Help with Derivation of Centrifugal Force Equation

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In summary, the conversation discusses the derivation of the centrifugal force equation and the confusion surrounding the coordinate transformation involved. The NASA article presents an equation of operators, which can be applied to any arbitrary function, and the second application of this transformation to obtain the expression for acceleration in the rotating reference frame. The confusion arises when trying to derive the equation, but it is eventually resolved by substituting the expressions for velocity in the rotating frame. The final result is the desired centrifugal force equation. However, the question still remains about the origin and justification of the coordinate transformation in the first place.
  • #1
lets_resonate
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Hello world,

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

  • http://observe.arc.nasa.gov/nasa/space/centrifugal/centrifugal6.html"
  • http://en.wikipedia.org/wiki/Rotating_reference_frame"

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.
From the NASA article

We wish to express Newton's second law in a reference frame that rotates uniformly and with angular velocity w (units of radians per second) relative to our inertial reference frame. This is accomplished by applying the coordinate transformation

[tex]\left( \frac {d} {dt} \right) _{i} = \left( \frac {d} {dt} \right) _{r} + \omega \times[/tex]

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, [tex]\times[/tex], represents the cross product.

2.
From the NASA article again

Upon applying the coordinate transformation a second time, to [tex]v_i[/tex], we obtain an expression for acceleration [tex]a_i[/tex] in the rotating reference frame:

[tex]a_i = a_r + 2 \omega \times v_r + w \times \left( \omega \times r \right)[/tex]

It appears to me that to obtain [tex]a_i[/tex], they simply took the derivative of [tex]v_i[/tex], the equation for which is:

[tex]v_i = v_r + \omega \times r[/tex].

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

[tex]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[/tex]

I simply differentiated the expression for [tex]v_i[/tex], 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.
 
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  • #2
1. is an equation of "operators". The idea is that it is true with any arbitrary function (say "f", "h", "r" or "[itex]v_i[/itex]") 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. [itex]a_i[/itex] is the rate of change (in the inertial frame) of the velocity (in the inertial frame). The problem is that [itex]a_r[/itex] 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 [itex]\alpha[/itex] = 0).
 
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  • #3
Thank you for the help. I managed to derive the equation properly:

[tex]\left( \frac{d}{dt} \right) _i = \left( \frac{d}{dt} \right) _r + \omega \times[/tex]

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

And then to [itex]v_i[/itex]:
[tex]\left( \frac{d}{dt} \left( v_i \right) \right) _i = \left( \frac{d}{dt} \left( v_i \right) \right) _r + \omega \times v_i[/tex]

Now substitute every [itex]v_i[/itex] on the right side of this equation with [itex]v_r + \omega \times r[/itex]:
[tex]\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)[/tex]

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

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:

[tex]\left( \frac{d}{dt} \right) _i = \left( \frac{d}{dt} \right) _r + \omega \times[/tex]

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.
 
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FAQ: Help with Derivation of Centrifugal Force Equation

What is the derivation of the centrifugal force equation?

The derivation of the centrifugal force equation involves using Newton's second law of motion and the concept of circular motion to calculate the magnitude of the centrifugal force acting on an object in circular motion. The equation is Fc = mv^2/r, where Fc is the centrifugal force, m is the mass of the object, v is the velocity, and r is the radius of the circular motion.

How does centrifugal force affect objects in circular motion?

Centrifugal force is a fictitious force that appears to act on objects in circular motion. It is directed away from the center of the circle and is proportional to the mass and speed of the object, as well as the radius of the circle. This force causes objects to move away from the center of the circle and maintain their circular path.

What are some real-life examples of centrifugal force?

Some common examples of centrifugal force include the feeling of being pushed outward when turning a car or riding a rollercoaster, the separation of cream from milk in a centrifuge, and the rotation of a washing machine drum.

Is centrifugal force the same as centripetal force?

No, centrifugal force and centripetal force are two different forces. Centripetal force is the actual force that acts towards the center of a circle to keep an object in circular motion, while centrifugal force is a fictitious force that appears to act outward on objects in circular motion.

Can the centrifugal force equation be applied to objects in non-circular motion?

No, the centrifugal force equation is only applicable to objects in circular motion. For objects in non-circular motion, the concept of centrifugal force does not apply and other equations must be used to calculate the forces acting on the object.

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