Conservation of Momentum in Collisions: Exploring Linear and Angular Momentum

[bobie]

Hi,
if a ball spinning on a rod hits another similar ball (m1=m2) what is conserved linear p or angular L momentum?
Thanks

 

[Drakkith]

What do you think and why?

 

[bobie]

I think the outcome should be the usual outcome of elastic collision: p/Ke conserved.

 

[Drakkith]

According to conservation of momentum, total momentum is always conserved. However, angular momentum and linear momentum can be converted between each other, so depending on how the two balls interact you could end up with different amounts of linear and angular momentum so long as total momentum is conserved.

 

[A.T.]

However, angular momentum and linear momentum can be converted between each other,...

No, they cannot be converted. They are different quantities with different units.

...so depending on how the two balls interact you could end up with different amounts of linear and angular momentum so long as total momentum is conserved.

No, they are conserved individually. The total linear momentum is conserved and the total angular momentum is conserved.

 

[bobie]

No, they are conserved individually. The total linear momentum is conserved and the total angular momentum is conserved.

Could you ,please, be more explicit AT?
suppose the spinning ball (m1=1) has a spoke .2m long and v 10 m/s: then KE=50, p=10, L = 2
m2 (at rest) has a spoke .1 m long
what happens

 

[UltrafastPED]

Could you ,please, be more explicit AT?
suppose the spinning ball (m1=1) has a spoke .2m long and v 10 m/s: then KE=50, p=10, L = 2
m2 (at rest) has a spoke .1 m long
what happens

You need to start with the correct equations; kinetic energy for a rotating body is not the same as for a body in linear translation:

http://hyperphysics.phy-astr.gsu.edu/hbase/rke.html

Angular momentum is always with respect to a specified axis; any fixed axis will suffice, but for a rotating object it is simplest to use its axis of rotation.

For an overview see http://hyperphysics.phy-astr.gsu.edu/hbase/circ.html#rotcon

 

[bobie]

You need to start with the correct equations; kinetic energy for a rotating body is not the same...

Could you explain that? suppose the two 'pendulums' are horizontal on a table: when m1 hits m2 , its vector is tangential and p=10, why should m2 not acquire p=10 tangentially and behave as if it were not fixed to a pivot and (move and then) rotate at 10 m/s (of course with a greater angular velocity)?

 

[jbriggs444]

Could you ,please, be more explicit AT?
suppose the spinning ball (m1=1) has a spoke .2m long and v 10 m/s: then KE=50, p=10, L = 2
m2 (at rest) has a spoke .1 m long
what happens

To be clear, this is not a "spinning ball". This is a 1 kg ball that is fixed to a massless spoke that is 0.2 m long. The spoke is in turn attached frictionlessly at the other end to an axle so that the ball is free to move in a circular arc of radius 0.2 m. The ball is small compared to 0.2 m.

The ball's kinetic energy as computed based on ordinary linear mechanics is KE = 1/2 mv^2. As you say, this comes to 50 J.

The ball's kinetic energy as computed based on rotational mechanics is KE = 1/2 Iω². The I is the moment of inertia of the ball-on-a-stick. Moment of inertia the integral of mass times the square of the distance from the selected axis. In this case, all of the mass is 0.2 m from the axis. So the moment of inertia is 0.04 kg m².

ω is the rotation rate measured in radians per second. For a ball rotating at 10 meters per second in a circle of radius 0.2 meters, that's 50 radians per second. 1/2 Iω² is then 1/2 * 0.04 * 50 * 50 = 50 J.

The numbers match -- as they should. The energy computed the one way matches the energy computed the other.

Now you have a second ball. Also 1 kg. This one is attached to a 0.1 meter long massless stick which is attached to a frictionless axle so that it is free to move in a circular arc of radius 0.1 meter. This ball is small compared to 0.1 m.

An elastic collision is arranged at the point where the two circular paths intersect, tangent to one another. The mental image I have in my head is that the two axles are 0.3 m apart. The two balls collide where the two circles touch between the two axles.

Start with an analysis from the perspective of ordinary linear momentum. The outcome is obvious. The ball on the 0.2 m stick stops dead and the ball on the 0.1 m stick starts moving at 10 m/s

New KE = OLD KE = 1/2 mv² = 50 J.
New momentum = Old momentum = mv = 10 kg m/sec.

What about from a rotational perspective?

The rotational KE of the ball on the the 0.1 m stick is given by 1/2 Iω². Again, that's the integral of mr². In this case r is a constant 0.1 meter from the axis. The moment of inertia of this ball around its axle is 0.01 kg m²

The ball has a tangential velocity of 10 m/sec on a radius of 0.1 m. That means that it is circling at 100 radians/sec. 1/2 Iω² is 1/2 * 0.01 * 100 = 50 J.

The KE still matches!

But what about angular momentum? That's supposed to be conserved, right? But we started with one ball orbitting its axle at 50 radians per second clockwise and ended with another ball orbitting its axle at 100 radians per second counter-clockwise. How could we possibly say that angular momentum is conserved?!

The simple reason is that angular momentum about any given axis is conserved. But these two angular momenta are computed about two different axes.

The axis moved by 0.3 meters and the composite system had linear momentum of 10 kg m/sec at right angles to the offset between the two axes. It follows that the angular momentum computed abound the one axis will be 3 kg m/sec different from the angular momentum computed about the other axis.

[Angular momentum is not just about the rotation of rigid bodies. Even object moving along straight lines have angular momentum with respect to axes that are not directly on the object's path. Angular momentum is formally defined as the integral of the cross product of the momentum of each bit of a system times the offset of that bit of the system from a chosen point of reference. With a little algebra you can easily show that that L' = L + momentum cross offset]

Let's check.

Angular momentum = Iω
The first ball had angular momentum of Iω = 0.04 * 50 = 2 kg m/sec
The second ball has angular momentum of Iω = 0.01 * 100 = 1 kg m/sec in the opposite direction.

Voila! The delta is 3 kg m/sec.

 

[bobie]

The simple reason is that angular momentum about any given axis is conserved. But these two angular momenta are computed about two different axes...
Voila! The delta is 3 kg m/sec.

Congrats, jbriggs, that's really a great post.

Since you are so smart, I hope you can spare some time to solve this intriguing problem I've been chasing in many threads, and nobody was able to solve. Here (at 0.45 ), I think we have a similar problem p and L are transferred to another system, what are the numbers that fit?
I figured out some basic data you can adjust them if you deem right:

a) the wheel: m = 2kg, r= .4 m, v = 10 m/s, p= 20, KE = 100, L = 8,
b) the radius of the platform (.4 wheel+ .1 girl ) R= .5 m, the mass of the girl+pedestal M = 50 kg, but, as the radius of M is about .15( =1/3 R), I imagine it counts for ≈ 18kg + wheel = 2 Kg = m₂ = 20
c) m2 rotates at about 1 rad/sec so the speed of the axis of the wheel is v = .5 m/s, the speed of the pedestal+girl is .15 m/s, p = ? ( ), KE = ? (1.3 J), L = 5

I have been told that the girl pushes her right hand down and the wheel responds with a clockwise(seen from above) torque/push. She opposes this force propping her feet to the pedestal and pushing anti-clockwise, transferring momentum to the platform that rotates clockwise (3rd law). Is that right?

If the full solution is complicated, just plug in arbitrary numbers, just to sho what's happening, the formulas, momenta, axes etc.
How do we find the work done by the girl ? from the momenta of the platform?

Thanks a lot for your time.

 

[jbriggs444]

Since you are so smart, I hope you can spare some time to solve this intriguing problem I've been chasing in many threads, and nobody was able to solve.

First, you need to try to work it yourself.

a) the wheel: m = 2kg, r= .4 m, v = 10 m/s, p= 20, KE = 100, L = 8,

Start with the "p=20" and see if you can identify anything wrong with that assertion.

 

[Drakkith]

No, they cannot be converted. They are different quantities with different units.


No, they are conserved individually. The total linear momentum is conserved and the total angular momentum is conserved.

Hmm, perhaps I need to refresh myself on momentum.

 

[UltrafastPED]

Hmm, perhaps I need to refresh myself on momentum.

Here is a good place to start: http://eduardo.physics.illinois.edu/phys582/582-chapter3.pdf

 

[bobie]

First, you need to try to work it yourself.
Start with the "p=20" and see if you can identify anything wrong with that assertion.

Thanks, jbriggs, I'd be grateful if you are really willing to help me solve (at last) this problem.
But, please, first confirm that what I think is right:

This case is similar to the previous, there is a transfer of momentum/a from the (bike) wheel to the pedestal. The only difference I see is that here the transfer is not direct but mediated by the girl. she stretches her body pushing on the gyroscope (with hands) and the pedestal (with feet).
In this sense it works the same way as if she were standing radially at the edge of the platform and punching/kicking a football tangentially. (in this case the tranfer would be linear to angular momentum). Is that right?

As to p = 20 , m = 2 v= 10, p = mv = 2*10 = 20 (k*m/s), what's wrong?

 

[sophiecentaur]

I could introduce the question as to whether the 'system' is really isolated. Potentially, the attachment of the rod to the axle is relevant. The collision needs to be tangential if you want to ignore this effect, I think, or radial momentum changes.

 

[Dale]

As to p = 20 , m = 2 v= 10, p = mv = 2*10 = 20 (k*m/s), what's wrong?

Is momentum a vector or a scalar? How do the momenta of different parts of the tire add up?

 

[bobie]

suppose the spinning ball (m1=1) has a spoke .2m long and v 10 m/s: then KE=50, p=10, L = 2
m2 (at rest) has a spoke .1 m long
what happens

I do not understand here I made p = 1*10 =10 and it was OK.

 

[jbriggs444]

I do not understand here I made p = 1*10 =10 and it was OK.

The momentum (p) of a point-like object is simple. It can be defined as the mass of the object multiplied by its velocity: p = mv.

Velocity (v) is a vector. It has magnitude and direction. Mass (m) is a scalar. It has magnitude but no direction. When you carry out the multiplication, direction is preserved. Momentum is a vector. It has magnitude and the same direction as the velocity.

The momentum of an extended object (one that is larger than a point) is not so simple. It can be defined in two ways. The two ways are equivalent to each other.

1. The momentum of an object is the mass of the object (m) multiplied by the velocity of its center of mass (v). The direction of the result is taken from the direction of the velocity.

2. The momentum of an object is the integral of the mass of each infinitesimal mass element times the velocity of that mass element. The direction of the result is derived naturally because the momenta of the mass elements add as vectors.

In the scenario with a 1 kg ball on a 0.1 meter spoke circling the axis at a tangential velocity of 10 m/sec the ball is point-like and the simple version applied. The momentum of the ball at any given instant was given by p = mv. Since m = 1 kg and v = 10m/sec (in a tangential direction), the result is that p = 10 kg m/sec (in a tangential direction).

Your stated result was numerically correct. It should have also included the units of measurement (kg m/sec).

In the scenario with a 2 kg wheel rotating with a tangential velocity of 10 m/sec we are no longer dealing with a point-like object. The velocity of the center of mass is zero. Can you see why the integral of the product of mass times velocity for the mass elements on the rim of the wheel is also zero?

 

[bobie]

we are no longer dealing with a point-like object. The velocity of the center of mass is zero. Can you see why the integral of the product of mass times velocity for the mass elements on the rim of the wheel is also zero?

I ' sorry, I thought that all the points can be considered (loosely on average ) at the same distance from the centre. Is also L wrong?

 

[jbriggs444]

I ' sorry, I thought that all the points can be considered (loosely on average ) at the same distance from the centre.

How is that relevant?

 

[Dale]

bobie, I would strongly recommend that, before you try to learn about complicated scenarios involving gyroscopes and significant amounts of angular momentum, that you first learn about simple scenarios involving linear momentum. If you cannot understand the mistake that you made here regarding linear momentum, then it is unlikely that you will be able to reason through any problem involving angular momentum.

Momentum is a vector quantity, meaning that it has a magnitude and a direction. A 1 kg object moving at 1 m/s in the x direction does not have the same momentum as a 1 kg object moving at 1 m/s in the y direction. In (x,y,z) coordinates the first object would have a momentum of (1,0,0) kg m/s and the second would have a momentum of (0,1,0) kg m/s. The sum of their momenta would be (1,1,0) kg m/s, which is a magnitude of $\sqrt{2}$ times the magnitude of each of the individual objects in a direction half-way between the x and y axes.

Now, given that, do you understand why the momentum of the spinning tire is 0? Consider what happens when you add the momentum of any given point on the tire with the momentum of the opposite point.

 

[bobie]

learn[/B] about simple scenarios involving linear momentum.
the momentum of the spinning tire is 0.

Thanks, Dalespam, I'd be grateful if you give me a few links, ouside wiki.

What is relevant here? in the first example we had a one-ball-pendulum , in the byke wheel we have many soldered pendulums, just suppose there is a jut on the wheel: when it collides (as in posts #6,8), the momentum transferred is not 10 but 20 (k*m/s).

Nobody has answered my repeated question (#10,14: "But, please, first confirm that what I think is right:This case is similar to the previous, there is a transfer of momentum/a from the (bike) wheel to the pedestal. The only difference I see is that here the transfer is not direct but mediated by the girl), so I assumed my descpription of what happens in the video is correct. The jut is here mediated by the body of the girl.

It would help me a lot if you gave me the correct description of the process and an (even approximate ) numerical correlation between the action on the wheel and the reaction of the platform.

Thanks for your kindness.

 

[bobie]

First, you need to try to work it yourself..

I gave you my (wrong) interpretation, could you tell me what is the real interpretation of the video?
Thanks a lot

 

[jbriggs444]

I gave you my (wrong) interpretation, could you tell me what is the real interpretation of the video?

Please refer to DaleSpam's post 21.

 

[Dale]

Thanks, Dalespam, I'd be grateful if you give me a few links, ouside wiki.

Here are a few links about momentum:
https://www.khanacademy.org/science...tum-tutorial/v/2-dimensional-momentum-problem
http://www.physicsclassroom.com/class/momentum/u4l2b.cfm
http://hyperphysics.phy-astr.gsu.edu/hbase/mom.html

What is relevant here? in the first example we had a one-ball-pendulum , in the byke wheel we have many soldered pendulums, just suppose there is a jut on the wheel: when it collides (as in posts #6,8), the momentum transferred is not 10 but 20 (k*m/s).

A pendulum is quite different from a spinning wheel. In a pendulum the center of mass is moving (system has net momentum), in a spinning wheel it is not (system has 0 net momentum). A pendulum requires a net external force to rotate about the axis, a spinning wheel does not. A spinning wheel posses axial symmetry, a pendulum does not.

The fact that you think they are equivalent is an indication of the importance of studying some more about momentum. Please focus on learning about linear momentum and vector addition before moving on to angular momentum.

Regarding the amount of momentum transferred, that depends on the details of the collision (elastic, inelastic, etc.). Your posts 6 and 8 provide insufficient details to determine the amount of momentum transferred, but once you have studied the above links and asked here for any required clarifications then you should be able to calculate the transfer.

Nobody has answered my repeated question (#10,14: "But, please, first confirm that what I think is right:This case is similar to the previous, there is a transfer of momentum/a from the (bike) wheel to the pedestal. The only difference I see is that here the transfer is not direct but mediated by the girl), so I assumed my descpription of what happens in the video is correct. The jut is here mediated by the body of the girl.

It would help me a lot if you gave me the correct description of the process and an (even approximate ) numerical correlation between the action on the wheel and the reaction of the platform.

I don't think any of that will be helpful until you understand linear momentum and vector addition.

 

[bobie]

A pendulum is quite different from a spinning wheel.

Thanks for the links,
In the first posts I called to a spinning pendulum the rotating wheel on a string/ spoke.

As to the example in the video, I already said that I just guessed some figures, but you can make up an example of your liking , and show me what is the numerical relation between a tilted wheel and a frictionless platform. That would be the best help you can give me. It's much easier to grasp a theoretical principle through a practical example.
I hope jbriggs is willing to do that.
Let me know at least one thing: if you know the angular momentum of the gyroscope, can you determine in advance the reaction of a platform of known characteristics, or that is impossible even to you?

Thank you for your time

 

[jbriggs444]

I hope jbriggs is willing to do that.

My patience has been exhausted. Good luck, Dale.

 

[Dale]

I already said that I just guessed some figures, but you can make up an example of your liking

I have no problem with your figures, but the linear momentum of the bike tire is clearly p=0, not p=20. Until you understand that, there is no point in worrying about the angular momentum of the tire nor anything at all about the girl or the platform.

Please review the links I sent. If it is obvious to you why p=0 after reading them then we can proceed. Otherwise you need to ask questions about that first. You absolutely must understand linear momentum and vector addition before you can hope to understand angular momentum.

 

[bobie]

Please review the links I sent.

Thanks for the links, I see that the Khan academy is a reliable site, I had usually avoided it .
Now, if you like to conclude this thread and its issue, there is a byke wheel of m=≈ 2k, r ≈ .4 spinning clockwise at ≈10 m/s. L, which is relevant here, is mvr = 2*10*.4 = 8 J*s. Is that right?
Could you now tell me, please,

- what happens when she turns the plane of rotation by 90°,
- how do you calculate the angular velocity ω of a platform of mass = M , arbitrarily chosen?Is this possible in the first place?
- what is the ratio between KE acquired by the platform and the work done by the girl, is it 50%?

Thanks for your time, Dalespam

 

[bobie]

My patience has been exhausted. Good luck, Dale.

Thanks, anyway, jbriggs, Dalespam has showed in many occasions great patience and comprehension.

 

[Dale]

Thanks for the links, I see that the Khan academy is a reliable site, I had usually avoided it .
Now, if you like to conclude this thread and its issue, there is a byke wheel of m=≈ 2k, r ≈ .4 spinning clockwise at ≈10 m/s. L, which is relevant here, is mvr = 2*10*.4 = 8 J*s. Is that right?

Do you understand why p=0? Do you understand vector addition?

The formula you used only applies for a point particle (although the final number is approximately correct for an idealized tire). To understand why goes back to the vector addition. Please let me know if you are comfortable with that concept.

 

[bobie]

. Please let me know if you are comfortable with that concept.

Sure, Dalespam, I have thoroughly studied collision:(https://www.physicsforums.com/showthread.php?p=4586458#post4586458). I was referring to the momentum in the case of a direct collision between the two wheels.
But what is relevant here is only L_w as there is no direct relation, and we can easily assume L_w = 6-8. Right?

Can you examine now what happens to the the platform? Must the angular momentum L_pof the platform be equal to L_w or what?
I'd really appreciate that. Nobody seems to know!

 

[Dale]

Sure, Dalespam, I have thoroughly studied collision:(https://www.physicsforums.com/showthread.php?p=4586458#post4586458). I was referring to the momentum in the case of a direct collision between the two wheels.

It doesn't matter if the collision is direct or not. The momentum of a spinning tire is 0 regardless. However, since you assert that you understand momentum and vector addition then I will proceed.

But what is relevant here is only L_w as there is no direct relation, and we can easily assume L_w = 6-8. Right?

Can you examine now what happens to the the platform? Must the angular momentum L_pof the platform be equal to L_w or what?

First, to calculate L you need to use the correct formula for the object of interest. The formula that you used previously, $L = r \times mv$, only applies for a point mass. For an extended object you need to use: $L = I \omega$ where I is the moment of inertia and $\omega$ is the angular velocity.

If we approximate the bike tire as a thin hoop rotated about its axle the moment of inertia is $I_W = r_W^2 m_W$. If we approximate the girl and the platform together as a cylinder rotated about its axis the moment of inertia is $I_P = \frac{1}{2} r_P^2 m_P$.

Now, if the wheel is initially rotating about the positive z axis with a magnitude of 8 Js and the platform is initially non-rotating then the total angular momentum is $L_T = L_{P0} + L_{W0} = (0,0,0) + (0,0,8) = (0,0,8)$. This total angular momentum is conserved if there is no external torque.

If the wheel is inverted then its new angular momentum is $L_{W1} = (0,0,-8)$, if the inversion is done without an external torque then conservation of angular momentum dictates that $L_{P1} = (0,0,16)$ so that $L_T=L_{P1}+L_{W1}=(0,0,8)$.

Since you understand vectors and vector addition, that should be clear now.

 

[bobie]

It doesn't matter if the collision is direct or not. The momentum of a spinning tire is 0 regardless. ...
If the wheel is inverted then its new angular momentum is $L_{W1} = (0,0,-8)$, if the inversion is done without an external torque then conservation of angular momentum dictates that $L_{P1} = (0,0,16)$ so that $L_T=L_{P1}+L_{W1}=(0,0,8)$.
.

Thanks, Dalespam.
As to p: suppose on the steel wheel there is a plate jutting out and that this hits another plate jutting out from another wheel at rest whose (with equal mass but radius 1/2) and we want to know at what speed the latter will start spinning (if the first one stops dead), do we need P=0 or P=20?

As to the video, of course -8 and +16 makes +8 and L is conserved.
- The tilting of the wheel is not considered an external torque as the system is girl-wheel-platform?

- According to your formula (if m=2, r=.4 and ω is 5 rps) L_w = 2*.4²* 5 = -1.6 , so L_p = 3.2
We know that r_p is slightly larger (.5m) but mass= 45 kg lays within .2 m radius + 5kg within .5m; we might roughly assume that I = (45* .2²/2 + 5*.5²/2) = 1.6, then (3.2/ 1.6) ω_p should be about 2 rps? right?

Now, if we find the KE of the platform (which is rather complicated, I suppose) can we find out what is the work done by the girl?
I hope I didn't make too many mistakes.

Thanks a lot, again

 

[Dale]

Thanks, Dalespam.
As to p: suppose on the steel wheel there is a plate jutting out and that this hits another plate jutting out from another wheel at rest whose (with equal mass but radius 1/2) and we want to know at what speed the latter will start spinning (if the first one stops dead), do we need P=0 or P=20?

The momentum is what it is regardless of how it interacts with some other object. If the momentum of the wheel is 0 then it is 0 regardless of whether it collides with a plate or a wheel or a fluid or anything else you could imagine.

If you have a balanced wheel (e.g. 2 plates jutting out on opposite sides), then regardless of how fast it is spinning the momentum is obviously 0. Do you understand why it is obviously 0? What principle makes it obviously 0?

It irritates me that you professed understanding of this point and continue to not understand it. If you don't understand an important point then don't pretend that you do. Stop and actually learn.

As to the video, of course -8 and +16 makes +8 and L is conserved.
- The tilting of the wheel is not considered an external torque as the system is girl-wheel-platform?

Typically a rotating platform is designed so that it cannot provide torque about one axis, but it can provide torque about the other two axes. So in this example the platform could provide torque in the x and y directions, but not the z direction.

- According to your formula (if m=2, r=.4 and ω is 5 rps) L_w = 2*.4²* 5 = -1.6 , so L_p = 3.2
We know that r_p is slightly larger (.5m) but mass= 45 kg lays within .2 m radius + 5kg within .5m; we might roughly assume that I = (45* .2²/2 + 5*.5²/2) = 1.6, then (3.2/ 1.6) ω_p should be about 2 rps? right?

Your moments of inertia are a reasonable approximation. But you clearly missed the point about vector addition. If the wheel is inverted then what is the angular momentum of the platform? Please use vector notation and vector addition to work the problem.