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Circular motion equations

  1. Jan 9, 2009 #1
    Is this derivation correct?

    I managed to derive [itex] v = r \omega[/itex] and (I think) [itex] a = \omega^2 r[/itex].

    I did [itex] a^2 = \ddot{x}^2 + \ddot{y}^2[/itex] to get eventually
    [itex]a^2 = r^2(\dot{\theta}^4 + \ddot{\theta}^2)[/itex]. Then I said that, for uniform circular motion, the angular velocity [itex]\dot{\theta}[/itex] is a constant [itex]\omega[/itex] meaning that [itex]\ddot{\theta}[/itex] is equal to zero. So then it came to [itex] a = \omega^2 r[/itex].

    But how do we know, from my derivation, that this is a centripetal acceleration toward the center? Is it because the tangential acceleration is the one I discarded and so this is the perpendicular component of acceleration? Is the formula valid where the angular velocity is not constant?
     
  2. jcsd
  3. Jan 9, 2009 #2
    For uniform circular motion the tangential speed is constant but the direction is changing, and this directional change is always pointed towards the center of rotation. To derive the acceleration vector showing the proper direction, one must start with the position vector in polar coordinates

    r = r <r> (θ) where <r> is the radial unit vector and bold letters indicate a vector. Taking the time derivative using the chain rule gives
    dr/dt = v = (dr/dt) <r> + r(d<r> /d θ) (d θ /dt)
    But d<r> /d θ = < θ > so,
    v = (dr/dt) <r> + r(d θ /dt) < θ >
    To find a, take the time derivative of v and use d< θ > /d θ = - <r> . This will give the components of a in the <r> and < θ > directions and will show the Coriolis acceleration.
     
  4. Jan 10, 2009 #3
    gansta, are you familiar with complex numbers? If so I suggest attempting the derivation that way, using polar form. If you only care about motion in a plane it's simpler than the vector approach suggested by chris, and the desired signs and direction of the acceration term comes out nicely.
     
    Last edited: Jan 10, 2009
  5. Jan 13, 2009 #4
    I haven't learned about complex numbers and polar co-ordinates yet. I find it interesting that complex numbers (which seem abstract and not that relevant in the real world) are good at describing circular motion. I thought they were only used in really advanced level physics.

    For uniform circular motion with radius r, the position vector is given by

    [itex]\mathbf{r} = r \begin{pmatrix} \cos \omega t \\ \sin \omega t \\ \end{pmatrix}[/itex]

    From this we can work out the acceleration vector to be

    [itex]\ddot{\mathbf{r}} = - \omega^2 \mathbf{r}[/itex]
    (that r is the position vector, not the radius).

    Does that negative tell us that it's toward the center? The omega doesn't make a difference because it's a scalar. I quite like this derivation because we can easily get the circular motion formulas by taking the magnitude of the velocity and acceleration vectors. Although it's only valid for constant angular velocity.
     
    Last edited: Jan 13, 2009
  6. Jan 13, 2009 #5
    yes, the negative means inwards facing (I presume that your second equals was supposed to be a minus sign).

    What you have is logically equivalent to the complex polar representation BTW.

    You can do non-constant angular velocity with your plane vector representation too if you want . Just include the omega derivatives too when you differentiate.
     
  7. Jan 13, 2009 #6
    But if we do it with vectors for non-constant angular velocity isn't it kind of the same as the way I did in the first post (take x and y separately and combine them with Pythagorean theorem)?
     
  8. Jan 13, 2009 #7
    it looks pretty similar, but in your first post you lost the direction.

    also note that you can do the non-constant radius with your vector approach pretty easily too.
     
  9. Mar 4, 2009 #8
    I've now learned a bit about polar co-ordinates. How does it work with them?
     
  10. Mar 4, 2009 #9
    Why don't you start with expressing your vector in polar coordinates as you've learned them, plus an attempt to take first derivatives.
     
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