# Help with rotation

1. Jun 16, 2007

### lets_resonate

Hello world,

I'm trying to understand the derivation of the centrifugal force equation, $$F_{centrifugal} = - m \omega \times \left( \omega \times r \right)$$. I used these pages to help me in my pursuit:

However, I lack the background to understand either of these articles. This last school year, I completed AP Calculus and first-year high school physics. I understood the material fairly well, but it does not appear to be enough to figure out the concepts discussed here. I will be taking AP Physics, multivariable calculus, and linear algebra next year.

So, off we go:

1.
Uhh... what? The derivative of what with respect to time? Omega times what? It appears this is some sort of a fill-in-the-blanks equation - just down below, they fill the blanks in with r. But can anyone explain the meaning of this equation? Why does it supposedly make sense?

By the way, the "times" symbol, $$\times$$, represents the cross product.

2.
It appears to me that to obtain $$a_i$$, they simply took the derivative of $$v_i$$, the equation for which is:

$$v_i = v_r + \omega \times r$$.

However, when I do it, I get a different result:

$$a_i = \frac {d} {dt} \left( v_i \right) = \frac {d} {dt} \left( v_r + \omega \times r \right) =a_r + \omega \times v_r + \alpha \times r$$

I simply differentiated the expression for $$v_i$$, using the product rule for the cross product. Alpha is angular acceleration, which is the derivative of angular speed, omega. What did I do wrong?

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For now, I'll be happy to just understand these 2 problems. I won't even begin about the Wikipedia article (any math-related Wikipedia article has a crushing effect on my self-esteem). If anyone can offer some help, I'll appreciate it highly.

Last edited: Jun 16, 2007
2. Jun 17, 2007

### cesiumfrog

1. is an equation of "operators". The idea is that it is true with any arbitrary function (say "f", "h", "r" or "$v_i$") placed on the right of all three terms. (At first such equations might make more sense if you always explicitly write in "f(x)" where appropriate, but the other notation is more efficient.)

2. $a_i$ is the rate of change (in the inertial frame) of the velocity (in the inertial frame). The problem is that $a_r$ is supposed to represent the rate of change in the rotating (not inertial) frame of the velocity in the rotating frame, so substitute your eq.(1) for the (inertial frame) differential operator you tried using in your last equation (and note that $\alpha$ = 0).

Last edited: Jun 17, 2007
3. Jun 17, 2007

### lets_resonate

Thank you for the help. I managed to derive the equation properly:

$$\left( \frac{d}{dt} \right) _i = \left( \frac{d}{dt} \right) _r + \omega \times$$

This "transformation" is first applied to r:
$$\left( \frac{dr}{dt} \right) _i = \left( \frac{dr}{dt} \right) _r + \omega \times r$$
$$v_i = v_r + \omega \times r$$

And then to $v_i$:
$$\left( \frac{d}{dt} \left( v_i \right) \right) _i = \left( \frac{d}{dt} \left( v_i \right) \right) _r + \omega \times v_i$$

Now substitute every $v_i$ on the right side of this equation with $v_r + \omega \times r$:
$$\left( \frac{d}{dt} \left( v_i \right) \right) _i = \left( \frac{d}{dt} \left( v_r + \omega \times r \right) \right) _r + \omega \times \left( v_r + \omega \times r \right)$$

After simplifying, I finally obtained the result I was looking for:
$$a_i = a_r + 2 \omega \times v_r + \omega \times \left( \omega \times r \right)$$

Phew!

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Well, everything makes sense, except for the very first step. I'm not sure how they obtained the original coordinate transformation in the first place:

$$\left( \frac{d}{dt} \right) _i = \left( \frac{d}{dt} \right) _r + \omega \times$$

I think I understand the concept of "equation of operators" now, but I don't understand why it is what it is. Why is it that a differential change (of something) in the inertial frame is equivalent to the same differential change (of the same something) in the rotating frame plus the angular velocity multiplied by that "something"?

Again, I would appreciate any help. Thank you in advance.

Last edited: Jun 17, 2007