# Calculus & velocity

1. Dec 17, 2007

### ranaroy

1. The problem statement, all variables and given/known data
omega_c = (nabla) cross (u)
where
omega_c is the rotational velocity of a particle
nabla refers to the gradient operator
u is the linear velocity of the particle

2. Relevant equations

what are the expressions of
omega_c_x =
omega_c_y =
i.e, in x- and y-cordinates only.

3. The attempt at a solution
is it ??
omega_c_x = du_dy - du_dx;
omega_c_y = dv_dy - dv_dy;

sorry i have tried alot..beats me..pls help.

2. Dec 17, 2007

### chickendude

If you want to find the velocity in the x and y directions, the problem is simply a matter of decomposing the linear velocity into x and y components.

Since you are given nabla (the gradient), you know which direction the linear velocity is pointing, so you should be able to split the linear velocity into components

3. Dec 17, 2007

### HallsofIvy

Staff Emeritus
The cross product is only defined in three dimensions- if you are working in the xy-plane you need to add "0" as the z component. You don't say what your understanding of the "cross product" is- there are several equivalent ways of calculating it. However one of the things you should know about the curl ($\nabla \cross v$, where v is in the xy-plane is perpendicular to the xy-plane! $\nabla \cross <u_x, u_y, 0>$ is
$<0, 0, d(u_y)/dx- d(u_x)/dy>$.

4. Dec 17, 2007

### what

Is the question resolve the x and y components of omega? If so then your answer is not correct, your x component for omega is close, and where did v come from?

First of all do you know how to take a determinant of a 3x3 matrix? If you do then the standard way to solve your problem without looking it up is to write a 3x3 matrix with the coordinate unit vectors as the first row and the first vector and second vector resolved into components as the 2nd and 3rd row respectively. Taking the determinant of this matrix gives you the resultant vector wrt your initial coordinate vectors. That’s all there is to it. If you aren't supposed to know how to take a determinant then just look up the formula for the curl of a vector.

btw, the gradient operator is called "del".

Edit:

I see now what the problem may have been, then you should focus on HallsofIvy's post; it should be more helpful to you.

Last edited: Dec 17, 2007
5. Dec 18, 2007

### ranaroy

hi all,

let me make my problem more clear.

my question is indeed to resolve the curl of u.
where, u = linear velocity resolved into x- and y- cartesian cordinates.
(del) cross (u) = curl(u) which measures a vector field's tendency to rotate about a point.
Now, let us represent this curl(u) by omega_c.

in a 3D system, this is simply the determinant of a 3x3 matrix i.e,
omega_c_x = dw_dy - dv_dz;
omega_c_y = du_dz - dw_dx;
omega_c_z = dv_dx - du_dy;

where du, dv, dw are nothing but u_x, u_y, u_z respectively.
omega_c = sqrt(omega_c_x*omega_c_x + omega_c_y*omega_c_y + omega_c_z*omega_c_z)

But for a 2D system, where i know only u_x and u_y, i am not able to understand how to build the omega_c_x and omega_c_y !!
If we go alongwith the expressions above, we dont have dw, dz terms.
so, dw_dy=0, dv_dz=0, du_dz=0, dw_dx=0..I hope I am correct..
then i am left with dv_dx and du_dy which have non-zero values.

in that case, what are the expresisons for omega_c_x and omega_c_y ??

pls help..

thanking you.

6. Dec 18, 2007

### HallsofIvy

Staff Emeritus
Did you see my previous post? The angular momentum vector is parallel to the axis of rotation. What direction is the axis of rotation for a rotation in the xy-plane?

7. Dec 18, 2007

### ranaroy

dear Mentor,

yes, i checked it now :)
so, the axis of rotation is z.

then i write omega_c_z=dv_dx - du_dy

so, i will not get two components ?
but why in 3D system, we get all 3 components of omega i.e., omega_c_x, omega_c_y, omega_c_z ?
sorry, if i sound foolish !

hope to get ur feedback

thanks a lot.

Last edited: Dec 18, 2007
8. Dec 18, 2007

### HallsofIvy

Staff Emeritus
What do you mean by "components"? < 0, 0, dv_dx- du_dy> HAS three components!

9. Dec 18, 2007

### ranaroy

extremely sorry..i meant that the final expression is
omega_c_x=0, omega_c_y=0, omega_c_z = dv_dx - du_dy
this means i have only i set of real value for omega_c in 2D, while in 3-D i have all real values for omega_c_x, omega_c_y, omega_c_z
am i right ?

10. Dec 18, 2007

### HallsofIvy

Staff Emeritus
My point was that "0" is a "real value"! And it doesn't really have anything to do with "3d" versus "2d". If you have something rotating in such a way that it's plane of rotation was parallel to the xz-plane, then its "rotational velocity vector" would be of the form < 0, $\omega$, 0>, since it would be pointing parallel to the y axis.

It is true that if you have rotation "in the plane" (i.e. xy-plane), then you KNOW that the angular velocity vector points in the z-direction, so if you are working in 2d, you don't need to mention that- you just give it length, the scalar angular velocity. If you are working in 3d, you can't be sure ahead of time in which direction the angular velocity vector points so you need to calculate both magnitude and direction: three components. Of course, after you have calculated it, it might happen that one or two of those components is 0.