Calculating speed and angular momentum

[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?

 

[Dale]

If the CoM is fixed on a frictionless fulcrum, it is easy to predict that the particle will bounce at 18 m/s

How do you get that?

 

[bobie]

How do you get that?

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

 

[Dale]

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.

 

[bobie]

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.

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

 

[Dale]

The result I gave comes from more equations,

What is the second equation?

 

[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

What is the second equation?

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?

 

[Dale]

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?

 

[bobie]

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?

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?

 

[Dale]

Are you saying that the energy acquired by the rod through the collision is not (80J) the same a ball of that mass would acquire?

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?

 

[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 $L_r = I\omega = 11 \rightarrow \omega = 12 * 11 = 132$

 

[jbriggs444]

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

 

[Dale]

Angular momentum, of course, is that conserved?

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.)

 

[bobie]

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.

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

 

[Dale]

Can you explain why you consider vfv_f an unknown?

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

 

[bobie]

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.
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

Let's start with 1):
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

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

.
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 $v_f$?

 

[jbriggs444]

if OP is wrong? isn't $v_f = -18$ ? what is $\omega$?

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?

 

[Dale]

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 $v_f$?

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.

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

You said:

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.

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!

 

[bobie]

You still were not clear, which is really frustrating, !

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?

 

[Dale]

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

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.

 

[bobie]

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. .

Thanks Dalespam, I feel bad because I irritate you when it is the last thing I'd like to do. You are really the nicest mentor at PF.

Now, we have ascertained that we have conservation of E (242) and $L_i$ ( is it 11 Nm?)
Suppose it is , I still cannot find $\omega$ if I do not know $v_f , L_f$ so , do I need a third equation?

 

[Dale]

OK, since we can use conservation of kinetic energy (elastic collision) and conservation of angular momentum (no external torque) then we can simply write out those equations. From conservation of kinetic energy we get:
$$\frac{i \omega _0^2}{2}+\frac{m v_0^2}{2}=\frac{i \omega _1^2}{2}+\frac{m
v_1^2}{2}$$

From conservation of angular momentum we get:
$$ i \omega _0+m r v_0=i \omega _1+m r v_1 $$

From the problem we are given:
$$i=\frac{l^2 M}{12},~r=\frac{l}{2},~l=1,~M=10,~m=1,~v_0=22,~\omega _0=0$$

Substituting those in we are left with only two unknowns: $v_1,~ \omega_1$.

Can you solve that?

Spoiler

$$ 242=\frac{v_1^2}{2}+\frac{5 \omega _1^2}{12},~11=\frac{v_1}{2}+\frac{5
\omega _1}{6} $$
$v_1=-11.84,~ \omega_1=20.31$

 

[bobie]

Can you solve that?

I read your post without the spoiler and I have tried for 2 days to find my mistake, I even started a thread: https://www.physicsforums.com/threads/solving-two-equations.776934/It can't be 20.3 because the energy of the rod would be 5*20.3^2/6 = 343.7 J, it is more than the whole E available 242J, and if $v_f = 11.84$ 71 J stay with the bouncing particle. Only 171 J are available for the rod.
Edit: now I realized it is 343/2. right
Now what happens if the rod is free to move ? I suppose the energy stays the same but some must go to Ke and I find it by conservation of linear momenum, right?

 

[Dale]

It can't be 20.3 because the energy of the rod would be 5*20.3^2/6 = 343.7 J

You are off by a factor of 2. Either you have forgotten the 1/2 from the KE formula or your moment of inertia is off.

 

[bobie]

we cross-posted, I realized that by myself, I was put off as they told me at HW that there was something wrong. Please read my edit

 

[bobie]

Then the rod would have v =.385 and Ke = 57.3 and $\omega = 16.58$ if that is right we spared at least 75 posts;)

 

[Dale]

Now what happens if the rod is free to move ? I suppose the energy stays the same but some must go to Ke and I find it by conservation of linear momenum, right?

Yes. Now the rod has a linear velocity, $V$, in addition to the angular velocity, $\omega$, that it had previously. That gives you a third unknown which you have to add into the formulas above. You also get a third equation from conservation of linear momentum. So you solve the system of three equations in three unknowns.

 

[Dale]

Then the rod would have v =.385 and Ke = 57.3 and $\omega = 16.58$ if that is right we spared at least 75 posts;)

Write your system of equations. That is more important than the numbers.

 

[bobie]

I have a few questions for you, probably I am asking too much, if you do not know just write the number and a dash, so i will not aske the again:

1) how do we know that the rod will react in the same way if there is or is not a fulcrum? I would think a free rod offers less resistance.
2) is there a way, a rule to predict what percentage of Energy will remain in the spin and how much will go as Ke?
3) I expected the rod on the fulcrum to offer 1/2 of the resistance, but it turned out that it is exactly 1/3 , does it depend on I = 1/12, or on what? Is there a parameter from which to deduce the response?

4)we had a discussion in another thread: if we want to stop a moving body that has KE = 50J, must we spend 50J in the opposite direction in any circumstance? Can it vary if we stay in front of it and stop it with our hand, if we pull it from behind with a string, or if it crushes into a wall and the wall crumbles or the wall stay put and the body is deformed?
5)here:http://hyperphysics.phy-astr.gsu.edu/hbase/carcr.html#cc1, they say clearly that work needed is -mv^2/2, but they said that it is not the tree that does the work, but the car itself. Is that realistic?

Thanks a lot Dalespam for your invaluable help

 

[Dale]

1) how do we know that the rod will react in the same way if there is or is not a fulcrum?

We don't know that. In fact, we know the opposite. We know that the rod will not react in the same way because we know that the equations (which I asked you to write and which you did not write) are different.

2) is there a way, a rule to predict what percentage of Energy will remain in the spin and how much will go as Ke?

Yes, solve the equations. The equations (which I asked you to write and which you did not write) are the rule and they are the way to predict the behavior.

3) I expected the rod on the fulcrum to offer 1/2 of the resistance, but it turned out that it is exactly 1/3 , does it depend on I = 1/12, or on what?

I am not sure what you mean by "resistance".

4)we had a discussion in another thread: if we want to stop a moving body that has KE = 50J, must we spend 50J in the opposite direction in any circumstance?

Energy doesn't have a direction. It doesn't make any sense to speak of spending energy in the opposite direction.

5)here:http://hyperphysics.phy-astr.gsu.edu/hbase/carcr.html#cc1, they say clearly that work needed is -mv^2/2, but they said that it is not the tree that does the work, but the car itself. Is that realistic?

Let's not pursue other scenarios at this time. The thread already has two scenarios, that is enough.

As I asked before, please write down the system of equations.

 

[bobie]

From conservation of kinetic energy we get:
$$\frac{i \omega _0^2}{2}+\frac{m v_0^2}{2}=\frac{i \omega _1^2}{2}+\frac{m
v_1^2}{2}$$
From conservation of angular momentum we get:
$$ i \omega _0+m r v_0=i \omega _1+m r v_1 $$
From conservation of linear momentum we get:
$$ mv_0 + mv_1 = Mv_r → 22 + 11.85 = 33.85 → v_r = 3.385 → E_k = 57.29 J $$

1) We know that the rod will not react in

.2)The equations are the rule and they are the way to predict the behavior.
I am not sure what you mean by "resistance".
.

By resistance I mean that the rod reacts like a ball of mass of 10/3
1) the rod has reacted in the same way 10/3 both with and without fulcrum.
2) I know I am asking too much, but as they found that a rod has I = 1/12, they might have studied that 1/12 corrsponds to 1/3 of the real mass in the collision. That would have made things easier and we might have got immediately $v_1 = 7/13 v_0 = 11.85 , v_r = 20/ (13[*M]) = 3.385$

 

[jbriggs444]

What about the contribution to kinetic energy due to the rod's final linear motion? What about the contribution to momentum due to the ball's final linear motion?

The naming convention for the variables needs thought as well. It was one thing to talk about $v_0$ and $v_1$ when we only had an initial and a final velocity of the ball. Now we have an initial and final velocity of the rod as well. I am a firm believer in documenting variable names.

 

[Dale]

From conservation of kinetic energy we get:
$$\frac{i \omega _0^2}{2}+\frac{m v_0^2}{2}=\frac{i \omega _1^2}{2}+\frac{m
v_1^2}{2}$$

You are forgetting the term for the translational KE of the rod. If we denote the velocity of the center of mass of the rod as $V$ (to go along with the mass of the rod as $M$) then that will be $\frac{1}{2} M V^2$. With $V_0$ before the collision and $V_1$ after the collision.

From conservation of linear momentum we get:
$$ mv_0 + mv_1 = Mv_r $$

This is not correct. The conservation of momentum relates the momentum before the collision (subscript 0) to the momentum after the collision (subscript 1). On the left hand side you should have two terms, one expressing the linear momentum of the particle (lower case m and v) before the collision and the other expressing the linear momentum of the rod (capital M and V) before the collision (subscript 0). On the right hand side you should have two terms, one expressing the linear momentum of the particle (lower case m and v) after the collision and the other expressing the linear momentum of the rod (capital M and V) after the collision (subscript 1). All of the subscript 0's should be on the left, and all of the subscript 1's should be on the right. There should be no subscript r anywhere.

By resistance I mean that the rod reacts like a ball of mass of 10/3

This concept of resistance is not a standard concept. Let's drop it. It is just a distraction and is not necessary for determining the physics.