# Calculating speed and angular momentum

1. Oct 15, 2014

### bobie

Suppose a rod of 1m length and 10 kg mass is lying on a frictionless surface or in vacuum.
Suppose now a point particle of mass 1 Kg hits one tip of the rod at speed 22 m/s in an elastic collision:

If the CoM is fixed on a frictionless fulcrum, it is easy to predict that the particle will bounce at 18 m/s and the rod will get 80J of Ke and spin consequently with angular velocity

ω =√ 2 * E (=80) / I (=1/12) = $\sqrt{24 * 80} =$ 43.8 and $\nu$ = 6.97 Hz

But if the rod is free to move, can we predict how much energy will remain as KE and the speed at which the rod will move forward while spinning?

Last edited: Oct 15, 2014
2. Oct 15, 2014

### Staff: Mentor

How do you get that?

3. Oct 15, 2014

### bobie

That is the result of an elastic collision between a body A (m = 1, v = 22, E = 242) and B (m =10)
$v_A = -18, E = 162, v_B = +4 , E = 80$

4. Oct 15, 2014

### Staff: Mentor

Elastic collision means KE is conserved. That gives you one equation. You have two unknowns, $v_a$ and $v_b$. You cannot solve two unknowns with one equation.

5. Oct 15, 2014

### bobie

The result I gave comes from more equations, and is the correct result:
E = 162 + 80 = 242, p = 40 -18 = 22 , energy and momentum are conserved

Last edited: Oct 15, 2014
6. Oct 15, 2014

### Staff: Mentor

What is the second equation?

Last edited: Oct 15, 2014
7. Oct 15, 2014

### bobie

Would you like to see the whole procedure by which I got the result? It is surely the correct result as both E = 242 and p = 22 and v = 22 are conserved
Do you wish to see how I got that result? It is surely correct as energy 162+80 = 242, velocity 4 - -18 = 22, and momentum 40-18 = 22 are conserved. Do you know if we can predict the result of the collison if the rod is free to move?

Last edited: Oct 15, 2014
8. Oct 15, 2014

### Staff: Mentor

Linear momentum is not conserved here. Linear momentum is only conserved if there is no external force acting on the system. Here, there is an external (and unknown) force acting on the system at the fulcrum. Therefore, linear momentum is not conserved.

Can you think of something else that is conserved which could be used to get your second equation?

9. Oct 15, 2014

### bobie

Could you help me get a clear picture saying if a free rod will spin and translate and if the speed of translation is predictable?

10. Oct 15, 2014

### Staff: Mentor

Correct.

Again, since linear momentum is not conserved, due to the external force at the fulcrum, can you think of some other quantity that is conserved?

11. Oct 15, 2014

### bobie

Angular momentum, of course, is that conserved? $L_p = m * v_i * r = 1 * 22 * 0.5 = 11$ right?

If that is what you mean , then $L_r = I\omega = 11 \rightarrow \omega = 12 * 11 = 132$

Last edited: Oct 15, 2014
12. Oct 15, 2014

### jbriggs444

Conservation of angular momentum does indeed give you a second equation. Can you write that equation down?

13. Oct 15, 2014

### Staff: Mentor

Yes.

So in the scenario with the fulcrum you have two unknowns, the velocity of the particle and the angular velocity of the rod. It also has two equations, conservation of energy and conservation of angular momentum. Two equations in two unknowns.

In the scenario without the fulcrum you have one additional unknown, the velocity of the center of mass of the rod. You also have an additional equation, conservation of linear momentum. Three equations in three unknowns.

(All of this assumes that the direction of the particle is known after the collision, otherwise you wind up with too many unknowns.)

Last edited: Oct 15, 2014
14. Oct 16, 2014

### bobie

OP stated that the particle will bounce back in the opposite direction $v_f$ = 18 m/s, therefore it will have Ke = 162 J and the rod 242-162 = 80 J, since energy is conserved. Linear momentum , of course is not conserved since mass/inertia is 1/12 of the nominal mass of the rod.

Can you explain why you consider $v_f$ an unknown? Since we know the energy of the spinning rod we can easily find its angular velocity = 43.8.

If we consider angular momentum, as I said : $L_i = 11, L_f = 9 \rightarrow L_r = 11 -9 = 2 \rightarrow \omega = 132 - 108 = 24$ we get near half the value, which means the E = 24 J and 56 J have been lost

15. Oct 16, 2014

### Staff: Mentor

Based on your responses. At some point, you need to choose what problem you want to solve and then describe it consistently.

I asked where it came from in post #2. You responded that it came from conservation principles, which means that it is unknown. You could have responded that it was given, in which case we could have used it to calculate how much energy was lost in the collision.

You cannot have it both ways. Either you are given conservation principles from which you can calculate the unknown velocities required to satisfy the conservation laws, or you are given the velocity, from which you can calculate how much energy and/or momentum is not conserved. Please choose and be clear. Assuming that we are given all of the initial conditions and assuming that the particle moves always along the same line.

For scenario 1:
Choose exactly two of the following as givens
1) Conservation of kinetic energy
2) Conservation of angular momentum
3) Rod final angular velocity
4) Particle final linear velocity

For scenario 2:
Choose exactly three of the following as givens
1) Conservation of kinetic energy
2) Conservation of angular momentum
3) Conservation of linear momentum
4) Rod final angular velocity
5) Particle final linear velocity
6) Rod center of mass final linear velocity

Last edited: Oct 16, 2014
16. Oct 17, 2014

### bobie

we know the masses and initial velocity (1,10,22) and E = 242, we know that the collision is elastic ergo E is conserved.
Now if it were a collision between balls the oucome would be as described in OP, and momentum is conserved too.

You said that
.
Could you answer this specific questions:
- what is the oucome of the collision, how do we find it, if OP is wrong? Linear momentum of the spinning rod is 0, therefore it is cannot conserved, but nevertheless:
- I need to know if the rod reacts to the collision like a ball of 10 Kg or not. If not then $vf$ is not -18.
- If it is not 18 I cannot calculate how much Ke is transferred to the rod and therefore I cannot find $\omega$ even if I Know L
- what is $\omega$? shall I find first this and then $v_f$?

Last edited: Oct 17, 2014
17. Oct 17, 2014

### jbriggs444

Units, please. You should be asking whether $v_f$ is equal to = -18 meters per second. You can check it yourself. If $v_f$ is equal to -18 meters per second and if the final rotation rate, $\omega$ is equal to 43.8 radians per second then what is the final total angular momentum about the axis to which the rod is fixed? How does that compare to the initial total angular momentum about that same axis?

18. Oct 17, 2014

### Staff: Mentor

You still were not clear, which is really frustrating, particularly when I expressly asked for clarity and made it as easy as possible for you to be clear.

You said:
So you chose 1) as a given, but you need to choose two of those four things. I would guess that you want to choose 2), but you have not been clear!

19. Oct 17, 2014

### bobie

I hope it will not take me 100 posts to make myself understood, this time, Dalespam: I cannot answer your question if you do not answer mine first and clarify what is not clear to me. I do not know if conservation of angular momentum is an option or is mandatory, or when it is or it is not, to begin with

I told you I do not know what to chose, and it really makes no difference to me. What is clear to me is only what happens if there is an elastic collision between a ball and another ball. If you need, choose any option and give me a first example so I can understand what is happening.

I have asked you many times to explain what is the difference when a rod is hinged on a fulcrum.
if the particle has v = 22 (E = 242) and hits the tip of the rod I haven't a clue about what happens. I know mass is 10 Kg but I do not know if I must consider mass = 5 since it is hinged at 50cm.
So I cannot chose, can you tell me please:
- if the particle will bounce back at -18m/s?
- if I calculated the initial L right at (22*1*.5 ) = 11 Nm?

Last edited: Oct 17, 2014
20. Oct 17, 2014

### Staff: Mentor

Conservation of linear momentum applies any time that the system has no external force. Similarly, conservation of angular momentum applies any time that the system has no external torque.

Since the first case has the fulcrum there is clearly an external force and therefore no conservation of linear momentum. You can use the violation of the conservation of linear momentum to calculate how big the external force is. Similarly, if there is an external torque then there is no conservation of angular momentum and you can use the violation of the conservation of angular momentum to calculate how big the external torque is.

You need to decide if you are envisioning a system where the external torque is 0. If so, then conservation of angular momentum is mandatory. Usually, a "frictionless fulcrum" would be used to indicate that there is no external torque, but you have pushed back so hard on this point that I am not sure of your intention.

Btw, you may be frustrated that I have not responded to your specific numbers. That is because, before we can determine the numbers, we must determine the equations that produce those numbers. The numbers come as solutions to the equations. Once we pin down the equations, then it is easy to solve them and get the numbers.

Last edited: Oct 17, 2014