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Acceleration in Plance Polar Coordinates

  1. Dec 15, 2015 #1
    I am looking to understand more about ##a=(\ddot{r}-r(\ddot{\theta})^2)\hat{r}+(r\ddot{\theta}+2\dot{r}\dot{\theta})\hat{\theta}##

    I understand the terms ##\ddot{r}## and ##r\ddot{\theta}## ,but why ##-r(\ddot{\theta})^2## has opposite direction to ##\hat{r}## and why ##2\dot{r}\dot{\theta}## is multiply by 2
     
  2. jcsd
  3. Dec 15, 2015 #2
    These terms came from the derivatives of the unit vectors. In polar coordinates the unit vectors are not constants. They have constant magnitude but direction changes from point to point.
    When you take the derivative of ##\dot{r}\hat{r}## (the radial part of the velocity) you need to use the chain rule:
    ##\ddot{r} \hat{r}+ \dot{r} \dot{\hat{r}}##
    Look up the derivatives of the unit vectors. (Hint: the derivative of ## \hat{r} ## is along ##\hat{\theta}## and viceversa
    After you do the same for the angular part and collect the terms you will get that formula.
     
  4. Dec 15, 2015 #3
    Yes, I understand that, I am just trying to get some intuition for the terms I mention
     
  5. Dec 15, 2015 #4
    The negative sign shows that the centripetal acceleration is opposite to ## \hat{r} ##.
    The term with 2 is the Coriolis acceleration.
     
  6. Dec 16, 2015 #5
    Also, that centripetal acceleration term should have ##\dot{\theta}##, not ##\ddot{\theta}##.
     
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