1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Kinetic energy and direction of rotation

  1. Dec 2, 2014 #1
    Hi,

    I would like to study these two cases:

    Case A
    1/ I turn counterclockwise the ring at w2, this need the energy [tex]\frac{1}{2}mr^2w_2^2[/tex] with m the mass of the ring
    2/ I turn the ring with arm clockwise at w1, this need the energy [tex]\frac{1}{2}md^2w_1^2[/tex] with d the length of the arm
    3/ I eject small part of ring when point are like green points, I eject all part of the ring, I recover kinetic energy [tex]\frac{1}{4}m((d+r)w_1-rw_2 ) ^2+\frac{1}{4}m((d-r)w_1+rw_2)^2[/tex]

    http://imagizer.imageshack.us/v2/800x600q90/673/ZITrPH.png [Broken]

    Case B
    1/ I turn clockwise the ring at w2, this need the energy [tex]\frac{1}{2}mr^2w_2^2[/tex]
    2/ I turn the ring with arm clockwise at w1, this need the energy [tex]\frac{1}{2}md^2w_1^2[/tex]
    3/ I eject small part of ring when point are like green points, I eject all part of the ring, I recover kinetic energy [tex]\frac{1}{4}m((d+r)w_1+rw_2)^2+\frac{1}{4}m((d-r)w_1-rw_2)^2[/tex]

    http://imagizer.imageshack.us/v2/800x600q90/912/lrRetE.png [Broken]

    The energy needed for turn the ring around itself and with the arm is the same in case A and case B. In the contrary, the energy recover in case A seems not be the same than in case B.

    Kinetic energy of the ring is:
    [tex]E_k=\frac{1}{2}mr^2\omega_2^2+\frac{1}{2}md^2\omega_1^2[/tex]

    With:

    d: the length of the arm
    r: radius of ring
    m: mass of ring
    w1: rotationnal velocity of the arm
    w2: rotationnal velocity of the ring around itself

    The ring is ejecting more and more when points are like that:

    http://imagizer.imageshack.us/v2/800x600q90/909/Jfn8ou.png [Broken]

    So, the expression of kinetic energy is false or there a force I forgot ?
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Dec 2, 2014 #2

    Bystander

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    The mass losses from the ring are discrete fractions of the disc mass you're ejecting when the rim of the disc intersects the radius extended along the arm?
     
  4. Dec 2, 2014 #3
    Is it a question ? If not I don't understand, could you explain more ? I would like to study a ring only because it's easier.
     
    Last edited: Dec 2, 2014
  5. Dec 2, 2014 #4

    Bystander

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    It's a question I'm asking to clarify to myself just when and where you're wanting to eject the mass. It's tangentially to the disc from positions where it intersects the radius along the arm? Am I correct?
     
  6. Dec 2, 2014 #5
    Yes, it's correct, the last image shows the ring with less and less parts.
     
  7. Dec 2, 2014 #6

    Bystander

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    So, what you want to do is calculate the velocities at the two ejection points for each case, and they will be different (recall ω2' from your previous "apparatus), and sum the products of the masses and squares of those velocities for both cases.
     
  8. Dec 2, 2014 #7
    At start, I need to give this energy:

    [tex]E_k=\frac{1}{2}mr^2\omega_2^2+\frac{1}{2}md^2\omega_1^2[/tex]

    This equation don't change if the direction of rotation is clockwise or counterclockwise. And it seems different of the energy recover from parts of ring when I ejected them.

    no ?
     
  9. Dec 2, 2014 #8

    Bystander

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Agreed
    Also agreed.
     
  10. Dec 2, 2014 #9
    If I give :

    [tex]E_k=\frac{1}{2}mr^2\omega_2^2+\frac{1}{2}md^2\omega_1^2[/tex]

    and I recover:

    [tex]\frac{1}{4}m((d+r)w_1+rw_2)^2+\frac{1}{4}m((d-r)w_1-rw_2)^2[/tex]
    or
    [tex]\frac{1}{4}m((d+r)w_1-rw_2 ) ^2+\frac{1}{4}m((d-r)w_1+rw_2)^2[/tex]

    like the energy is conserved, where is the energy I forgot ?
     
  11. Dec 2, 2014 #10

    Bystander

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    That's the first of seven conservation laws --- they are self-enforcing. You haven't forgotten anything, and your bookkeeping on energy came out perfect.
     
  12. Dec 3, 2014 #11
    It's not possible, the energy is conserved (it's a basic device), you're sure about my calculations ? I done a numerical example and I recover more energy than I gave.
     
    Last edited: Dec 3, 2014
  13. Dec 3, 2014 #12

    Bystander

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Now you're making me go back through and read your work in detail ---
    The masses you eject are moving at (r+d)ω2' and (r-d)ω2'. I should have caught that --- I just glanced at your expression and thought you had included ω2' = ω1 ± ω2.
     
  14. Dec 3, 2014 #13
    You're right I was wrong in my velocities. With several rings all around the big trajectory. If I'm ejecting parts like that:

    http://imageshack.com/a/img661/5961/DizFnx.png [Broken]

    Arrows = velocities, I can find the energy gave at start ? Green points can be link when there are ejecting. And vectors seems to cancel part of themselves, at least with w2.
     
    Last edited by a moderator: May 7, 2017
  15. Dec 3, 2014 #14

    Bystander

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    You're having fun with this "device," aren't you. I think you've pretty much got a handle on how to handle the various set-ups now. It's mostly a matter of paying attention to details.
     
  16. Dec 3, 2014 #15
    Yes, I find this device fun ! It's not possible to link 2 green parts from 2 different rings ? or my velocities are not correct ?
     
  17. Dec 3, 2014 #16

    Bystander

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Add the vectors just the way you've drawn them, and it should work --- as I said, I'm reasonably confident you've got a full understanding of the principles you have to apply.
     
  18. Dec 3, 2014 #17
    Sorry to just jump in (I'm new here) but I was wondering about this expression; if ω2' were 0 ie ω1 = ω2 for case B or ω1 = -ω2 for case A, wouldn't that mean the masses would have 0 kinetic energy when they are ejected? This would obviously be incompatible as the energy of the fragments would be 0.

    I understand the question that Del8 asked but not the solution unfortunately.
     
  19. Dec 3, 2014 #18
    If the ring is turning at w2 and the arm is turning at w1 with w2 < w1, I take 2 external blue objects (free to move in space) that are turning at w1, like the ring turns counterclockwise in the arm reference, it's possible to eject these 2 objects and apply a clockwise torque to the ring in the same time. The two external objects don't lost the energy from the rotation w1 and win energy from the ring. Like this:


    http://imagizer.imageshack.us/v2/800x600q90/537/aZD6z5.png [Broken]

    So, here I don't see where the energy is lost. It must be the velocity of blue object.

    @-Stradivarius: with case A, if w1=w2, one mass is ejected at 0 if d=r, no ? For the other mass I ask myself, maybe w2. But with even w2, the energy is not conserved, the device recover 1/4 of input.
     
    Last edited by a moderator: May 7, 2017
  20. Dec 4, 2014 #19
    First, for reply to the question of -Stradivarius257,

    The expression of small parts of ring ejecting is not:

    CaseA:
    outer: dw1-rw2
    inner: dw1+rw2

    CaseB:
    outer: dw1+rw2
    inner: dw1-rw2

    If d=r and w1=w2, with the case A, the external object has a velocity at 0. For the inner, the velocity is at 2w1 like that the energy is conserved, is it correct ? For me, it's a little strange that at the inner the velocity is at 2w1.

    -------------------------------------------------------------Second---------------------------------------------------------

    For my case with 2 free external objects (object are turning at w1) shocked from the ring (the ring keep all its mass), for me the velocity of free object is increasing from the ring, because velocities are like that:

    http://imageshack.com/a/img540/6472/x5VeY9.jpg [Broken]

    The part Vw1 is the velocity that come from w1, the part Vw2 is the velocity that come from w2, like velocities are perpendicular, how Vw2 can affect Vw1 ? Or the ring don't receive a torque ?
     
    Last edited by a moderator: May 7, 2017
  21. Dec 4, 2014 #20

    Bystander

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    If you say, "Two free external objects separated by a distance of 2r moving in a straight line tangent to the circle described by the center of the ring at V1 = ω1d," I'll be happier. You've indicated straight line motion, and it's tricky to constrain them with another piece added to the apparatus (probably possible, but I'm not ready to try it yet).

    You then have elastic collisions of the ring, whatever its mass, with two masses (equal?), and some specified fraction or multiple of the ring mass for each, in which both energy and momentum are conserved. And ....... now I'm really in over my head (not that I wasn't earlier) handling the momenta, and we may have to go back and look at the previous case. I'm thinking we'll have to drag in trig functions to handle the "angular momenta" of the two blue masses relative to the center pivot (time dependent), and that means we may have tripped over the same problem previously, releasing the green masses from the disc.

    I'm gonna have to think about this. I appreciate your patience.
     
  22. Dec 4, 2014 #21
    Yes, in a straight line because if masses are turning (around the red axis) the vector of velocity is not like I drawn. If masses are turning around red axis the velocity decreases with shock. I calculated the trajectory in horizontal axis for verify if there is more at left than at right for the highest mass and more at right than at left for the lowest mass and I think it's correct. I find:

    [tex]rsin((w2-w1)t)[/tex]

    compare at:

    [tex]d-\sqrt(r^2+d^2)cos(artg(\frac{r}{d})-w_1t)[/tex]

    For example, with:

    r=1
    d=3
    w2=0
    w1=10
    t=1e-5

    For the highest mass, the result is :

    9.9999e-5 at left
    9.9985e-5 at right

    Take care of yourself Bystander, I can wait, you're friendly.

    Edit: for the highest mass it seems correct but no for the lowest. But with only the highest mass there is a torque on the ring, and the arm don't receive a torque (the force is in the axis).
     
    Last edited: Dec 4, 2014
  23. Dec 5, 2014 #22
    I did a video for show the trajectory of a point (w2=0):



    It's very difficult to see if the trajectory is correct or not for increase V without a torque on the arm, but even in this position:

    http://imagizer.imageshack.us/v2/800x600q90/537/RJ4x3G.png [Broken]

    I think the trajectory of the free object must change all the time, no ?
     
    Last edited by a moderator: May 7, 2017
  24. Dec 7, 2014 #23

    Bystander

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    For the case in post #19, you're adding components to the velocity vectors of your "free masses" that are at right angles to the pre-collision velocities, and that are parallel to the instantaneous tangential velocities of the disc --- shouldn't require any new tricks to do the bookkeeping.

    The case in post #22 isn't clear. You might want to try something intermediate; "case 19b" would be collision of your falling masses with the ring/disc such that the collision forces act only along the vertical axis (torque on r through the disc center).
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook




Loading...