Conservation of momentum

1. May 2, 2014

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

Last edited: May 2, 2014
2. May 2, 2014

Staff: Mentor

What do you think and why?

3. May 2, 2014

bobie

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

4. May 2, 2014

Staff: Mentor

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.

5. May 2, 2014

A.T.

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.

6. May 2, 2014

bobie

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

7. May 2, 2014

UltrafastPED

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

8. May 2, 2014

bobie

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)?

9. May 2, 2014

jbriggs444

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.

10. May 2, 2014

bobie

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.

Last edited by a moderator: Sep 25, 2014
11. May 2, 2014

jbriggs444

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.

Last edited: May 2, 2014
12. May 2, 2014

Staff: Mentor

Hmm, perhaps I need to refresh myself on momentum.

13. May 2, 2014

UltrafastPED

14. May 3, 2014

bobie

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?

15. May 3, 2014

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.

16. May 3, 2014

Staff: Mentor

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

17. May 3, 2014

bobie

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

18. May 4, 2014

jbriggs444

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?

Last edited: May 4, 2014
19. May 4, 2014

bobie

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?

20. May 4, 2014

jbriggs444

How is that relevant?