower said, we are all in agreement that a massless object cannot exert a force, if it is not anchored by something else that can exert a force. But this cannon is anchored to the asteroid, so it can exert forces, willy nilly. The sole constraint applied by its masslessness is that it must never experience a net force, so any force it exerts, which will of course create an equal and opposite force on the cannon, must be "passed on" to some other object so that the net force on the cannon may always be zero. Massless objects that exert forces in this way are perfectly mundane in physics, have no not heard of a "massless string" or a "massless pulley"? They appear in many places in physics textbooks, and as swansont put it, it is just an example of "physics license". Is a massless cannon really possible? Of course not, but physics commonly makes idealizations of that nature.

I see that if the cannon were massive then it would recoil along with the asteroid, but because it is massless, the initial motion of the asteroid will be toward the cannonball.You lost me there. One can take the limit as the mass of the cannon gets less and less, and in all cases the motion of the asteroid will be completely the same, ergo we pass to the limit of a "massless" cannon with no difficulty.
Or are you going to continue to assert that a massless object can exert a force?Yes, this is quite common in physics, in virtually every chapter of an introductory textbook.


Better to call the cannon massive and analyze the way Hornblower and Ken G say, then to call it massless and erroneously analyze it that way you two say.You are of course welcome to write your own textbook, as we would all allow you that there are no massless cannons.

m74z00219
2009-Mar-25, 09:58 AM
Ok, Thanks Ken and Hornblower. I was wrong to think that to think the cannon wouldn't exert a force. Upon further thought and comparison, I realize that the massless cannon will exert a force on the asteroid provided it is firmly attached. I see that is simply serves as a means to transport force from the cannonball to the asteroid without further complication.

I do see now that the OP's question is not like my setup. Which was equivalent to setting a ball with an initial velocity tangential to a spherical asteroid that has zero initial velocity. And, placing that ball at the surface of the asteroid. But, this simply isn't the case and so I fully concede to the analysis offered by Ken and Hornblower.

Ken G
2009-Mar-25, 06:21 PM
Thank you for that graceful clarification.

blah22
2009-Sep-12, 01:32 AM
Old thread, I know, blame google.

I have a question for Ken G, and it concerns the last part of your post here, which makes an interesting point.



Note also that if we are neglecting gravity, another interesting point to choose is the point of the initial location of the cannon ball. Around that point, the angular momentum of the cannon ball is always zero (if we say it is shot with no spin of its own), so the angular momentum of the planet will also be zero about that point. That allows us to connect the spin of the planet to its linear momentum, as the latter will show up as orbital angular momentum around that point (whether the system is bound or otherwise, and indeed here I'm neglecting gravity so it is unbound).

So if you choose the original location of the cannon ball to measure angular momentum, as you say the ball has zero (if it doesn't spin) but the planet has a non zero angular momentum. How does that work?

It seems that the planet's spin must cancel out its translational 'angular' momentum about the ball's original position. But can you add angular momentum taken from two different reference points? What am I missing here?

Also,



take heed: some of the force would have to go through the center of mass to have the asteroid move other than spin.

stick your arm straight out with open fist. Now, I'll punch your hand. Your body rotates about the center of mass - it does not move translationally. Just like the bicycle wheel doesn't move, it only spins.

Or, have a kid on a swing. You push the kid, and he/she moves in a pivot around the bar above. The bar doesn't move, and yet you pushed the kid tangentially. Why doesn't the bar move?

I thought this was kind of hilarious.

Ken G
2009-Sep-12, 10:36 PM
So if you choose the original location of the cannon ball to measure angular momentum, as you say the ball has zero (if it doesn't spin) but the planet has a non zero angular momentum. How does that work? The planet would also have zero angular momentum, it just has two compensating contributions-- spin and linear motion (the latter has angular momentum because of the "moment arm").


It seems that the planet's spin must cancel out its translational 'angular' momentum about the ball's original position. But can you add angular momentum taken from two different reference points? Angular momentum is just momentum times the lever arm, added up over the whole planet. The spin means that parts of the planet will actually be going in the same direction as the cannon ball. That surprising result may be what you are missing.

DrRocket
2009-Sep-12, 11:32 PM
Angular momentum is just momentum times the lever arm, added up over the whole planet. The spin means that parts of the planet will actually be going in the same direction as the cannon ball. That surprising result may be what you are missing.


Right.

Conservation of angular momentum comes from conservation of linear momentum.

Linear momentum is mv where v is to be measured in an inertial reference frame. Angular momentum of a body about a point is simply linear momentum crossed with the vector from that point to the body. It is easily seen to be conserved in the presence of forces obeying Newton's third law. See thumbnail for details.

Both linear and angular momentum (about any point) are conserved in a closed system.

DRZion
2010-Mar-09, 07:29 AM
I think I can decidedly prove that linear momentum can be transformed into angular momentum.

Imagine you have a vacuum chamber.
Suspended from the ceiling of this chamber you have a windmill. Pointed at one of the vanes of the windmill is a laser.

The photons will carry linear momentum to the windmill and the windmill will spin (angular momentum). If it does not spin, then the whole windmill is pushed. But since the windmill is attached to the ceiling by a string, any movement of the windmill will be angular in nature.

The string can only pull, it cannot push, and it cannot pull the ceiling in any direction other than down if it is perpendicular to the ceiling. If the string pulls at the ceiling in any direction other than down then angular motion is present. If it does not pull at the ceiling then the windmill must be spinning, if any momentum is conserved.

Ken G
2010-Mar-09, 01:27 PM
Linear momentum does not even need to be "transformed" into angular momentum, as these are not two mutually exclusive things. Angular momentum is defined with respect to some point of our own choosing, and equals the linear momentum times the "lever arm" (a distance perpendicular to the line of motion) to that point. So linear momentum appears in the very definition of angular momentum.

tashirosgt
2010-Mar-09, 07:50 PM
When I shop for massless cannons, I like to get the kind with a very long barrel. That way, when the cannonball gets about halfway down the barrel, the asteroid has already begin to turn so the barrel is, so to speak, trying to fling the ball in a direction that is different from where the gun was originally pointed.

Ken G
2010-Mar-09, 10:02 PM
Interestingly, there was once a thread that showed why much of the recoil happens when the ball leaves the barrel, not when it first starts to accelerate. Apparently, a lot of the momentum goes into hot gas!

DrRocket
2010-Mar-09, 10:26 PM
Interestingly, there was once a thread that showed why much of the recoil happens when the ball leaves the barrel, not when it first starts to accelerate. Apparently, a lot of the momentum goes into hot gas!

A lot of the momentum does go into the hot gas. The gas is an appreciable fraction of the mass of the projectile and its exit speed at the muzzle, sonic velocity for the hot gas, is quite a bit higher than the speed of the projectile.

Ken G
2010-Mar-10, 08:57 PM
Indeed, I suspect that is why some cannon muzzles have holes in the sides-- to let the hot gas escape laterally as the shell leaves the muzzle, reducing the recoil.

cjl
2010-Mar-10, 10:35 PM
Indeed, I suspect that is why some cannon muzzles have holes in the sides-- to let the hot gas escape laterally as the shell leaves the muzzle, reducing the recoil.
Correct. Some are designed to redirect the gases backwards to further reduce recoil (the same is true with some of the more powerful rifles). It's known as a muzzle brake.

This image shows its operation quite well:

http://upload.wikimedia.org/wikipedia/commons/f/f7/4-14_Marines_in_Fallujah.jpg

Ken G
2010-Mar-10, 10:54 PM
That's quite a pic, it does indeed show it quite well.