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Central force w1/w2=√r2/r1

  1. Feb 24, 2015 #1
    • Member warned about posting without the template and with no effort
    Hi
    we are going through cirkular centralforce and Im complete stuck...
    I cant find the derivation to why w1/w2=√r2/r1 is correct



    w1/w2=√r2/r1


    sorry im lost...
    best regards
    Fred
     
  2. jcsd
  3. Feb 24, 2015 #2

    Dick

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    Maybe you could explain what those symbols mean?
     
  4. Feb 24, 2015 #3
    yes of course , sorry
    There are two objects that are spinning on a plate and they fall off, and the physical reasoning is that the angular velocity (ω) has to meet the condition
    ω1/ω2=√r2/r1

    ω1= angular velocity objekt 1
    ω2= angular velocity objekt 2
    r2= objet 2 position to the center of the plate
    r1= objet 1 position to the center of the plate

    Best regards
     
  5. Feb 24, 2015 #4

    Dick

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    You should probably the objects will fall of when the acceleration exceeds that which can be produced by the frictional force holding them on. Do you know that the acceleration is given by the expression ##v^2/r## where ##r## is distance from the center and ##v## is rotational speed? What's the relation between ##v## and angular velocity?
     
  6. Feb 24, 2015 #5
    I Think that the relation is V=ωr correct? (and the friktionskoefficient is the same for both the objects)
     
  7. Feb 24, 2015 #6

    BvU

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    That is correct. Now we need an equation to express 'staying on the plate' versus 'flying off' a bit more in physics terms using the variables in our exercise.
    The template had an item for that; unfortunately it has disappeared (how ?, strange !). So here is a copy:

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

    2. Relevant equations

    3. The attempt at a solution
     
  8. Feb 24, 2015 #7
    1. The problem statement, all variables and given/known data

    There are two objects that are spinning on a plate and they fall off, and the physical reasoning is that the angular velocity (ω) has to meet the condition
    ω1/ω2=√r2/r1

    ω1= angular velocity objekt 1
    ω2= angular velocity objekt 2
    r2= objet 2 position to the center of the plate
    r1= objet 1 position to the center of the plate

    2. Relevant equations
    ω1/ω2=√r2/r1
    V=ωr

    3. The attempt at a solution

    √r2/r1 * ω1/ω2=0

    √((ω1r2)/(ω2r1))=0

    hmm this doenst feel right..
     
  9. Feb 24, 2015 #8

    Dick

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    You haven't really done anything except for make an algebra mistake. Start from the physics. What condition will make an object fall off the plate?
     
  10. Feb 24, 2015 #9

    BvU

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

    Object 1 lies on a spinning plate at a distance ##r_1## from the axis. It flies off at angular speed ##\omega_1## (at lower speeed it is held in orbit by friction)
    Object 2 lies on the same plate at a distance ##r_2## from the axis. It flies off at angular speed ##\omega_2##.
    Friction coefficients are the same for both objects.
    Masses of objects may or may not be the same.

    Show that ##\omega_1/\omega_2=\sqrt{r_2/r_1}##


    2. Relevant equations

    Friction force required to stay in circular orbit F = ...
    Maximum centripetal acceleration friction force can provide aat fly-off speed =


    3. The attempt at a solution

    ...
     
  11. Feb 24, 2015 #10
    2. Relevant equations

    Friction force required to stay in circular orbit F = m·v2/r
    Maximum centripetal acceleration friction force can provide a at fly-off speed ac =v2/r

    on the right track?
     
  12. Feb 24, 2015 #11

    Dick

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    Right track. Now replace the ##v## with an expression involving angular velocity.
     
  13. Feb 24, 2015 #12
    3. The attempt at a solution
    F = m·v2/r = (m* (ωr)2)/r
    ac =v2/r =(ωr)2/r
     
  14. Feb 24, 2015 #13

    Dick

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    Ok, now simplify that expression. ##a_c## doesn't depend on the mass, agree? So for any two objects ##a_c## is the same.
     
  15. Feb 24, 2015 #14
    agree
    (ωr)2/r = 2ω3r2
     
  16. Feb 24, 2015 #15

    Dick

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    I don't recognize the algebra you did you get that. Can you explain?
     
  17. Feb 24, 2015 #16
    Ohh sorry calculation fault
    I dont know the name in English (kvadreringsregler) (a+b)2 = a2+2ab+b2

    this one is wrong! se next one..
     
    Last edited: Feb 24, 2015
  18. Feb 24, 2015 #17
    or the potens rule axbx=(ab)x
     
  19. Feb 24, 2015 #18
    3. The attempt at a solution

    I am overseeing the weight i F.
    ω²r=ω²r
    ω²/ω²=r/r
    ω/ω=√r/r

    am I right ??
     
  20. Feb 24, 2015 #19

    Dick

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    Right idea. It would look much better if you'd distinguish the two values of ##\omega## and ##r##. Start from ##\omega_1^2 r_1 = \omega_2^2 r_2##.
     
  21. Feb 24, 2015 #20
    3. The attempt at a solution

    ω21r122r2
    ω2122=r2/r1
    2122)=(r2/r1)
    ω12=(r2/r1)

    I Think i got it right ?!
     
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