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Contradiction between equations of angular momentum

  1. Oct 26, 2012 #1
    1. The problem statement, all variables and given/known data
    Two equal, parallel and opposite forces at at both sides of a horizontal disk that lies on a smooth table, according to the picture.
    The mass is m and the moment of inertia is: kmR2
    Angular momentum round the center point A:
    [tex]2FR=kmR^2 \cdot \alpha[/tex].
    Angular momentum round the point B on the outer edge:
    [tex]2FR=mR^2(k+1) \cdot \alpha[/tex].
    It is clear i will get 2 different angular acceleration [itex]\alpha[/itex], how come?

    2. Relevant equations
    M=I[itex]\alpha[/itex]
    Shteiner's theorem of the parallel axis:
    Ib=Ic+mb2

    3. The attempt at a solution
    It is clear that the second equation is wrong, since the first one is right, since it is round a static point.
    Maybe i have to compensate, when calculating round point B, for it's acceleration?
    How? maybe with D'alamber's sentence?
    But then, can i solve only from the point of view of the accelerating system?
    I want to solve from the static, inertial system.
     

    Attached Files:

  2. jcsd
  3. Oct 26, 2012 #2

    tiny-tim

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    Hi Karol! :smile:

    LP = IPω only works if P is the centre of mass or the centre of rotation.

    From the PF Library on angular momentum

    about point P, where [itex]\mathbf{v}_P[/itex] is the velocity of the part of the body at position P:

    [tex]\mathbf{L}_{P}\ =\ \tilde{I}_{P}\mathbf{\omega}\ +\ m(\mathbf{r}_{c.o.m.}-\mathbf{r}_P)\times\mathbf{v}_P[/tex]​
     
  4. Oct 27, 2012 #3
    And what if the center of rotation is itself under acceleration?
    For example the new picture here.
    Can i use L=I[itex]\alpha[/itex] around point B?
    A rope is wound around a falling cylinder.
     

    Attached Files:

  5. Oct 27, 2012 #4

    tiny-tim

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    that equation is for a fixed point P,

    and vP is the velocity of the (changing) point on the body that happens to be at P

    (so no question of P accelerating :wink:)
    (you mean L = Iω) yes :smile:

    but a moment later, B will no longer be the centre of rotation, so you'll have problems finding dL/dt …

    it'll be much easier to use the centre of mass! :wink:
     
  6. Oct 27, 2012 #5
    I want to make clear when to use equation:
    [tex]\mathbf{L}_P\ =\ I_{c.o.m.}\,\mathbf{\omega}\,+\, \mathbf{r}_{c.o.m.}\times m\mathbf{v}_{c.o.m.}[/tex]
    You have a disk rotating round point A, as in the picture.
    Round points A, the center and point B:
    [tex]L_A=mR^2(k+1) \cdot \omega[/tex]
    [tex]L_{Center}=kmR^2 \cdot \omega[/tex]
    [tex]L_B=kmR^2 \cdot \omega \,+\, mR\cdot\sqrt{2}R \cdot \omega[/tex]

    About B i am confused, since i don't understand the transition between the equations:
    [tex]\mathbf{L}_{P}\ =\ \tilde{I}_{P}\mathbf{\omega}\ +\ m(\mathbf{r}_{c.o.m.}-\mathbf{r}_P)\times\mathbf{v}_P[/tex]
    And:
    [tex]=\ \tilde{I}_{c.o.m.}\mathbf{\omega}\ +\ m(\mathbf{r}_{c.o.m.}-\mathbf{r}_P)\times\mathbf{v}_{c.o.m.}[/tex]
    In the first we use the moment of inertia and velocity of the moving point P, and in the second-of the center of mass.
     

    Attached Files:

    Last edited: Oct 27, 2012
  7. Oct 27, 2012 #6

    tiny-tim

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    my computer won't open that .bmp, it says it's corrupt :redface:
     
  8. Oct 27, 2012 #7
    I replaced
     
  9. Oct 30, 2012 #8

    tiny-tim

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    Hi Karol! :smile:

    I'm sorry for the delay … I've been working this out! :redface:

    You're right :smile:, that first equation is far too general

    it only applies if P is the centre of rotation and if the axis of rotation stays parallel to a principal axis of the body.

    (So this applies, for example, to a sphere or a cylinder rolling over a step, but not to a cone rolling on a plane, or a wheel rolling on a curved rail.)

    Thanks for pointing this out. :smile:

    I've now corrected the Library entry (and given you the credit!).​
     
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