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Rotational motion

  1. Mar 30, 2005 #1
    I calculated the equations for circlular motion as fallows...

    \vec{r}= p \hat{e_p} + z \hat{e_z}[/tex] where e_p = unit vector in the radial direction, and so on

    \frac{\partial{\vec{r}}}{\partial{t}} = \dot{p}\hat{e_p} + p\dot{\theta}\hat{e_{\theta}} + \dot{z}\hat{e_z} [/tex]

    how do i show that the velocity is perpendicular to the radius, show me my mistake, but...
    [tex] \vec{r} \bullet \vec{\dot{r}} = p \dot{p} + z \dot{z} [/tex] which isn't obvious to me that that is 0 so what to do?
  2. jcsd
  3. Mar 30, 2005 #2

    Doc Al

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    Staff: Mentor

    Maybe I'm missing the point of what you are doing, but if you want to deduce something about circular motion, you must impose some conditions. What you've described so far is just the use of cylindrical coordinates to describe any motion. For circular motion, set [itex]\dot{p} = \dot{z} = 0[/itex]; then all terms in the velocity drop out except [itex]p\dot{\theta}\hat{e_{\theta}}[/itex].
  4. Mar 30, 2005 #3
    i agree the dp/dt would be zero, however if circle is oriented at some angle to a xyz coord system, then wouldn't dz/dt be non-zero

    bare with the picture the blue lines are there to help u reference where in the plane the point is

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  5. Mar 31, 2005 #4

    Doc Al

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    Staff: Mentor

    Sure. But if your circle is oriented at an angle to your z-axis, you've chosen an inconvenient coordinate system to describe the circular motion. The advantage of cylindrical coordinates is only apparent if the circular motion is centered around the z-axis.
  6. Mar 31, 2005 #5
    i agree it is inconvenient however should it not be able to be calculated, not sure what u mean bye the "advantage of cylindrical coordinates is only apparent if the circluar motion is centered around the z-axis"
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