From cylindrical velocity to cartesian

Question

I am struggling with this. I need to convert this velocity in cylindrical coordinates to Cartesian coordinates:

$$ V(r, \theta, z) \: \mathrm{to} \: V(u,v,w) $$

$$ V_{r} = 10; \: V_{\theta}= 0; \: V_{z}=0; $$

How can I find the $u,v,w$ components?

Edit: Will it be?:

$$ u = V_{r} \cos(\theta) - V_{\theta} \sin(\theta) $$

$$ v = V_{r} \sin(\theta) + V_{\theta} \cos(\theta) $$

$$ w = V_{z} $$ Best regards

Answer 1

Put in terms of derivatives, $V_r=\frac{dr}{dt},$ $V_\theta=\frac{d\theta}{dt},$ and $V_z=\frac{dz}{dt}.$

Since $u=r\cos\theta,$ $v=r\sin\theta,$ and $w=z,$ then the Chain and Product rules get us $$V_u=\frac{du}{dt}=\frac{dr}{dt}\cos\theta - r\sin\theta\frac{d\theta}{dt}=V_r\cos\theta-rV_\theta\sin\theta,$$ $$V_v=\frac{dv}{dt}=\frac{dr}{dt}\sin\theta + r\cos\theta\frac{d\theta}{dt}=V_r\sin\theta+rV_\theta\cos\theta,$$ and $$V_w=V_z.$$

Consequently, in general, we need to know more than just the cylindrical velocities, but also the cylindrical coordinates. In this case we only need to know $\theta,$ as substitution gets us $V_u=10\cos\theta,$ $V_v=10\sin\theta,$ and $V_w=0.$

Answer 2

You are correct. The velocity in cylindrical coordinates is $$\vec v=\dot r\hat e_r +r\dot\theta \hat e_\theta+\dot z\hat e_z$$ Now identify $\dot r=V_r,\dot\theta =V_\theta,\dot z=V_z$, substitute the basis vectors $\hat e_r,\hat e_\theta, \hat e_z$ and you are done.

Answer 3

Transform from cylindrical coordinates $(r, \theta, z)$ to Cartesian coordinates $(u, v, w)$ is given by:

$$ \begin{aligned} u &= r\cos(\theta) \\ v &= r\sin(\theta) \\ w &= z \\ \end{aligned} $$

Then,

$$ \begin{aligned} V(u, v, w) &= V(r\cos(\theta), r\sin(\theta), z) \Rightarrow \\ &\left. \begin{aligned} V_r &= V_u\cdot \cos(\theta) + V_v\cdot \sin(\theta) + V_w\cdot 0 \\ V_{\theta} &= V_u\cdot \left(-r\sin(\theta)\right) + V_v\cdot r\cos(\theta) + V_w\cdot 0 \\ V_z &= V_u\cdot 0 + V_v\cdot 0 + V_w\cdot 1 \end{aligned} \right\} \Rightarrow \\ \begin{pmatrix} V_r \\ V_{\theta} \\ V_z \end{pmatrix} &= \begin{pmatrix} \cos(\theta) & \sin(\theta) & 0 \\ -r\sin(\theta) & r\cos(\theta) & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} V_u \\ V_v \\ V_w \end{pmatrix} \Leftrightarrow \\ \begin{pmatrix} V_u \\ V_v \\ V_w \end{pmatrix} &= \begin{pmatrix} \cos(\theta) & \sin(\theta) & 0 \\ -r\sin(\theta) & r\cos(\theta) & 0 \\ 0 & 0 & 1 \end{pmatrix}^{-1} \begin{pmatrix} V_r \\ V_{\theta} \\ V_z \end{pmatrix} \end{aligned} $$

So, you have to find an inverse matrix of $\begin{pmatrix} \cos(\theta) & \sin(\theta) & 0 \\ -r\sin(\theta) & r\cos(\theta) & 0 \\ 0 & 0 & 1 \end{pmatrix}$ and multiply it by $\begin{pmatrix} V_r \\ V_{\theta} \\ V_z \end{pmatrix} = \begin{pmatrix} 10 \\ 0 \\ 0 \end{pmatrix}$.