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Centripetal Force and proportionality

  1. Mar 6, 2005 #1
    Hi,

    can someone please point me in a direction. I need to explain WHY "The centripetal force is proportional to the mass and to the radius and proportional to the square of the frequency."

    We did the spinning lab, and now I need to do a discussion.

    Thanks.
     
  2. jcsd
  3. Mar 6, 2005 #2

    Andrew Mason

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    You have to work out the change in velocity as a function of its tangential speed, [itex]v [/itex] or [itex] \omega r[/itex].

    Draw a diagram of the velocity vector of the mass at time 0. Then draw its velocity vector after a time dt. The mass turns through an angle [itex]d\theta = ds/r = \frac{vdt}{r}[/itex] in that time.

    Also remember that [itex]v = 2\pi r/T = \omega r[/itex] and [itex]d\theta = \omega dt[/itex] where [itex]\omega = 2\pi f[/itex] is the angular frequency in radians/sec.

    Now, the new velocity vector at t=dt is the same length as at t=0 but pointed [itex]d\theta[/itex] to the original. The difference is the change in velocity or dv and is directed toward the centre of the circle along the radius. You can see from the diagram that:

    [itex] dv = vsin(d\theta)[/itex] which approaches the limit of [itex] dv = vd\theta[/itex] as [itex]d\theta \rightarrow 0[/itex].

    This means: [tex]dv = vd\theta = \omega r d\theta = \omega^2r dt[/tex] so

    [tex]dv/dt = a_{centripetal} = \omega^2r[/tex] so:

    [tex]F_{centripetal} = m\omega^2r[/tex]

    AM
     
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