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.