Vector product question in cylindrical coordinates

In summary: Yes, there might be a mistake in the calculation. If ω is expressed in terms of the ψ and z-components instead of the ψ and r-components, then you would get the incorrect result of v=[0,0,ω].
  • #1
GarethB
15
0
I am trying to work the following problem;
A rigid body is rotating about a fixed axis with a constant angular velocity ω. Take ω to lie entirely on th z-axis. Express r in cylindrical coordinates, and calculate;
a) v=ω × r
b)∇ × v
The answer to (a) is v=ψωρ and (b) is ∇ × v = 2ω

Firstly, I am not sure if we need r=[ρcosψ, psinψ, z] or simply [p, ψ, z]. Secondly, are they saying that ω is simply [0, 0, ω]? if so then I don't get the result so I am doing something wrong. Any help? I am sure that tackling this problem will deepen my understanding of curvilinear coordinates.
 
Physics news on Phys.org
  • #2
the first expression ist good ... r=[ρcosψ, psinψ, z] ...keep going with that ;)
next describe omega as a vector ... and calc. velocity with cross product :>
 
  • #3
GarethB said:
I am trying to work the following problem;
A rigid body is rotating about a fixed axis with a constant angular velocity ω. Take ω to lie entirely on th z-axis. Express r in cylindrical coordinates, and calculate;
a) v=ω × r
b)∇ × v
The answer to (a) is v=ψωρ and (b) is ∇ × v = 2ω

Firstly, I am not sure if we need r=[ρcosψ, ρsinψ, z] or simply [ρ, ψ, z]. Secondly, are they saying that ω is simply [0, 0, ω]? if so then I don't get the result so I am doing something wrong. Any help? I am sure that tackling this problem will deepen my understanding of curvilinear coordinates.

I think the answer to (a) should be: ##\boldsymbol{\vec v}={\boldsymbol{\hat \psi}}\omega\rho##, where ##\boldsymbol{\hat \psi}## is the local unit vector in the ψ direction.

Similarly the answer to (b) should be: ##\nabla \times \boldsymbol{\vec v}=2 \boldsymbol{\vec \omega}##, where ##\boldsymbol{\vec \omega}## is the vector in the z direction with length ##\omega##.

I expect that they want you to express r as [ρcosψ, ρsinψ, z], so you can calculate the curl in cartesian coordinates.
(Alternatively, you could simply use r=[ρ, ψ, z] and use the relevant formula from: http://en.wikipedia.org/wiki/Del_in_cylindrical_and_spherical_coordinates)

And yes, they are saying that ##\boldsymbol{\vec \omega}## is simply [0, 0, ω].

You say you didn't get the same result they do, so what did you try?
 
  • #4
I get the answer you describe for part (a) (that v=[0,ρω,0] which seems the obvious answer) if I use r=[ρ,0,0] not with r=[pcosψ,ρsinψ,0] is ω=[0,0,ω] unless I am really dumb and can't comput vector products (which is likely the problem b.t.w!) I am guessing that if I use the cartesian representation for r then ω has to be re-expressed in some way to resolve this.
The answer that v=[0,ρω,0] seems to give a curl of 2ω in one of the components I can't remember which (making use of the cylindrical coordinate version of curl).
Regarding the problem for (a) I guess there is a typo in the textbook?
 
  • #5
GarethB said:
I get the answer you describe for part (a) (that v=[0,ρω,0] which seems the obvious answer)
Regarding the problem for (a) I guess there is a typo in the textbook?

Not a typo, but I'm missing vector notations for things that are vectors and, in particularly, unit vectors.
if I use r=[ρ,0,0] not with r=[pcosψ,ρsinψ,0] is ω=[0,0,ω] unless I am really dumb and can't comput vector products (which is likely the problem b.t.w!) I am guessing that if I use the cartesian representation for r then ω has to be re-expressed in some way to resolve this.

The challenge would be in the curl in cylindrical coordinates.
Can you express v in cartesian coordinates (x,y,z)?
Then take the curl, followed by a conversion to cylindrical coordinates?
The answer that v=[0,ρω,0] seems to give a curl of 2ω in one of the components I can't remember which (making use of the cylindrical coordinate version of curl).

The curl of 2ω is in the z-direction, which is the same direction as ω points.
That is: ##2ω\boldsymbol{\hat z}=2\boldsymbol{\vec \omega}##, where ##\boldsymbol{\hat z}## is the unit vector in the z-direction.
 
Last edited:
  • #6
Can be deleted
 
Last edited:
  • #7
So I know this problem is old, but it is cut & paste one of my HW problems with which I am having a lot of trouble. If I need to start a new thread I can, my apologies if I've broken a rule, I looked and didn't see one.

Anyways,

a) I have been letting:

[itex]\vec{r}=[\rho*cos(\psi),\rho*sin(\psi),0] [/itex]

[itex]\vec{\omega}=[0,0,\omega][/itex]

Then I get [itex]\vec{v}=\vec{\omega}\times\vec{r} [/itex] [itex]=[-\rho\omega*sin(\psi),\rho\omega*cos(\psi),0] [/itex]

If I have performed this correctly, then I don't see how I get to the asnswer: [itex][0,\omega\rho,0][/itex]?

If [itex]\psi=0[/itex] then I get the answer, but I think that is just a coincidence. Does anyone see anything wrong with my logic.
 

Related to Vector product question in cylindrical coordinates

1. What is the vector product in cylindrical coordinates?

The vector product, also known as the cross product, is a mathematical operation that takes two vectors as input and produces a third vector that is perpendicular to both input vectors. In cylindrical coordinates, the vector product is given by the following formula:

A x B = (AρBz - AzBρ)ρ̂ + (AzBφ - AφBz)φ̂ + (AφBρ - AρBφ)ẑ

where A and B are the input vectors and ρ̂, φ̂, and ẑ are unit vectors in the ρ, φ, and z directions, respectively.

2. How do you find the magnitude of a vector product in cylindrical coordinates?

The magnitude of a vector product in cylindrical coordinates can be found using the following formula:

|A x B| = √((AρBz - AzBρ)2 + (AzBφ - AφBz)2 + (AφBρ - AρBφ)2)

This gives the magnitude of the resulting vector from the vector product.

3. Can the vector product in cylindrical coordinates be negative?

Yes, the vector product in cylindrical coordinates can be negative. The sign of the vector product is determined by the right-hand rule. If the resulting vector points in the opposite direction of the right-hand thumb, then the sign is negative. If the resulting vector points in the same direction as the right-hand thumb, then the sign is positive.

4. How do you calculate the direction of the resulting vector in cylindrical coordinates?

The direction of the resulting vector from the vector product in cylindrical coordinates can be determined using the right-hand rule. Place your right hand on the two input vectors with your thumb pointing in the direction of the first vector and your fingers pointing in the direction of the second vector. The resulting vector will point in the direction of your right-hand palm.

5. What are some real-world applications of the vector product in cylindrical coordinates?

The vector product in cylindrical coordinates has various real-world applications, such as in engineering, physics, and robotics. Some examples include calculating the torque on a rotating object, determining the direction of magnetic fields, and controlling the movement of robotic arms. It is also used in fluid dynamics to calculate the vorticity of a fluid flow.

Similar threads

  • Calculus and Beyond Homework Help
Replies
4
Views
2K
  • Calculus and Beyond Homework Help
Replies
2
Views
672
  • Calculus and Beyond Homework Help
Replies
12
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
2K
  • Calculus and Beyond Homework Help
Replies
5
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
2K
  • Differential Equations
Replies
1
Views
425
  • Calculus and Beyond Homework Help
Replies
17
Views
4K
  • Calculus and Beyond Homework Help
Replies
7
Views
1K
Back
Top