Centripetal And Centrifugal Forces

• danago
In summary, Dan said that the centrifugal force does not exist, and that it is just the centripetal force acting on an object in a circular motion. He also said that the centrifugal force is simply the laws of inertia in action.f

danago

Gold Member
Hi. This isn't a homework problem, but I am asking for the sake of adding to my understanding of the topic.

In class, we have just begun uniform circular motion on the horizontal plane.

Now, i understand that for an object to be accelerating (changing direction in this case), there needs to be a net force acting on it, in the direction of the acceleration. When referring to circular motion, this is the centripetal force, which acts directly towards the center of the circle.

I have heard of a centrifugal force. I've searched it on the internet, and most of the things I've read talk about how it doesn't exist. Its a force that people assume to exist, when actually, it is just the laws of intertia kicking in.

Im a bit confused. What exactly is the centrifugal force?

Dan.

The centrifugal force does not exist. Suppose we are in a car, traveling round an arc of a circle to the left. Then, we will appear to feel a force pulling us to the right of the car. However, this is not a "real" force, but is simply our body wanting to continue traveling forward in a straight line.

This is one of those cases where searching on the internet may be dangerous, since there will undoubtably be incorrect information on websites out there written by someone who knows nothing about Physics!

The centrifugal force does not exist. Suppose we are in a car, traveling round an arc of a circle to the left. Then, we will appear to feel a force pulling us to the right of the car. However, this is not a "real" force, but is simply our body wanting to continue traveling forward in a straight line.

This is one of those cases where searching on the internet may be dangerous, since there will undoubtably be incorrect information on websites out there written by someone who knows nothing about Physics!

Thanks for that. Thats what most sources told me, and they also used the exact same example.

One website told me that centripetal/centrifugal were action-reaction force pairs. Guess that's an example of one of those inaccurate websites

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Hmmm so they are really the same thing then? Just depending on the frame of reference?

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Let's look at the mathematical situation. Consider two frames of reference S and S', with S' rotating with constant angular velocity ω. Let r denote the displacement vector of the particle in S, and r' the displacement in S'. Now, I'm sure you'll agree that we can use Newton's second law in frame S, to give us $$\bold{F}=m\frac{d^2\bold{r}}{dt^2}$$ (1)

Now, let's switch to the rotating frame. Using a general result, we have that the differential operators in S and S' are related by $$\frac{d}{dt}=\left(\frac{d}{dt}\right)'+\bold{\omega}\times$$(2)

Differentiating the RHS of (1):$$\frac{d^2\bold{r}}{dt^2}=\frac{d}{dt}\left(\frac{d\bold{r}}{dt}\right)'+\frac{d}{dt}(\bold{\omega}\times \bold{r})=\left(\frac{d^2\bold{r}}{dt^2}\right)'+\bold{\omega}\times\left(\frac{d\bold{r}}{dt}\right)'+\bold{\omega}\times\frac{d\bold{r}}{dt}$$ (note that since ω is constant, the term involving its derivative vanishes).

So, we have Newton's second law in the rotating frame: $$\bold{F}=m\left[\left(\frac{d^2\bold{r}}{dt^2}\right)'+2\bold{\omega}\times\left(\frac{d\bold{r}}{dt}\right)'+\bold{\omega}\times(\bold{\omega}\times{\bold{r}})\right]$$, where we have obtained the last term by applying (2) on the corresponding term in the above.

Now, the final term on the RHS is commonly called the "centrifugal force," however we note that it is not a force, but merley arises when converting the acceleration of the particle in S to the acceleration in S'.

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