Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Circular motion

  1. Feb 1, 2014 #1
    I have seen the derivation of the centripetal acceleration formula a=v^2/r by saying r= rcosθi+isinθj=rcosωti+rsinωtj and differentiating twice. Since ω is constant we get a=-ω[itex]^{2}[/itex]r.

    I've started looking at non-uniform circular motion where there is also the tangential acceleration vector component. I'm told that the component of acceleration directed towards the center of the circle has magnitude v^2/r, but I don't believe the original proof works because we assumed ω is constant, and now it isn't. Can this type of proof be made to work still?
     
  2. jcsd
  3. Feb 1, 2014 #2

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    Sure. Then you write
    [tex]\vec{r}(t)=r \cos [\theta(t)] \vec{i} + r \sin[\theta(t)] \vec{j}[/tex]
    and differentiate twice (assuming [itex]r=\text{const}[/itex], because it's supposed to be a circular motion). Then you get
    [tex]\vec{v}(t)=\dot{\vec{r}}(t) = -r \dot{\theta} \sin \theta \vec{i} + r \dot{\theta} \cos \theta \vec{j}=r \dot{\theta} \hat{\theta},[/tex]
    [tex]\vec{a}(t)=\dot{\vec{r}}(t)=-r \ddot{\theta} \sin \theta \vec{i} - r \dot{\theta}^2 \cos \theta \vec{i} + r \ddot{\theta} \cos \theta \vec{j} - r \dot{\theta}^2 \sin \theta \vec{j} =r \ddot{\theta} \hat{\theta}-r \dot{\theta}^2 \hat{r}.[/tex]
    Here I have used the orthonormal basis of the polar coordinates,
    [tex]\hat{r}=\cos \theta \vec{i} + \sin \theta \vec{j}, \quad \hat{\theta}=-\sin \theta \vec{i}+\cos \theta \vec{j}.[/tex]
    It's clear that [itex]\hat{r}[/itex] points perpendicularly outward from the circle (normal of the curve) and [itex]\hat{\theta}[/itex] along the circle (tangent of the curve).

    That shows that the tangential acceleration is
    [tex]a_{\theta} = \hat{\theta} \cdot \vec{a}=r \ddot \theta[/tex]
    and the normal acceleration is indeed the momentaneous centripetal acceleration as in uniform circlar motion,
    [tex]a_{r}=\hat{r} \cdot \vec{a} = -r \dot{\theta}^2.[/tex]
    It's easy to calculate that also then
    [tex]a_{r}=-\frac{\vec{v}^2}{r}.[/tex]
     
  4. Feb 1, 2014 #3
    In uniform circular motion ω is constant which means centripetal acceleration ac2R is constant .

    In case of non uniform circular motion centripetal acceleration is still given by ac2R ,but as you have rightly said ω is changing ,which means centripetal acceleration is also varying (magnitude changing with time).

    In addition there is also tangential acceleration aT = dv/dt .

    The net acceleration of the object is given by the vector sum of the two i.e ac and aT .

    Since they are at right angles the magnitude is given by aTotal = √(ac2 + aT2)
     
  5. Feb 1, 2014 #4
    Thanks a lot!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Circular motion
  1. Circular Motion (Replies: 53)

  2. Circular Motion (Replies: 1)

  3. Circular motion (Replies: 3)

Loading...