Conservation of Momentum in Collisions: Exploring Linear and Angular Momentum

[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?

 

[A.T.]

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.

You can apply a force to a moving object, without doing work.

The girl is moving, rotating a spinning wheel so she is transferring some Ke to the wheel,

You can apply a torque to a rotating object, without doing work.

 

[bobie]

You can apply a force to a moving object, without doing work.
You can apply a torque to a rotating object, without doing work.

Sure, but here the situation is different, AT,
If an object is not moving and then moves is it impossible that nobody did work.
If an object is moving in one direction and then moves in two directions work must have been done. If you throw a ball with spin you do work both for translation and rotational energy.
The wheel is spinning on one plane and the girl rotates it in another direction in which it was not moving, does that motion come free?
But the main argument on my side is that that is the only action the girl is doing, and the motion of the platform can be originated only by the work she is doing on the wheel.
There is no other source of energy around. Am I wrong?


Thanks for your help, could you please answer my post #53, it probably escaped your attention

 

[A.T.]

If an object is moving in one direction and then moves in two directions work must have been done.

Wrong. You can change the direction of the velocity vector without doing work. That has been explained several times to you already.

Thanks for your help, could you please answer my post #53, it probably escaped your attention

See my post #46.

 

[bobie]

You can change the direction of the velocity vector without doing work. That has been explained several times to you already..

You said that many times, but that is an abstract principle, could you apply it here?

Can you, please, answer directly the practical,simple questions:
who does the work that sets the platform spinning? how? what is the value of the work done? how you derive it from the given parameters?
That is the only way to make me understand. In many threads, when I asked these simple questions I got only abstract, apodictical principles that are surely clear to you scientists.

As to post 46-53 I gave you a quote from wiki that says that angular velocity is given by ω, what is wrong?

Thanks a lot for your patience

 

[A.T.]

You said that many times,...

And you just keep contradicting it. So what is the point of telling you something.

 

[bobie]

And you just keep contradicting it. So what is the point of telling you something.

I am not contradicting it, I am only asking you to show me how it works in practice.
I am saying, all right, the girl does no work, who does it?
The outcome of the video seems to contradict that, I do not know what is happening, ( a scientist as great as Laithwaithe raised the issues and he was not an ignorant)
if you know what is happening , just tell me, if you wish.
Just repeating that the girl tilting the wheel does no work does not explain anything of what we see in the video.
Thanks for your posts.

 

[A.T.]

I am not contradicting it

Yes, you are. You said: "If an object is moving in one direction and then moves in two directions work must have been done." which simply doesn't follow from the definition of work.

Laithwaithe raised the issues and he was not an ignorant

On this topic he was. His misconceptions are explained here:

https://www.youtube.com/watch?v=tLMpdBjA2SU

 

[bobie]

On this topic he was. His misconceptions are explained here:

I had seen that video (and hundred of others) and it does not explain misconceptions. Of course he was wrong when he went as far as to question the validity of the laws of motion, but most of issues still have no explanations, I hoped that the issue in the OP video had found a full description in these last 40 years.

What happens here is that force to lift the wheel is applied to accelerate it horizontally and that acceleration is shifted by 90° , i.e. upwards, by the gyroscope propriety. You should note that, when he starts to roteate it, his own weight increases as he is pushing/leaning on his foot to push the wheel around.
You surely know that if you accelerate the horizontal motion the gyroscope goes up (seems lighter) if you slow it it gets 'heavier' and falls down. He feels less effort as (a) it is easier to do work horizontally than vertically, (b) he is diluting his work, that is all.
For other reasons even when you swing a stone on a string it goes up.

If you are interested in this intriguing issues I'd be glad to discuss them with you,
see here , for example ( at 1:45/49)
why does the horizontal rotation stops immediately when the weight is lifted?

 

[A.T.]

I had seen that video (and hundred of others) and it does not explain misconceptions.

Works fine for me.

 

[Dale]

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, I never said anything of the sort. Momentum and energy are separate quantities and cannot be transformed into each other. They are each individually and separately conserved for an isolated system.

I know that. If I hold up a ball I do effort, no work.

When you make a mistake and are corrected, it accomplishes nothing to say "I know that". If you make mistakes then obviously you don't know it well enough to avoid the mistakes.

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.

"Win resistance" is not a physics term. It doesn't mean anything that I am aware of.

If you apply a force on an object and that object's mechanical energy increases then you have done work on the object. If you apply a force on an object and that object's energy does not change then you have done no work on the object, regardless of the fact that you exerted a force on the object and regardless of the fact that the object has inertia and regardless of whatever energy you (as an inefficient machine) may have expended.

The girl is moving, rotating a spinning wheel so she is transferring some Ke to the wheel,

How much is "some" KE? Use the formula I provided above and calculate the KE in the beginning and ending and then tell me exactly how much KE was transferred to the wheel.