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Homework Help: What is the angular velocity that the rod will rotate

  1. Sep 12, 2014 #1

    MMS

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    Hello!
    I'm pretty much translating this question word by word, so if you have problem understanding something off of it, please, let me know and I'll try explaining it better.

    1. The problem statement, all variables and given/known data

    Two masses of similar mass M are connected by a rod of length 2a. the rod is rotating on a horizontal plane in an angular velocity of ω, when its center of mass (point A) is at rest. The table is horizontal and friction is neglected. (the rod is not wired to the table).
    At some given time t, we locate a mass of 2M so that it sticks to one of the masses.

    Question:What is the angular velocity that the rod will rotate in around its new center of mass after the collision?
    Z2Oxpgs.png
    2. Relevant equations




    3. The attempt at a solution
    What I did was this:
    1. I calculated the location of the new center of mass (the after frame) and it turns out to be a/2 towards the 3M.

    2. I looked at the frame just a moment before collision so that the mass 2M is really close to the mass M (one of them) but yet to collide with one. I calculated the angular momentum before collision by I*ω around the new center of mass that I stated above:

    [m*(3a/2)^2 + m*(a/2)^2]*ω = 2.5m*(a^2)*ω

    3. Now, I looked at the frame just a moment after the collision so the masses pretty much made no movement and again I calculated the angular velocity after collision by I*ω around the new center of mass:

    [m*(3a/2)^2 + m*(a/2)^2 + 2m*(a/2)^2]*ω' = 3m*(a^2)*ω'

    4. Of course there is conservation of angular momentum, therefore I equalized the two expressions to get that the new angular velocity is ω' = (5/6)*ω and not (2/3)*ω as the correct answer is.

    Here's the thing that's bothering me other than that answer is wrong. I tried applying RxMV on the before system with the location of the new center of mass and it turns out to be right!
    I really don't understand where have I mistaken in the solution I've given and why in A LOT of cases when I do I*ω and RxMV in this type of questions (conservation of angular momentum) I get different results. I thought they're equal...

    Anyway, I'd really appreciate it if you guys told me at which point of the solution I did an incorrect step and also answer my question in the last paragraph.

    THANK YOU!
     
  2. jcsd
  3. Sep 12, 2014 #2

    haruspex

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    I agree with your answer.
    Regarding cases where the two methods disagree, can you post an example?
     
  4. Sep 12, 2014 #3

    MMS

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    Hi and thanks for the reply.

    What answer do you agree with? the ω'?
     
  5. Sep 12, 2014 #4

    haruspex

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    5ω/6
     
  6. Sep 12, 2014 #5

    MMS

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    Oh alright. But it's wrong so... :P
     
  7. Sep 12, 2014 #6

    nrqed

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    The omega they give (the omega "before") is with respect to the center of mass located at the center (distance of "a" for both masses), not at the new position.
    And using [itex] I \omega [/itex] or [itex] | \vec{r} \times \vec{p} | [/itex] has, of course, to give the same answer in any given problem. You should show us an example.
     
  8. Sep 12, 2014 #7

    haruspex

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    A rotation rate is a rotation rate. It will take the same time to complete one revolution regardless of where you take the axis to be.
     
  9. Sep 12, 2014 #8

    nrqed

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    But if you have an object rotating with respect to some axis and you change the axis, the value of omega will change.

    I did not want to explain everything in details to let the OP do some work but since you missed my point, let me be more detailed: the calculation of the angular momentum depends on where the axis of rotation is. You won't get the same value of [itex] I \omega [/itex] if you use a different point. Since the [itex] \omega [/itex] they give is specified with an axis of rotation at the center, it is that axis of rotation that one must use to calculate the angular momentum.
     
  10. Sep 12, 2014 #9

    haruspex

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    The angular momentum changes, of course, but not the rotation rate.
     
  11. Sep 12, 2014 #10

    nrqed

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    Like I mentioned in my previous post, since they say that the initial rotation is with respect to the center of the rod, you must use that information to find the initial angular momentum. Therefore it is

    [itex] (m a^2 + m a^2) \omega = 2 m a ^2 \omega [/itex]

    Set this equal to your final angular momentum and you will get the correct answer.
     
  12. Sep 12, 2014 #11

    Orodruin

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    The total momentum is zero. Thus the angular momentum will be the same around any point. The momentum of inertia the OP is using would be fine if the object was fixed at that point, which is not the case. Seen from the CoM after the collision, the two masses would have different angular velocities at the collision time (they have the same velocity and one is further away). This does not mean the object´s rotation has a different angular frequency, just that the angular velocities seen from the post-collision CoM are different.
     
  13. Sep 12, 2014 #12

    MMS

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    I have noticed that if I do it with respect to the center of the rod on the before angular momentum I get the right answer. But notice what you're doing here. You're actually applying conservation of momentum with respect to two different axes. The angular momentum before with respect to the center of the rod and the angular momentum after with respect to the new location of the center of mass which is completely invalid.

    Also, I believe that the angular velocity is the same at any point on the rod.
     
  14. Sep 12, 2014 #13

    MMS

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    I really want to type it down but I'm on my phone and it's completely annoying.

    Pleasw, try calculating Iw with respect to the location of the new center of mass and after that calculate it by rxmv and see the results for yourself.
     
  15. Sep 13, 2014 #14

    haruspex

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    You've taken my post out of context. Nrqed was claiming the rate of rotation of an object depends on the axis with respect to which you measure it. My response was that the angular momentum (in general) depends on the choice of axis, but the rotation rate does not. You are right that in the present case the angular momentum does not depend on it either.
     
  16. Sep 13, 2014 #15

    haruspex

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    At last I see where we both went wrong. We calculated the prior angular momentum wrongly. We did it as though the rod were rotating about the system CoM. Instead, look at the linear velocities. Each mass has speed aω, so the moment is maω(3a/2 + a/2) = 2maω2.
     
  17. Sep 13, 2014 #16

    MMS

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    I don't believe it's right to say that each has a a linear velocity of what you said. The linear velocity of each mass is v=ωxr where r is the distance from the location of our new center of mass to where the body is and then you apply rxmv where r, again, is the distance mentioned above.
    What you've done is calculate the linear velocity of each mass with respect to the point they're rotating around before colliding with the 3rd mass.
     
    Last edited: Sep 13, 2014
  18. Sep 13, 2014 #17

    ehild

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    Nrqrd is right.

    The initial linear velocities were v and -v, where v=aω, as the rod rotated about its centre.

    The angular momentum of a point mass about any fixed point is mrxv. That of the system of the rotating equal masses, with respect to the centre of the rod, was 2mav=2ma2ω.
    The angular momentum in the new frame of reference was (3/2 a ) v m + (1/2 a) v m= 2avm again as orodruin pointed out in Post #11.
    The 2m mass was added to the system. Originally in rest, it had zero contribution to the angular momentum. After it stuck to the bottom mass, the angular momentum became IΩ where the moment of inertia I=m(3/2 a)2+3m (1/2 a)2= 3 a2m .

    The angular momentum is conserved:3 a2m Ω =2ma2ω.
     

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    Last edited: Sep 13, 2014
  19. Sep 13, 2014 #18

    MMS

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    First off, thanks for the solution you've posted.

    I've posted a question about the linear velocities. Can you please take a look at my last post (#16)?

    Thanks.
     
  20. Sep 13, 2014 #19

    ehild

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    My post was the answer to your Post #16. Just before the collision, the linear velocities of the masses were v and -v, in any rest frame of reference. The velocities do not change if you change the point of reference. v=rω is valid if r is measured from the centre of rotation, and it was at the middle of the stick before the collision.

    ehild
     
  21. Sep 15, 2014 #20

    nrqed

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    I don't even see what you mean by saying that a rate of rotation is independent of the axis of rotation. We can't even talk about rotation of an object without defining the axis around which it rotates. Let's take two masses M connected by a rigid rod of negligible mass. If I tell you that it is rotating at 20 rad/second, what does that mean?!? It does not mean anything unless you know around what axis it is rotating. If it does, then explain to me what this means without specifying an axis of rotation.

    The rod could be held at the center, at one extremity or it could be some other rotation entirely. It is meaningless to give a rotation rate without specifying the axis of rotation (and it is even less meaningful to do any calculation without knowing about the axis of rotation).

    Regards,

    Patrick
     
    Last edited: Sep 15, 2014
  22. Sep 16, 2014 #21

    haruspex

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    Remember that our point of disagreement does not concern an actual change of axis, but the choice of axis for calculating angular momentum..
    Any pair of points of the body defines a direction in space. If I rotate a body such that every such pair defines a line parallel to the line they originally defined then I have not changed the orientation of the body, but I may have changed its location.
    If I rotate it steadily about a fixed axis, after some period of time T it will be back at its original orientation. Every line defined by points in the body will similarly be back in its original orientation. Thus the period of rotation (and hence the rate) is independent of which axis I consider the rotation to be about.
    Another way to think of this is to consider a different inertial frame, moving at constant velocity wrt to the original frame. The axis of rotation will appear to be different, but the rate the same.
     
  23. Sep 16, 2014 #22

    nrqed

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    But it is a mistake to say that if the period of rotation is the same, the rate is necessarily the same. This is obvious to see if you consider an explicit example.

    Consider a rod rotating at constant ##\omega ## with respect to its center. Then the extremities have an angular velocity that is constant and equal to ## \omega ##. Now take another reference point, somewhere at a certain distance from the rod (not within the disk covered by the rod). The ##\omega## of of the masses will clearly not be the same as the ##\omega ## with respect to the center. The ##\omega## measured from this other point even changes sign!!
     
  24. Sep 16, 2014 #23

    haruspex

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    Time to move this discussion to a private thread.
     
  25. Sep 16, 2014 #24

    ehild

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    And time to define ##\omega##.

    ehild
     
  26. Sep 16, 2014 #25

    nrqed

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    Yes. I just want to post a last time in order to make sure that there is no confusion for others who might happen to read this thread so let me again emphasize my point: the value of [itex] \omega [/itex] of a certain object ## DOES ## depend on what axis of rotation is it measured from.

    Let me be very precise. Consider an object in rotation, for example the wheel of a car being driven at constant speed on a road. What is the ## \omega ## of the wheel? The first key point is that this question is ambiguous, from a physics point of view. Of course, the intuitive answer is that it has a unique and precise value but when people jump to this conclusion it is because they are ## implicitly ## thinking about the angular speed with respect to the center of the wheel and they are therefore already choosing an axis of rotation (even if they might not want to admit it ;-) )

    So let's be very precise and careful. First, in order to define an ## \omega ##, we must pick a specific point mass. Let's pick a small piece of rubber at the edge of the wheel.

    Now we want to calculate ## \omega ##. What is the definition of ## \omega ##? It is

    [tex] \omega = \frac{ d \theta (t)}{dt} [/tex]

    So we need to obtain the function ## \theta (t) ## giving the angular position of the piece of rubber at the edge of the wheel as a function of time. And now comes the obvious point: this function depends on where we set our axis of rotation!


    Let's say you set the axis of rotation at the center of the wheel and wait a time equal to half the period of rotation, then the angle changes by ## \pi ## radians in that time. Now use a different axis of rotation, fixed let's say at a point on the ground a certain distance from the car, then it is clear that in the same amount of time the point on the wheel will not cover the same angle. Therefore the function ## \theta (t) ## depends on the reference point used for the axis of rotation and ## \omega## also depends on the axis of rotation chosen.
     
    Last edited: Sep 16, 2014
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