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Circular motion-centripetal acceleration

  1. Apr 26, 2006 #1
    hi

    This latex code is giving me some problems. I write one thing, it displays something completely different

    In circular motion velocity only changes direction but not size

    change of velocity - [tex]\Delta v[/tex]
    Change of angle - [tex]\Delta T[/tex]
    Velocity - V
    Centripetal acceleration - a

    delta(V) = V * delta(T)

    When delta(T) approaches its limit (goes to zero), change of velocity has same direction as acceleration vector?

    We compute the magnitude of velocity change with :

    Delta(v) = v * Delta(T)

    I see this being true when change of angle approaches its limit ( goes to zero ), since then length of circular arc ( with radius begin velocity vector ) equals [tex]\Delta v[/tex]. But that is not true if delta(T) is not approaching limit. So how can we use formula

    delta(v) = V * delta(T)

    in cases were delta(T) is not approaching zero, since I assume length of circle arc is quite different than delta(T) if delta(T) doesn't go to zero?

    cheers
     
    Last edited: Apr 26, 2006
  2. jcsd
  3. Apr 26, 2006 #2

    mezarashi

    User Avatar
    Homework Helper

    I think your original assumption may be incorrect. If you want to relate the linear velocity of the object on the circle circumference with the change in angle, it would be:

    [tex]s = r\theta[/tex]
    [tex]\frac{ds}{dt} = r\frac{d\theta}{dt}[/tex]
    [tex]v = r\frac{d\theta}{dt}[/tex]

    where s is the circle arclength, r is the circle radius, and theta is the angle. The [tex]\frac{d\theta}{dt}[/tex] would be equal to your [tex]\Delta T[/tex].
     
  4. Apr 27, 2006 #3
    I don't get it. First of all, [tex]v = r\frac{d\theta}{dt}[/tex] is not same formula as delta(v)=v*delta(theta)

    I still don't understand why delta(v) = v * delta(theta) would give us correct result when delta(theta) is anything but [tex]d\theta[/tex] ?
     
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