# Circular Force Conversions

1. Apr 9, 2008

### Lxhook

How would I go about the following:

A set of forces are pushing an object in a circle. These forces are recorded in the x and y direction in a 2D Cartesian coordinate system. I would like to convert the x and y force components to radial and tangential components.

Is this possible? If so, how?

All of the information I can find deals with using particle kinematics in such a conversion but I want a kinetic conversion.
Thanks!

2. Apr 9, 2008

### Lxhook

Sorry if the above question seems like a homework problem or trivially possible/impossible. I study biomechanics and human motor control and and using the conversion to study central nervous system control strategies in handwriting. It's been awhile since I've looked at this sort of thing, so any help is much appreciated.

3. Apr 15, 2008

### WhiteFox

Say $$f_x$$ and $$f_y$$ are your force components in cartesian coordinates (along x and y respectively), $$f_r$$ is the radial component of the force and $$f_t$$ is the tangential component of the force. And say that $$\theta$$ is the angle of the line segment between the center of the circle and the point where the force is applied (see attached diagram). (the angle is measured from the X axis (horizontal))

Your problem consists simply in expressing your force vector in another reference frame, which can be done by multiplying it by a rotation matrix. Using the definitions given above (and in the diagram), you have the equations:
$$\left(\begin{array}{c} f_r \\ f_t \\ \end{array} \right) = \left( \begin{array}{cc} cos \theta & sin \theta \\ -sin \theta & cos \theta \\ \end{array} \right) \left(\begin{array}{c} f_x \\ f_y \\ \end{array} \right)$$
and, inversly,
$$\left(\begin{array}{c} f_x \\ f_y \\ \end{array} \right) = \left( \begin{array}{cc} cos \theta & -sin \theta \\ sin \theta & cos \theta \\ \end{array} \right) \left(\begin{array}{c} f_r \\ f_t \\ \end{array} \right)$$

I hope this helps!

If you want to know more about rotation matrices, I suggest taking a look at Wolfram's page on the subject (mathworld.wolfram.com/RotationMatrix.html), which is presented in a more elegant fashion than its Wikipedia counterpart (en.wikipedia.org/wiki/Rotation_matrix).
(Sorry, apparently I'm too new here to be allowed to put direct URL links in my replies )

#### Attached Files:

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Last edited: Apr 15, 2008
4. Apr 15, 2008

### Lxhook

Thanks. I was way over thinking it. I appreciate your help a lot!

5. May 18, 2008

### jayakkumar

Hi, I have some problem in transformation, if the point is lying on the r=0 axis. (i.e. Z axis), can any one have any idea for this?

6. May 18, 2008

### pzlded

The magnitude F of the centripetal force is equal to the mass m of the body times its velocity squared v$$^{2}$$ divided by the radius r of its path: F=mv$$^{2}$$/r.

The x and y components of centripetal force within an X/Y plane are:

Fx$$^{2}$$ + Fy$$^{2}$$ = centripetal force$$^{2}$$

7. May 18, 2008

### pzlded

Perhaps by "set of forces are pushing an object in a circle," you are looking for torque. Foot-pounds of torque is the number of pounds force applied to the end of a handle times the length of the handle.

Torque is a tangential force with no radial component, centripetal force is a radial force with no tangential component.

The influence of a force that has both a tangential and a radial component depends on the details of the system. For example: If only circular motion is possible, application of radial force will not influence angular velocity of the target object.

8. May 18, 2008

### WhiteFox

I'm not certain I understand your question. Maybe you could explain the context of your problem.

Nonetheless, let me just say that expressing a force in terms of radial and tangential components only makes sense with respect to a specific point (i.e. the center of rotation). In other words, radial and tangential components of an applied force are only relevant in the contexte of a rotation around a given point.

However, if a force is applied AT the center of rotation, it will not induce any rotation. This is why I do not understand why you would want to express such a force in radial/tangential components.

9. May 19, 2008

### jayakkumar

Thanks for the response.

Let me explain my question in detail.

My question about transformation is not for force components but displacements and stresses at a point in cylindrical coordinate frame.

I have displacement and stress tensors in Cartesian coordinate frames (XYZ) for some points and I would like to transfrom those data in Cylindrical coordinate frame (RTZ). I used the same formula given above for force transformation for R and Theta components and z component is same as cartesian Z component.

Here the formula has the cosine and sine theta, where the points lying in the z axis, singularity occurs in finding the RTZ of a point in Cylindrical Coordinate from Cartesian coordinate XYZ. I used the following formula to convert the XYZ to RTZ. R = SQRT (x*x + y*y) and theta = tan-1(x/y) and Z=Z.

Here I have problem with the points lying on Z axis where R=0 and theta is indeterminate.

Can anyone help me.?

10. May 19, 2008

### WhiteFox

Ok, this clarifies a few things.

First, for the sake of clarity for anyone who may be reading this, I'd like to point out that my original answer (and the original question) did not involve polar coordinates (which describe a point in terms of angle 'theta' and arc length 'r'). It was rather a matter of rotating cartesian coordinates for a given angle 'theta'.

Now, on the subject of converting from XYZ to RTZ, there is in fact an indetermination for 'theta' at R = 0, and unfortunatly I do not have a magic answer.

I am by no means an expert on the subject and I have never worked with displacement and stress tensors (therefore I can't really grasp the context of your problem), but if it can help at least a bit, my advices would be:
1 - If this is for an analytical problem, and if you can work in cartesian coordinates (or generalized coordinates (not an expert at that either) or another singularity-free coordinate system) it might allow you to have 'cleaner' equations, albeit possibly more complex;
2 - If you are working on an algorithm, you probably will have to resort to a 'hack' for the R=0 case (e.g. if 'theta' is of little importance, set it to 0 or to a random value)

You might also want to check the litterature on your subject (although you probably already did). Chances are someone faced the same problem and found a way around.

Finally, you can always hope that someone more knowledgeable will drop by! ;)

11. May 20, 2008

### jayakkumar

Thanks for the suggestion White Box. I'll study the literature and get it clarified. Thanks again for responsible suggestion.

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