- #1

vwishndaetr

- 87

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

[itex]P(x,y,z) = (0,R,0)[/itex]

[tex]\overrightarrow{\omega}= Ct^2\hat{k}[/tex]

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

Ans: Now I know that,

[tex]P(r,\theta,\phi,) = (R,\frac{\pi}{2},\frac{\pi}{2})[/tex]

But I don't think that helps much.

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

[tex]\hat{\phi}[/tex]

I have:

[tex]\overrightarrow{r}= R\ \hat{r}+\frac{\pi}{2}\ \hat{\theta}+\ \ \ \ \ \ \ \ \hat{\phi}[/tex]

The last term is giving me issues.

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

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

But I don't think it is legal to just integrate [tex]\omega[/tex] 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.