Centripetal And Centrifugal Forces

The centrifugal force does not exist. Suppose we are in a car, travelling 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 travelling 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!

 

Does anyone wish to comment on this?
http://www.badastronomy.com/bablog/2006/08/30/when-i-say-centrifugal-i-mean-centrifugal/

 

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 [tex]\bold{F}=m\frac{d^2\bold{r}}{dt^2}[/tex] (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 [tex]\frac{d}{dt}=\left(\frac{d}{dt}\right)'+\bold{\omega}\times[/tex](2)

Differentiating the RHS of (1):[tex]\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}[/tex] (note that since ω is constant, the term involving its derivative vanishes).

So, we have Newton's second law in the rotating frame: [tex]\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][/tex], 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'.

 

The author is perfectly correct. Here's what I tell my students:

"To explain these apparent forces, consider Figure 27.8.3.a, which tracks a particle moving at a constant velocity with respect to an observer in a fixed basis. The time-displacement graph is a straight line. If the observer now moves at a constant velocity in the same direction as the particle (Figure 27.8.3.b), the measured velocity of the particle will be reduced but it will still appear to be constant. However, if the observer accelerates in the direction as the particle (Figure 27.8.3.c) then its apparent velocity will change. A change in velocity is an acceleration and to produce an acceleration requires a force. The accelerating observer thus perceives a force on the particle.

Inertia is an example of an apparent force and the non-inertial behaviour of two cars during a head-on collision is the reason why the kinetic energy calculated by a roadside bystander is different from what would be calculated by either of the drivers. Many authors disdain the use of inertial forces, including centrifugal force, regarding them as ‘fictitious’ but even diehards would agree that gravity is a ‘real’ force. However that disappears in the non-inertial situation called free-fall (incorrectly called ‘weightlessness’). This is the whole basis of general relativity."

You'll get the full discussion in Chapter 27, Section 8 at
http://moodle.gla.ac.uk/eng/moodle/course/view.php?id=40 .

...and try telling the occupants of a cornering car that it's not a 'real' force.