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Centripetal/angular acceleration

  1. Mar 1, 2005 #1
    I was doing a physics problem and realized that the formula for angular acceleration was the same as the formula for centripetal acceleration (in terms of angular speed)

    They both are [tex]\omega^2r[/tex]
    where w is angular speed and r is the radius

    Why is that so? When I tried to derive this I got...

    [tex]a_{centripetal}=\frac{v^2}{r}[/tex]
    [tex]a_{centripetal}=\frac{v}{r}v[/tex]
    since v/r=w then...
    [tex]a_{centripetal}=\omega v[/tex]

    how are they equal? [tex]\omega v=\omega^2r[/tex]
     
  2. jcsd
  3. Mar 1, 2005 #2
    You've got it wrong ,check your book

    true centripetal acceleration is v^2 /r BUT ANGULAR acceleration is the something spinning up faster or spinning down slower. Angular acceleration is [tex] \alpha = \frac{\Delta \omega}{\Delta t} = \frac{v}{rt} [/tex]
     
  4. Mar 1, 2005 #3
  5. Mar 1, 2005 #4
    [tex] \omega [/tex] is the ANGULAR VELOCITY (or frequency) and [tex] \omega = \frac{v}{r} [/tex]
    the CENTRIPETAL ACCELERATION is a = v^2 /r

    since omega = v/r
    a = omega r
     
  6. Mar 1, 2005 #5
    do you mean omega v? That's what I have on my first post... does this mean my book was wrong?
     
  7. Mar 1, 2005 #6

    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 a unit 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]

    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]

    AM
     
  8. Mar 1, 2005 #7
    so the formula for angular acceleration is the same as the formula for centripetal acceleration?
     
  9. Mar 1, 2005 #8
    what is wrong with my method? I subbed in v/r for omega but got r*omega
    I understand that if I subbed v=omega*r then the equation would come out correct
     
  10. Mar 1, 2005 #9

    Andrew Mason

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    No. They are two distinct concepts; two quite different vector quantities with different directions.

    For a mass moving in a curved path, centripetal acceleration is radial - toward the centre. Tangential acceleration - in the direction of travel gives rise to non-zero angular acceleration.

    The centripetal acceleration ([itex]a_c = -\omega^2r[/itex]) is always non-zero if there is circular motion.

    AM
     
    Last edited: Mar 1, 2005
  11. Mar 1, 2005 #10
    yes I understand the concepts are different, but both equations can be expressed as [tex]\omega^2 * r[/tex]

    is that correct?
     
  12. Mar 2, 2005 #11

    Andrew Mason

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    No. Angular acceleration has nothing to do with [itex]\omega[/itex]. It depends on torque not angular speed, just as acceleration is a function of force not velocity.

    The definition of angular accelaration is [itex]\alpha = a/r = f/mr = fr/mr^2 = \tau/mr^2[/itex]. So [itex]\tau = m\alpha r^2[/itex]

    AM
     
    Last edited: Mar 2, 2005
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