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Circular motion and (maybe) pendulum

  1. May 27, 2007 #1
    1. The problem statement, all variables and given/known data
    an object with mass m=2kg is placed on a horizontal smooth surface. it's tied to a rigid and unextensible rope long L=1m. this object moves around in a circular motion without lifting from the surfuce. the maximum tension of the rope is T=1000N. find the angular velocity [tex] \omega_{max}[/tex]

    2. Relevant equations
    [tex] \omega= \frac{d \theta}{dt}[/tex] or [tex]a_{centre}=\frac{v^2}{r}[/tex]
    [tex]\vec N=mg[/tex]


    3. The attempt at a solution

    I really don't know where to start from
    There's no friction so the forces on the object are N, the normal force going up along the positive axis y, the T force (though I don't know the angle between the surface and the L). Well ... that's all that comes to my mind right now. Anything right in my blabbing?
     
    Last edited: May 27, 2007
  2. jcsd
  3. May 27, 2007 #2

    Kurdt

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    The tension in the rope is the clue. The force experienced during circular motion can have a maximum value of what? It should be easy to determine the maximum angular velocity from there.
     
  4. May 27, 2007 #3
    sorry can't follow you. What is the force during circular motion equal to?
    Anything to do with sin & cos ?
     
  5. May 27, 2007 #4
    how do I calculate the force which pushes the object in a circular motion?
    [tex]\sum F= \vec T+ \vec N + \vec P [/tex] ?

    or is it something like this

    [tex]\theta=\theta_0 + \omega (...t)[/tex]

    (I'm not sure of this last formula, I remember something like that)
     
    Last edited: May 27, 2007
  6. May 27, 2007 #5

    Kurdt

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    You're not really close at all. In circular motion the object is subject to an acceleration (centripetal acceleration) toward the centre of the circular motion. Thus there is also a force proportional to that acceleration toward the centre of the circular motion. Think of Newton's second law.
     
  7. May 27, 2007 #6
    F=m*a (N.'s 2nd law);
    T is a form of force so maybe T=m*a;
    centrip. acc is [tex]a_c=\frac{v^2}{R}[/tex]

    where R is the radius of the circle drawn by the object in motion and "v" is the velocity.
    I know from theory that

    [tex]\vec v= \omega R[/tex]

    if I substitute in the previous formula

    [tex]a_c=\frac{v^2}{R}= \frac{\omega^2 R^2}{R}=\frac{\omega^2}{R} [/tex]

    ok, now where to fit L? in place of R? why?
     
  8. May 27, 2007 #7

    Kurdt

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    Yes the length of the rope will be the radius of the circular motion. if you drw a diagram it will be easy to convince yourself why the rope length is the radius of the circular motion. Essentially the rope will be at full extension whil the mass is travelleing at its maximum angular velocity.

    EDIT: Just noticed you made a slight mistake in your derivation so [tex]a=\omega^2R[/tex]
     
  9. May 27, 2007 #8
    ok, thanks a lot!
     
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