Calculating Rotational Motion: Proving Velocity is Perpendicular to Radius

[Phymath]

I calculated the equations for circlular motion as fallows...

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

<br /> \frac{\partial{\vec{r}}}{\partial{t}} = \dot{p}\hat{e_p} + p\dot{\theta}\hat{e_{\theta}} + \dot{z}\hat{e_z}

how do i show that the velocity is perpendicular to the radius, show me my mistake, but...
\vec{r} \bullet \vec{\dot{r}} = p \dot{p} + z \dot{z} which isn't obvious to me that that is 0 so what to do?

 

[Doc Al]

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 \dot{p} = \dot{z} = 0; then all terms in the velocity drop out except p\dot{\theta}\hat{e_{\theta}}.

 

[Phymath]

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

 

[Doc Al]

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

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.

 

[Phymath]

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"