What is conserved in collisions: linear momentum or angular momentum?

[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.

 

[bobie]

If you have a balanced wheel (e.g. 2 plates jutting out on opposite sides), then ..

1) - ...The vectors mutually cancel out. I know P=0, but what about my example? when the 2 plates clash, how do we calculate the speed of rotation of the second wheel? what kind of momentum is transferred to the second wheel?

2)- If the wheel is rotated by 90° anticlockwise is the correct notation $L_{W1} = (-8, 0, 0)$, and $L_{P1} = (8, 0, 8)$ ?

In the video the axis of rotation is -x and it is turned to z and then to -z ?$L_{W} = (-3.2, 0, 0)$, and then$L_{W1} = (0, 0, +/-3.2)$ ?

Thanks

 

[Dale]

1) - ...The vectors mutually cancel out. I know P=0, but what about my example? when the 2 plates clash, how do we calculate the speed of rotation of the second wheel? what kind of momentum is transferred to the second wheel?

OK, if you know that p=0 then don't go back and try to put in p=20 or anything else that you know to be incorrect.

The amount of momentum transferred to the second object depends on the details of the collision. Without specifying the details you cannot know. What you can know is e.g. that if the momentum of the second object changes by Δp=(1,6,-3) as a result of the collision, then the momentum of the wheel will change to p=(-1,-6,3). The more the wheel changes the second object's momentum the more the second object changes the wheel's momentum (in the opposite direction).

2)- If the wheel is rotated by 90° anticlockwise is the correct notation $L_{W1} = (-8, 0, 0)$, and $L_{P1} = (8, 0, 8)$ ?

If the wheel were being rotated in outer space, yes. However, the platform can exert a torque about x and y and is subject to the constraint that the x and y components of angular velocity are 0. So the 90° anticlockwise rotation would give $L_{W1}=(-8,0,0)$ and $L_{P1}=(0,0,8)$.

Angular momentum is not conserved in x and y due to the external torque from the platform, it is only conserved in z due to the lack of external torque from the platform.

In the video the axis of rotation is -x and it is turned to z and then to -z ?$L_{W} = (-3.2, 0, 0)$, and then$L_{W1} = (0, 0, +/-3.2)$ ?

Yes. Does that make sense now?

 

[jbriggs444]

what kind of momentum is transferred to the second wheel?

Both angular momentum and linear momentum are transferred.

The amount of ordinary linear momentum that is transferred can be computed as the product of the force of impact multiplied by the duration of the impact. If you want to get picky, it would be the integral of force over time.

For short duration impacts it is often difficult to get accurate measurements of force sufficiently quickly. So rather than quantifying force or time you simply quantify "impulse" -- the amount of momentum that is transferred.

If the impact force is applied at a point that is not located at the axis of rotation, the interaction will also involve a transfer of angular momentum. The amount of angular momentum that is transferred can be computed as the product of impulse and and the offset between the impact location and the axis of rotation. If you want to get picky, in three dimensions, it is the vector cross product of the impulse and the distance from the selected point of reference.

In the problem that you posed:

"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)"

If both wheels are just sitting on an ice-covered table top then there is NO POSSIBLE WAY to bring the one wheel to a dead stop with a simple impact on its paddle. An impact that causes the moving wheel to stop spinning would also cause it to start moving linearly.

If both wheels were attached to frictionless axles so that their linear momentum were constrained to be zero then we could arrange an impact that would bring the moving wheel's angular momentum to zero. But you've been going on for pages about how the moving wheel's linear momentum is non-zero.

 

[bobie]

If both wheels are just sitting on an ice-covered table top ...[/B]

I thank you and Dalespam for the lavish and interesting explanations. I was posing a problem similar to the one in OP and in #8 that you answered in #9 .The only difference is that in OP there were two single balls/pendulums and here there are two set of balls/pendulums soldered in a circle. Why is the response different? That escapes me

 

[bobie]

Angular momentum is not conserved in x and y due to the external torque from the platform, it is only conserved in z due to the lack of external torque from the platform.
...
Yes. Does that make sense now?

I regret giving the impression that it did not make sense to me.

Thanks for your help, now, the most important issue:
- the Ke acquired by the platform is work made by the girl?
- is it 50% of total work , or what part of it?

Thanks.

 

[Dale]

Thanks for your help, now, the most important issue:
- the Ke acquired by the platform is work made by the girl?
- is it 50% of total work , or what part of it?

The platform does no work. 100% of the work is done by the girl.

 

[bobie]

The platform does no work. 100% of the work is done by the girl.

So 100% of the energy done by the girl goes to the platform, no inertia/resistance from the gyroscope?.

Now the last bit, the girl is only changing the direction of angular momentum, tilting the axis from -x to +/-z, in a direction normal to the plane of rotation.
Probably I misundertood in another thread when they told me that it takes no work to change the direction of the plane of rotation.
What's happening here
- why does the girl spend so much energy that she sets a platform of 50kg spinning?
- where does 3rd law of motion come into play?

Thanks again for this great thread!

 

[A.T.]

Probably I misundertood in another thread when they told me that it takes no work to change the direction of the plane of rotation.

I clarified it here:
https://www.physicsforums.com/showthread.php?p=4731572#post4731572

why does the girl spend so much energy

See here:
https://www.physicsforums.com/showthread.php?p=4728945&postcount=13

 

[Nugatory]

Now the last bit, the girl is only changing the direction of angular momentum, tilting the axis from -x to +/-z, in a direction normal to the plane of rotation.
Probably I misundertood in another thread when they told me that it takes no work to change the direction of the plane of rotation.
What's happening here
- why does the girl spend so much energy that she sets a platform of 50kg spinning?
- where does 3rd law of motion come into play?

The torque the girl applies to the gyroscope does no work, but the girl is also applying a torque to the platform, and that torque does do work as it starts the platform rotating.

 

[bobie]

You do no work, if you apply a torque perpendicular to angular velocity. But if the wheel spins while the wheel axis is rotating around some other axis, then the net angular velocity is not parallel to the wheel axis. So while holding the axis bearings, you can apply torques that have a component parallel to angular velocity, and thus do work.

Can you explain what you mean by net angular velocity?

 

[A.T.]

Can you explain what you mean by net angular velocity?

The angular velocity vector of an object, representing its total rotation.

 

[jbriggs444]

I thank you and Dalespam for the lavish and interesting explanations. I was posing a problem similar to the one in OP and in #8 that you answered in #9 .The only difference is that in OP there were two single balls/pendulums and here there are two set of balls/pendulums soldered in a circle. Why is the response different? That escapes me

The two scenarios that I understand us to be distinguishing are:

1. Two [equally massive, small] single balls are attached via massless spokes to frictionless vertical axles. The spoke on ball 1 has length r. The spoke on ball 2 has length r/2. The axles are separated by distance 3r/2.
The first ball is moving at some velocity. It is obviously constrained to a circular path. The second ball is at rest at a point in the path of ball 1. The first ball strikes the second ball elastically.

The claim is that this stops the first ball dead in both a linear and a rotational sense.

2. Two [equally massive, uniform, thin] bicycle wheels are on a frictionless table top. Wheel 1 has radius r. Wheel 2 has radius r/2. Their centers separated by distance 3r/2 (with enough clearance so that the wheels do not rub). A paddle of negligible mass is attached to each wheel. The first wheel is spinning in place at some rate. The second wheel is motionless with its paddle pointing toward the first wheel. The paddle on the first wheel strikes the paddle on the second wheel. Neither wheel is attached to a fixed axle.

The claim is that the collision cannot stop the first wheel dead in both a linear and a rotational sense.

In case 1, a collision can cause the first ball to stop dead because the ball has both linear momentum and angular momentum that "match". The impulse that is delivered in the collision cancels the linear momentum of the first ball. It also exactly cancels its angular momentum.

In case 2, a collision could not cause the first wheel to stop dead (in the absence of a fixed axle) because the wheel has zero linear momentum and non-zero angular momentum. Any single impulse that is applied at the rim of the wheel to cancel the angular momentum would impart linear momentum.

Do you understand that a wheel that is spinning in place has zero linear momentum?

[Edit: added some language clarifying that the balls and wheels have the same mass]

 

[Dale]

So 100% of the energy done by the girl goes to the platform, no inertia/resistance from the gyroscope?.

In this case the energy of the gyroscope is unchanged, so yes, but it is not a general principle, it is specific to this scenario.

Now the last bit, the girl is only changing the direction of angular momentum, tilting the axis from -x to +/-z, in a direction normal to the plane of rotation.
Probably I misundertood in another thread when they told me that it takes no work to change the direction of the plane of rotation.

For rotational motion $W = \int \mathbf{\tau} \cdot \mathbf{\omega} \; dt$. In this case, the torque on the wheel and the angular velocity vector of the wheel are perpendicular, so the dot product is 0 and therefore no work is done on the wheel. The platform is more complicated. The angular velocity in x and y is 0, so the x and y torques do no work on the platform, but the angular velocity in z is non-zero, so the z torque does work on the platform.

- why does the girl spend so much energy that she sets a platform of 50kg spinning?

Because the laws of physics require it, as calculated above. Other than that, I am not sure what you are asking here. Do you want to know why the laws of physics are the way they are, or did you not follow the calculations?

- where does 3rd law of motion come into play?

The torque exerted by each object is equal and opposite to the torque exerted on each object.

 

[bobie]

In case 2, a collision could not cause the first wheel to stop dead (in the absence of a fixed axle) ]

Probably this is the misunderstanding, I am referring to a fixed axle as in the first case.
It might help you see what I meant if you consider (instead of a second wheel) a single ball of mass equal to the first wheel. When the 'paddle' hits the ball will stop dead and the ball will get v=10 ,m =2 so it'll have P = 20.
It all came from the wheel, probably I was wrong to call it P, but that is what I have been trying to tell you

Thanks

 

[bobie]

. In this case,... the torque on the wheel and the angular velocity vector of the wheel are perpendicular, so the dot product is 0 and therefore no work is done on the wheel.

This is the point that is obscure to me:
you say that the girls does no work pushing her right hand down. No work means no energy spent, no effort on her muscles (I have learned that you can make effort/ spend energy and do no work), but I suppose that no-work implies no-energy/effort. Is this correct?

Now, wiki says clearly that a spinning wheel has inertia that opposes a change of the plane of rotation, and any change of that implies a perpendicular torque. You need energy/work to win inertia, so hou do I reconcile the two notions?
Many videos on the web show that it's hard to win that inertia.
Thanks, Dalespam for your help.

 

[bobie]

The angular velocity vector of an object, representing its total rotation.

What is total or net ω, isn't ω always the net velocity? if the bykewheel in the video is spinning at 5 rps, what is net velocity?

 

[A.T.]

What is total or net ω, isn't ω always the net velocity?

"ω" is what you define "ω" to be. I didn't even use that symbol, so I don't know why you ask me.

 

[bobie]

"ω" is what you define "ω" to be. I didn't even use that symbol, so I don't know why you ask me.

Sorry, but that is what I studied:

In physics, the angular velocity is defined as the rate of change of angular displacement and is a vector quantity (more precisely, a pseudovector) which specifies the angular speed (rotational speed) of an object and the axis about which the object is rotating... Angular velocity is usually represented by the symbol omega (ω).

Were you not referring to that, anyway when is 'that' net?

 

[Dale]

No work means no energy spent, no effort on her muscles

Absolutely not. No work means no energy transferred to another system. Efficiency is the ratio of work done over energy spent, and it is not generally equal to 1. Unless you are dealing with a machine with 100% efficiency you cannot equate work done with energy expended.

The human body is a remarkably inefficient machine. In many cases its efficiency is 0 meaning that energy is spent without any work done.

 

[A.T.]

The human body is a remarkably inefficient machine.

For example: Muscles are not optimized for applying static forces. Therefore some birds can lock their wing joints for soaring, and relax the muscles which would have been used otherwise to stabilize the joint.

 

[Dale]

some birds can lock their wing joints for soaring, and relax the muscles which would have been used otherwise to stabilize the joint.

That is cool! I didn't know that.

 

[Dale]

bobie, if you are interested in the KE of a rotating object, then you can use this formula: $KE = \frac{1}{2}I \omega^2$. Together with the other formulas I provided, you can calculate all of the details if you like. I would recommend starting with a highly simplified scenario.

 

[A.T.]

That is cool! I didn't know that.

Prime example is the Albatross:
http://faculty.weber.edu/rmeyers/meyers-albatross.pdf

It stays airborne for weeks without flapping its wings once, using dynamic soaring. Having the muscles tensioned all this time would be a tremendous waste of energy.

 

[bobie]

bobie, if you are interested in the KE of a rotating object, then you can use this formula: $KE = \frac{1}{2}I \omega^2$. I would recommend starting with a highly simplified scenario.

My first scenario was simple: a bikewheel (m= 2, r= .4, v = 10) 'paddle' hits a ball at rest (m= 2 or another 'paddle'), are you saying that is not momentum that is tranferred as in ordinary collisions, but KE and that this is tranformed into P=20?

No work means no energy transferred to another system.

I know that. If I hold up a ball I do effort, no work. If I push a ball on the floor, the ball has inertia, I have to win resistance, I give KE to move it , it moves, I have done work.
What you say on birds is interesting, but, please explain how it applies here:
The girl is moving, rotating a spinning wheel so she is transferring some Ke to the wheel, the wheel has inertia when it is at rest, wiki says that a spinning wheel has extra inertia and resists changes, how can you conclude that she has done no work?or you mean that the work on the wheel is being transferred to the platform?

 

[bobie]

. The platform is more complicated. The angular velocity in x and y is 0, so the x and y torques do no work on the platform, but the angular velocity in z is non-zero, so the z torque does work on the platform.

The girl does no work on the wheel but does work on the platform. correct?

The formula for work is F*d(rad), could you show me how you calculate the work done on the platform? what we need here is only L_w = 1.6 Js, or do we need the details of m,r,ω ? How do we find out the value of F in this example?