Relation between r ,ω and θ for rotation around fixed axis.

relation between r ,ω and θ for rotation around fixed axis.

$$\frac{d\textbf {r}}{dt} = \textbf {ω}$$

$$\frac{dθ}{dt} = ω$$

$$\lvert\frac{d\textbf {r}}{dt}\rvert = \frac{dθ}{dt}$$

bold means vector. Is this right?

Last edited:

tiny-tim
Homework Helper
Hi AakashPandita! (type \left and and \right before two ordinary ||s, and they resize to fit! )

$$\frac{dθ}{dt} = ω = |\textbf{ω}|$$
ω (the angular velocity vector) is along the axis of rotation (ie, out of the page)
$$\left|\frac{d\hat{\textbf {r}}}{dt}\right| = r\frac{dθ}{dt} = rω$$
and
$$\frac{d\hat{\textbf {r}}}{dt} = rω\hat{\textbf{θ}}$$
where ##\hat{\textbf {r}}## and ##\hat{\textbf {θ}}## are the unit vectors in the radial and tangential directions

vanhees71
Gold Member
2021 Award
Let the rotation be around the $z$ axis of a Cartesian coordinate system. Then
$$\vec{r}(t)=r(t) \begin{pmatrix} \cos[\alpha(t)] \\ \sin[\alpha(t)] \\ 0 \end{pmatrix}.$$
This gives
$$\vec{v}(t)=\dot{r}(t) = \dot{r}(t) \begin{pmatrix} \cos[\alpha(t)] \\ \sin[\alpha(t)] \\ 0 \end{pmatrix} + r(t) \dot{\alpha}(t) \begin{pmatrix} -\sin[\alpha(t)] \\ \cos[\alpha(t)] \\ 0 \end{pmatrix} = \dot{r} \hat{r} + r \omega \hat{\theta}.$$
If the mass is fixed to a circle, then you have $\dot{r}=0$ and you can write
$$\vec{v}=\vec{\omega} \times \vec{r} \quad \text{with} \quad \vec{\omega}=\omega \vec{e}_z=\dot{\theta} \vec{e}_z.$$

ZapperZ
Staff Emeritus
$$\frac{d\textbf {r}}{dt} = \textbf {ω}$$

$$\frac{dθ}{dt} = ω$$

$$\lvert\frac{d\textbf {r}}{dt}\rvert = \frac{dθ}{dt}$$

bold means vector. Is this right?

You should already know that something is not quite right here, because the dimensions are all wrong! Always check that first, because that is your first "line-of-defense"!

Zz.

ω (the angular velocity vector) is along the axis of rotation (ie, out of the page)
$$\left|\frac{d\hat{\textbf {r}}}{dt}\right| = r\frac{dθ}{dt} = rω$$

how did you get this?

Let the rotation be around the $z$ axis of a Cartesian coordinate system. Then
$$\vec{r}(t)=r(t) \begin{pmatrix} \cos[\alpha(t)] \\ \sin[\alpha(t)] \\ 0 \end{pmatrix}.$$
This gives
$$\vec{v}(t)=\dot{r}(t) = \dot{r}(t) \begin{pmatrix} \cos[\alpha(t)] \\ \sin[\alpha(t)] \\ 0 \end{pmatrix} + r(t) \dot{\alpha}(t) \begin{pmatrix} -\sin[\alpha(t)] \\ \cos[\alpha(t)] \\ 0 \end{pmatrix} = \dot{r} \hat{r} + r \omega \hat{\theta}.$$
If the mass is fixed to a circle, then you have $\dot{r}=0$ and you can write
$$\vec{v}=\vec{\omega} \times \vec{r} \quad \text{with} \quad \vec{\omega}=\omega \vec{e}_z=\dot{\theta} \vec{e}_z.$$

i dont understand matrices

D H
Staff Emeritus