Sperical coords: Position vector of spinning disk.

[vwishndaetr]

I posted this in the Intro physics sections, but then realized that spherical coords might be a bit complex for introductory physics. This has been bothering my head for a couple days now. Any help is appreciated.

Given: A wheel of radius R rotates with an angular velocity. The wheel lies in the xy plane, rotating about the z-axis.

P(x,y,z) = (0,R,0)

\overrightarrow{\omega}= Ct^2\hat{k}

Ques: What is the position vector of point P in spherical coordinates?

Ans: Now I know that,

P(r,\theta,\phi,) = (R,\frac{\pi}{2},\frac{\pi}{2})

But I don't think that helps much.

For the position vector, I can't figure out the term for:

\hat{\phi}

I have:

\overrightarrow{r}= R\ \hat{r}+\frac{\pi}{2}\ \hat{\theta}+\ \ \ \ \ \ \ \ \hat{\phi}

The last term is giving me issues.

Now I know that \phi changes with time, so the term must depends on t.

I also know that \omega is rad/s, which can also be interpreted as \phi/s.

But I don't think it is legal to just integrate \omega to get position. Is it?

Since the angular velocity is quadratic, that means the disc is accelerating. So the position should be third order correct?

I'm being really stubborn here because I know it is something minute that is keeping me from progressing.

 

[Inferior89]

Why wouldn't it be legal to integrate? You have already established that |\mathbf{\omega} (t)| = \dot{\phi}(t)

So \phi (t) = Ct^3/3 + \pi / 2

Maybe I am missing something here but it looks straightforward..

 

[vwishndaetr]

I don't know, just seemed out of place. Since in sperical coords the formula from position to velocity pics up an R for the phi-term, it throws me off.

Edit: Just realized that what I just said relates to tangential velocity. So it would pick up an R.

Thank you for clarifying this for me. I lacked the confidence to make that jump.

 

[Inferior89]

Yes
\mathbf{v} = \mathbf{\hat{r}}\dot{r} + \mathbf{\hat{\theta}} r \dot{\theta} + \mathbf{\hat{\phi}}r\dot{\phi}\sin{\theta}

In your example
\dot{r} = 0, \quad r = R, \quad \dot{\theta} = 0, \quad \theta = \pi / 2 , \quad \dot{\phi} = |\mathbf{\omega}|
so
\mathbf{v} = R\mathbf{\hat{\phi}}\dot{\phi} = R\mathbf{\hat{\phi}}|\mathbf{\omega}|

 

[vwishndaetr]

Thanks again!

Appreciate it.