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Angular speed calculation after an inelastic collision

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Homework Statement


A disk [m=0.1 Kg; R=0.1 m] rotates about its center of mass [w=40 rad/s], on a smooth floor. A bar [m=0.1 Kg; lenght=R=0.1 m] moves on the floor with a speed Vb=4 m/s. At one point, the bar hits disk's edge in an inelastic collision, and they start rotating together.
A)[Fixed axis passing through disk's center of mass and perperdicular to the floor]
-Find the new angular speed [w'] of the system disk+bar]
B)[System can translate after the collsion]
-Find the new angular speed [w''] of the system disk+bar];
-Find the CM's speed



Homework Equations





The Attempt at a Solution


A)
When the bar hits disk's edge, same forces on the axis through the disk's CM prevent the system start rotating about the new CM. The momentum of these forces is zero because they have the same direction of the vector r, so the angular momentum (P) is costant:
[I choose disk's CM as pole for the momentum, so I'll calculate the moment of inertia from that pole]
-P1=I1*w1=1/2*m*R^2*w1
-P2=I2*w2=(I(disk)+I(bar))*w2=(1/2*m*R^2+1/12*m*R^2+m*(R+1/2R)^2)w2=17/6*m*R^2*w2

-P1=P2
- 1/2*m*R^2*w1=17/6*m*R^2*w2--------->w2=0.5*w1*6/17--->w2=0.5*40*6/17=7.06 rad/s
[Right solution]

B) In this case the rotation axis isn't fixed so, after the collision, the system starts translating and rotating about the new CM. Also, since the axis is not fixed, there are no external forces, so the traslational momentum [Q] is costant.


New CM's coordinates:
-DISK:
SUM(m(i)*x(i)^2)/m=0------>SUM(m(i)*x(i)^2)=0
-BAR:
SUM(m(i)*x(i)^2)/m=3/2*R------>SUM(m(i)*x(i)^2)=3/2*R*m

-DISK+BAR
Xcm=SUM(m(i)*x(i)^2)/2m=(0+3/2*R*m)/2m=3/4*R
[Ycm=0]
CM=(3/4*R;0);

CM's speed
Q=costant
Qi=m*Vb
Qf=Mcm*Vcm=2*m*Vcm
Qi=Qf
m*Vb=2*m*Vcm---->Vcm=Vb/2---->Vcm=2m/s [Right solution]

Angular speed of the system (w'')
Initially I thought that w'' was the same of w', which I found in the previous point.
Also in this case the momentum of external forces is zero [there are no external forces bisides weight, actually] so P is costant. If I choose the same pole for the angular momentum I find the previous equation. However book's solution is different, since it calculate the angular momentum about the CM.
I tryed to motivate this solution in this way: in general, the angular momentum about a generic pole O (in this case disk's CM) can be written and the sum of two terms: the angular momentum about the CM's of the system and the term r^(M*Vcm) wher Vcm is the CM's speed aboout the pole O. So:

Po=Pcm+r^(M*Vcm)
In this case r parallel to Vcm so that r^(M*Vcm)=0 and Po=Pcm

My finally question:
In point A, I tried to calcute P using Po=Pcm+r^(M*Vcm), and I found the same result of calculating the moment of inertia about the disk's CM and using P=I*w both before and after the collision.
In point B P=I*w can be used only before the collision.
So, I don't understand why it can't be used also after the collision and why in the case A
Po=Pcm+r^(M*Vcm) and Po=Io*w provides the same result, so when I can use both of theme
 

Answers and Replies

  • #2
haruspex
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A bar [m=0.1 Kg; lenght=R=0.1 m] moves on the floor with a speed Vb=4 m/s. At one point, the bar hits disk's edge
Is there a diagram? I can interpret that description in a number of ways. Was the bar moving parallel to its own length or transversely? If the second, what part of the bar hit the disk? Did it strike the disk full on or tangentially?...
 
  • #3
kuruman
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The information you provide is incomplete. It is important to know the relative velocity between the bar and the disk's edge and how the bar sticks to the disk. See below for just some possibilities that affect the system's new CM and moment of inertia.

DiskandBar.png
 

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  • #4
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I think I would need a diagram. It is important where the bar’s initial vector points relative to the disks rotation axis, and also I cannot tell the orientation of the bar after the collision. In fact it is difficult to determine the shape of the bar from this post much less how it sticks to the disk
 
  • #5
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Ha! Three simultaneous posts saying the same thing. Must be something to them!
 
  • #6
kuruman
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Ha! Three simultaneous posts saying the same thing. Must be something to them!
Not quite simultaneous. It must be that @haruspex and I (depending on when you refresh your screen) are moving at the same velocity relative to you.
 
  • #7
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Not quite simultaneous. It must be that @haruspex and I (depending on when you refresh your screen) are moving at the same velocity relative to you.
A simple question of precision. We all made our posts in the same year.

Actually I was distracted between starting and finishing and the thread didn’t refresh until I hit post.
 
  • #8
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The information you provide is incomplete. It is important to know the relative velocity between the bar and the disk's edge and how the bar sticks to the disk. See below for just some possibilities that affect the system's new CM and moment of inertia.

View attachment 230084
Sorry guys and thanks a lot for the help. I tried to explain it in the text but I haven't been clear enough. The new system is the second one (the system in the middle).
PS: How did you create those diagrams?
 
  • #9
haruspex
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The new system is the second one (the system in the middle).
But which way was the bar moving?
 
  • #10
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Sorry guys and thanks a lot for the help. I tried to explain it in the text but I haven't been clear enough. The new system is the second one (the system in the middle).
PS: How did you create those diagrams?
And is the bar’s velocity vector pointed directly at the disk’s axis? Also is there any more information regarding the shape of the bar? You can assume it is thin compared to R and the shape won’t matter much. However the post wasn’t clear.
 
  • #11
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So in this problem you need to conserve angular momentum and linear momentum. Case a is easiest. The fixed axle ( force of constraint). means you only need to conserve angular momentum. The axle also gives you no choice what point to use as an axis.
 
  • #12
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Sorry but I don't speak english very well, so I have some difficulties to explain the problem. The bar is unidimensional (length) and, yes, the velocity vector is pointed directly at the disk's axis.
 
  • #13
kuruman
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PS: How did you create those diagrams?
Microsoft PowerPoint.
 
  • #14
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I’m not sure I explained well my difficulties: my results agree with book’s solutions. However I would like to know if the method I used is right and why I can’t calculate the angular momentum in the same way of the case A and so when I can use Po=Io*w (where Io is the monent of inertia about a pole O).
 
  • #15
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I’m not sure I explained well my difficulties: my results agree with book’s solutions. However I would like to know if the method I used is right and why I can’t calculate the angular momentum in the same way of the case A and so when I can use Po=Io*w (where Io is the monent of inertia about a pole O).
In case B where there are no external forces, Newton's laws must apply regardless of choice of coordinate system. That is to say there are no external or unaccounted for forces so momentum must be conserved for all choices of origin. In particular, angular momentum will be conserved no matter which pivot point you choose. There may be a choice which is more convenient for calculation, but the answer will still be correct for all choices.

In case A where the axle of the disk is fixed to the floor, there are external forces you do not know about. The floor will act through the axle to keep the disk from translating, and you do not have information as to what those forces are. If there are external forces you do not know about, you cannot apply Newton's laws. In particular momentum will not be conserved. To get around this you choose the axle as the pivot point. That way the unknown forces are acting at the pivot point. They cannot impart a torque about that point, so at least you know that angular momentum about that point must be conserved. However, you can't say that about any other point, so you are not free to choose your pivot.
 
  • #16
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Also I can't explain very well, sorry. okay I understand that in case A, I can choose only the the axis through disk's CM beacause force of constraint havo no momentum about that axis, while in case B I can choose any pole beacause there are no external forces. However when I calculate the angular momentum (in case B) about disk's CM (instead of the new system's CM) I can use the two eqautions, I wrote above:
-Po=Io*w [used in case A to calculate the angular momentum about the same pole, but that in this case doesn't work and provides the same equation's system of the case A and so the same w]
-Po=Pcm+r^(M*Vcm) [in this case the term r^(M*Vcm) is zero and so Po=Pcm. Now I'm a bit confused because, according to this equation I have to calculate the angular momentum above system's CM and then add the second term but as the method used in the case A , I can calculate the angular momentum from a pole O, calculating the mometum of inertia I above that pole and using the equation Po=Io*w. It's clear that there is a contradiction because Pdisk=Pcm----> Idisk=Icm but it's impossible.
 
  • #17
haruspex
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in case B I can choose any pole beacause there are no external forces.
You still have to be careful how you choose your axis in rotational kinetics. The general rule is the you can use either a point fixed in the lab frame (the initial position of the disk's centre would be good) or the mass centre of the rigid body.
Not sure whether you used the disk centre in terms of its initial position or as fixed in the disk's frame. Your working is not clear to me. Please define all your variables.

By the way, in my experience it is rarely wirth the effort of finding the common mass centre of a complex body. It is usually much easier just to sum the momenta etc. of its simple components,
 
  • #18
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I don't know what part of my work isn't enough clear because I think I explained each passage of my reasoning and I reported all the given data.
I'll try to clarify this passage:
Angular speed of the system (w'')
Initially I thought that w'' was the same of w', which I found in the previous point.
Also in this case the momentum of external forces is zero [there are no external forces bisides weight, actually] so P is costant. If I choose the same pole for the angular momentum I find the previous equation. However book's solution is different, since it calculate the angular momentum about the CM.
I tryed to motivate this solution in this way: in general, the angular momentum about a generic pole O (in this case disk's CM) can be written and the sum of two terms: the angular momentum about the CM's of the system and the term r^(M*Vcm) wher Vcm is the CM's speed aboout the pole O. So:

Po=Pcm+r^(M*Vcm)
In this case r parallel to Vcm so that r^(M*Vcm)=0 and Po=Pcm
When I calculate w'' (which is the angular speed in the case B), initially I tried to use the axis of disk's CM, using the equation P=I*w.
Before the collision the bar has not angular momentum about the chosen pole because its speed is pointed directly to the axis, while the disk rotates with anguar speed w=40 rad/s. So Pi=I(system)*w=(Idisk+Ibar)*w=Idisk*w+Ibar*w. The term Ibar*w is zero beacuse the bar doesn't rotate so Pi=Idisk*w=1/2*m*R^2*w-------->Pi=0.5*0.1*0.1^2*40=0.02 m*kg*m/s [same result of case A]

Afet the collision the disk and the bar rotate together with the same angular speed w''. So Pf=(Idisk+Ibar)*w''
Idisk=1/2*m*R^2
Ibas=1/12*m*R^2+m*(R+R/2)^2=7/3*m*R^2
Pf=(1/2*m*R^2+7/3*m*R^2)*w''=17/6*m*R^2*w'' [same result of case A]

Pi=Pf------>1/2*m*R^2*w=17/6*m*R^2*w''------------->w''=0.5*6/17*w''-------->w''=0.5*6/17*40=7.06 rad/s [same result of A, but wrong solution]
 
  • #19
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25s27o8.jpg


**In case B- before collision the disk rotate with w
 

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  • #20
kuruman
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In case (b), after the collision the final angular momentum is ##L_{final}=I_{cm}\omega_{final}##, where ##I_{cm}## is the moment of inertia about the two-body system's center of mass. I trust you can find ##I_{cm}.## At this point you have to be careful. The angular momentum just before the collision must also be calculated about the point where the CM is located just after the collision, not about the center of the disk.

On edit: If you choose to calculate the angular momentum about the center of the disk, you have to be consistent and include the angular momentum of the CM about the center of the disk in the expression for the final angular momentum. You cannot use the center of the disk for the initial angular momentum and the center of the mass for the final angular momentum.
 
Last edited:
  • #21
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In case (b), after the collision the final angular momentum is ##L_{final}=I_{cm}\omega_{final}##, where ##I_{cm}## is the moment of inertia about the two-body system's center of mass. I trust you can find ##I_{cm}.## At this point you have to be careful. The angular momentum just before the collision must also be calculated about the point where the CM is located just after the collision, not about the center of the disk.

On edit: If you choose to calculate the angular momentum about the center of the disk, you have to be consistent and include the angular momentum of the CM about the center of the disk in the expression for the final angular momentum. You cannot use the center of the disk for the initial angular momentum and the center of the mass for the final angular momentum.
Yes, In fact I applied Huygens-Steiner theroem to calculate the moment of inetia from the disk’s CM and so the angular momentum.

Indeed I don’t know how I can calculate the angular momentum about the disk’s CM calculating the momentum about the CM of the system before the collision and then add the momentum of the system’s CM beacuse I should obtein the same result but I don’t understand how.

However I think this is not the error of my resolution which, I repeat, it’s right but I don’t understand why one method doesn’t work
 
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