Simple cylindrical coords problem

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In summary, the conversation discusses a rigid body rotating about a fixed axis with constant $\omega$ along the z axis. The position vector $\vec{r}$ is expressed in cylindrical coordinates and used to find (a) $\vec{v}=\omega \times \vec{r}$ and (b) $\nabla \times \vec{v}$. The solution for (a) is a typo in the book, as it should be $\hat{\phi} \omega \rho$ instead of just $\phi \omega \rho$. For (b), the notation indicates finding the Curl of $\vec{v}$, which should be a vector, so the correct answer includes the unit vector $\hat{z}$ as well
  • #1
ognik
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Rigid body rotating about a fixed axis with constant $\omega$ along the z axis. Express position vector $\vec{r}$ in cyl. circ. cords and using cyl. circ. cords find (a) $\vec{v}=\omega \times \vec{r}$ (b) $\nabla \times \vec{v}$

So $ \vec{r} = \vec{\rho}\rho + \vec{z}z $

(a) = $\begin{vmatrix}
\hat{\rho}&\hat{\phi}&\hat{z}\\0&0&\omega\\\ \rho&\ 0&z
\end{vmatrix} = \vec{\phi}\omega \rho $

But the solution shown is $ \vec{\phi} \omega \rho $? Shouldn't it be the unit vector $\hat{\phi}$?

(b) $ \nabla = \left( \partial_{\rho}, \frac{1}{\rho} \partial_{\phi}, \partial_z \right) $

$ \therefore \nabla \times \vec{v} =
\begin{vmatrix}
\hat{\rho}&\hat{\phi}&\hat{z}\\\partial_{\rho}, & \frac{1}{\rho} \partial_{\phi}, &\partial_z \\\ 0&\ \omega \rho & 0
\end{vmatrix}$
which is obviously not going to give me the answer in the book ($2\omega$), so what am I missing please? Also isn't the book's answer missing a direction?
 
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  • #2
ognik said:
Rigid body rotating about a fixed axis with constant $\omega$ along the z axis. Express position vector $\vec{r}$ in cyl. circ. cords and using cyl. circ. cords find (a) $\vec{v}=\omega \times \vec{r}$ (b) $\nabla \times \vec{v}$

So $ \vec{r} = \vec{\rho}\rho + \vec{z}z $

(a) = $\begin{vmatrix}
\hat{\rho}&\hat{\phi}&\hat{z}\\0&0&\omega\\\ \rho&\ 0&z
\end{vmatrix} = \vec{\phi}\omega \rho $

But the solution shown is $ \vec{\phi} \omega \rho $? Shouldn't it be the unit vector $\hat{\phi}$?

(b) $ \nabla = \left( \partial_{\rho}, \frac{1}{\rho} \partial_{\phi}, \partial_z \right) $

$ \therefore \nabla \times \vec{v} =
\begin{vmatrix}
\hat{\rho}&\hat{\phi}&\hat{z}\\\partial_{\rho}, & \frac{1}{\rho} \partial_{\phi}, &\partial_z \\\ 0&\ \omega \rho & 0
\end{vmatrix}$
which is obviously not going to give me the answer in the book ($2\omega$), so what am I missing please? Also isn't the book's answer missing a direction?

Hi ognik,

I am not sure what your problem is with the first part. The solution you have obtained; $\vec{\phi} \omega \rho $ seems to be correct.

For the second part the notation $\nabla \times \vec{v}$ seems to indicate you have to find the Curl of $\vec{v}$ not the gradient.
 
  • #3
Hi, thanks - must be a typo in the book for part a, left out the \hat, good to have confirmation.

Part b - oops, yes curl, which should be: $\frac{1}{\rho}\begin{vmatrix}
\hat{\rho}&\rho\hat{\phi}&\hat{z}\\ \partial_{\rho}&\partial_{\phi}&\partial_z \ \\0&\rho \omega \rho &0
\end{vmatrix}$ for $ \vec{v}=\hat{\phi}\omega \rho $

I get $ \nabla \times \vec{v} = \hat{z}2\omega $, the book again appears to have a typo - no $\hat{z}$, assume mine is OK? Thanks for the help.
 
  • #4
ognik said:
Hi, thanks - must be a typo in the book for part a, left out the \hat, good to have confirmation.

Part b - oops, yes curl, which should be: $\frac{1}{\rho}\begin{vmatrix}
\hat{\rho}&\rho\hat{\phi}&\hat{z}\\ \partial_{\rho}&\partial_{\phi}&\partial_z \ \\0&\rho \omega \rho &0
\end{vmatrix}$ for $ \vec{v}=\hat{\phi}\omega \rho $

I get $ \nabla \times \vec{v} = \hat{z}2\omega $, the book again appears to have a typo - no $\hat{z}$, assume mine is OK? Thanks for the help.

Well, the Curl should always be a vector. If the book only asks for the magnitude of Curl the answer is $2\omega$ otherwise the vector component should be included. :)
 

Related to Simple cylindrical coords problem

1. What is a simple cylindrical coordinates problem?

A simple cylindrical coordinates problem is a mathematical problem that involves converting a point or a set of points from Cartesian coordinates to cylindrical coordinates. This is done by using the distance from the origin, the angle of rotation, and the height of the point from the x-y plane.

2. Why do we use cylindrical coordinates for certain problems?

Cylindrical coordinates are useful for problems that involve circular or cylindrical shapes, such as cylinders, cones, and spheres. They are also helpful for problems that involve rotational symmetry, as the angle of rotation remains constant while the point moves along the surface.

3. How do you convert from Cartesian coordinates to cylindrical coordinates?

To convert from Cartesian coordinates (x, y, z) to cylindrical coordinates (r, θ, z), we use the formula r = √(x² + y²), θ = tan⁻¹(y/x), and z = z. The first step is to find the distance from the origin (r) by using the Pythagorean theorem. Then, we find the angle of rotation (θ) by using inverse tangent. Finally, we keep z the same as it is in Cartesian coordinates.

4. Can you provide an example of a simple cylindrical coordinates problem?

One example of a simple cylindrical coordinates problem is finding the volume of a cylinder with a height of 5 units and a radius of 2 units. To solve this, we first convert the radius and height to cylindrical coordinates (2, 0, 5). Then, we use the formula V = πr²h to calculate the volume, which would be 20π cubic units.

5. What are the advantages of using cylindrical coordinates over Cartesian coordinates?

Cylindrical coordinates are advantageous because they simplify problems that involve cylindrical or circular shapes, as well as those with rotational symmetry. They also allow for easier visualization and representation of these shapes, which can aid in problem-solving and understanding. Additionally, cylindrical coordinates can be used to solve problems in three dimensions without needing to use a third variable.

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