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Does centrifugal force exist?

For several weeks now I have been planning to write a blog about centrifugal force, mainly prompted by seeing a post by John Gribbin on Facebook of the xkcd cartoon about it. In the cartoon James Bond is threatened with torture on a centrifuge. Here is a link to the original cartoon.

The xkcd cartoon about centrifugal force involves James Bond being tortured on a centrifuge

I have taught mechanics many times to physics undergraduates, and they are often confused about centripetal force and centrifugal force, and what the difference is between them. Some have heard that centrifugal force doesn’t really exist, just as Bond states in this cartoon. What is the real story?

Rotating frames of reference

Everyone reading this (apart from a few “flat-Earth adherents” maybe) knows that we live on the surface of a planet which is rotating on its axis once a day. This means that we do not live in an inertial frame of reference (an inertial frame is one which is not accelerating), as clearly being on the surface of a spinning planet means that we are experiencing acceleration all the time; as we are not travelling in a straight line. That acceleration is provided by the force of gravity, and it stops us from going off in a straight line into space!

Because we are living in a non-inertial frame of reference we need to modify Newton’s laws of motion to properly describe such a non-inertial frame (which I am going to call a “rotating frame” from now on, although a rotating frame is just one example of a non-inertial frame but it is the one relevant to us on the surface of a rotating Earth).

Let us consider our usual Cartesian coordinate system. The unit vector in the x-direction is usually written as $\hat{\imath}$, the one in the y-direction as $\hat{\jmath}$, and the one in the z-direction as $\hat{k}$. We are going to consider an object rotating about the $\hat{k}$ (z-axis) direction.

We will consider two reference frames, one which stays fixed (the inertial reference frame), denoted by $(\hat{\imath},\hat{\jmath},\hat{k})$, and a second reference frame which rotates with the rotation, denoted by $(\hat{\imath}_{r} ,\hat{\jmath}_{r} ,\hat{k}_{r})$, where the subscript $r$ reminds us that this is the rotating frame of reference.

For the derivation below I am going to assume that we are considering motion with a constant radius $r$. I want to illustrate how centrifugal force arrises in a rotating frame such as being on the surface of our Earth. Our Earth is not spherical, but at any given point the size of the radius does not change, so this is a reasonable simplification.

As I showed in this blog on angular velocity, we can write the linear velocity $\vec{v}$ of an object moving in a circle as

$\vec{v} = \frac{ d \vec{r} }{ dt } = \vec{\omega} \times \vec{r}$

where $\vec{r}$ is the radius vector and $\vec{\omega}$ is the angular velocity.

Writing $\vec{r}$ in terms of its x,y and z-components in our inertial (non-rotating) frame, $\vec{r}=(\hat{\imath},\hat{\jmath},\hat{k})$, so in general we then have

$\vec{v} = \frac{ d \vec{r} }{ dt } \rightarrow \frac{ d \hat{\imath} }{dt} = \vec{\omega} \times \hat{\imath}, \; \; \frac{ d \hat{\jmath} }{dt} = \vec{\omega} \times \hat{\jmath} , \; \; \frac{ d \hat{k} }{dt} = \vec{\omega} \times \hat{k}$

Let us consider the specific case of a small rotation $\delta \theta$ about the $\hat{k}$ axis, as shown in the figure below. As the figure shows, in our inertial (fixed) frame of reference, the new direction of the x-axis is now $\hat{\imath} + \delta \hat{\imath}$, and the new direction of the y-axis is $\hat{\jmath} + \delta \hat{\jmath}$. The direction of the $\hat{k}$ axis is unchanged.

We are going to rotate about the z-axis ($\hat{k}$ direction) by an angle $\delta \theta$

Because we are rotating about the $\hat{k}$ axis, the angular velocity is in this direction, and so we can write (using the right-hand rule for vector products as I blogged about here)

$\vec{\omega} \times \hat{\imath} = \omega \hat{\jmath}, \; \; \vec{\omega} \times \hat{\jmath} = -\omega \hat{\imath}, \; \; \vec{\omega} \times \hat{k} =0$

Let us now consider some vector $\vec{a}$, which we will write in the rotating frame of reference as

$\vec{a} = a_{x} \hat{\imath}_{r} + a_{y} \hat{\jmath}_{r} + a_{z} \hat{k}_{r}$

If we now look at the rate of change of this vector in the rotating frame we have

$\left( \frac{d \vec{a} }{dt} \right)_{r} = \frac{d}{dt}(a_{x}\hat{\imath}_{r}) + \frac{d}{dt}(a_{y}\hat{\jmath}_{r}) + \frac{d}{dt}(a_{z}\hat{k}_{r})$

In the rotating frame of reference, $\hat{\imath}_{r}, \hat{\jmath}_{r}$ and $\hat{k}_{r}$ do not change with time, so we can write

$\left( \frac{d \vec{a} }{dt} \right)_{r} = \frac{ da_{x} }{dt} \hat{\imath}_{r} + \frac{ da_{y} }{dt} \hat{\jmath}_{r} + \frac{ da_{z} }{dt} \hat{k}_{r}$

In the inertial frame of reference $\hat{\imath}_{r}, \hat{\jmath}_{r}$ and $\hat{k}_{r}$ move, so

$\left( \frac{d \vec{a} }{dt} \right)_{i} = \frac{d}{dt} (a_{x} \hat{\imath}_{r}) + \frac{d}{dt} (a_{y} \hat{\jmath}_{r}) + \frac{d}{dt} (a_{z} \hat{k}_{r})$

$\left( \frac{d \vec{a} }{dt} \right)_{i} = \frac{ da_{x} }{dt}\hat{\imath}_{r} + \frac{ da_{y} }{dt}\hat{\jmath}_{r} + \frac{ da_{z} }{dt}\hat{k}_{r} + a_{x} \frac{d \hat{\imath}_{r} }{dt} + a_{y} \frac{d \hat{\jmath}_{r} }{dt} + a_{z} \frac{d \hat{k}_{r} }{dt}$

But, we can write (see above) that

$\frac{d\hat{\imath}_{r} }{dt} = \vec{\omega} \times \hat{\imath}_{r}, \; \; \frac{d\hat{\jmath}_{r} }{dt} = \vec{\omega} \times \hat{\jmath}_{r}, \; \; \frac{d\hat{k}_{r} }{dt} = \vec{\omega} \times \hat{k}_{r}$

and so

$\left( \frac{d \vec{a} }{dt} \right)_{i} = \frac{ da_{x} }{dt}\hat{\imath}_{r} + \frac{ da_{y} }{dt}\hat{\jmath}_{r} + \frac{ da_{z} }{dt}\hat{k}_{r} + a_{x} \vec{\omega} \times \hat{\imath}_{r} + + a_{y} \vec{\omega} \times \hat{\jmath}_{r} + + a_{z} \vec{\omega} \times \hat{k}_{r}$

$\left( \frac{d \vec{a} }{dt} \right)_{i} = \frac{ da_{x} }{dt}\hat{\imath}_{r} + \frac{ da_{y} }{dt}\hat{\jmath}_{r} + \frac{ da_{z} }{dt}\hat{k}_{r} + \vec{\omega} \times \vec{a}$

$\boxed{ \left( \frac{d \vec{a} }{dt} \right)_{i} = \left( \frac{d \vec{a} }{dt} \right)_{r} + (\vec{\omega} \times \vec{a}) }$

A fixed point on the Earth’s surface

Let us now consider the point $\vec{a} = \vec{r}$, where $\vec{r}$ is a fixed point on the Earth’s surface. We can write

$\left( \frac{d \vec{r} }{dt} \right)_{i} = \left( \frac{d \vec{r} }{dt} \right)_{r} + (\vec{\omega} \times \vec{r})$

But, in the rotating frame of reference this point does not change with time, so

$\left( \frac{d \vec{r} }{dt} \right)_{r} = 0$

and so

$\left( \frac{d \vec{r} }{dt} \right)_{i} = (\vec{\omega} \times \vec{r}) = \omega r \sin(\theta)$

where $\theta$ is the angle between the Earth’s rotation axis and the latitude of the point (so $\theta = 90^{\circ} - \text{ latitude}$).

Let us now calculate the acceleration in an inertial frame in terms of acceleration in a rotating frame. Writing $\vec{a}$ as $\vec{r}$ as above, we now have

$\left( \frac{d \vec{r} }{dt} \right)_{i} = \left( \frac{d \vec{r} }{dt} \right)_{r} + (\vec{\omega} \times \vec{r})$

To make things easier to write, we will re-write

$\left( \frac{d \vec{r} }{dt} \right)_{i} = \frac{d \vec{r}_{i} }{dt} \text{ and } \left( \frac{d \vec{r} }{dt} \right)_{r} = \frac{d \vec{r}_{r} }{dt}$

so

$\frac{d \vec{r}_{i} }{dt} = \frac{d \vec{r}_{r} }{dt} + (\vec{\omega} \times \vec{r})$

$\vec{v}_{i} = \vec{v}_{r} + (\vec{\omega} \times \vec{r})$

If we now differentiate $\vec{v}_{i}$ with respect to time, we will have the acceleration in the inertial frame

$\left( \frac{d \vec{v}_{i} }{dt} \right)_{i} = \left( \frac{d \vec{v}_{i} }{dt} \right)_{r} + (\vec{\omega} \times \vec{v}_{i})$

But, $\vec{v}_{i} = \vec{v}_{r} + (\vec{\omega} \times \vec{r})$

so

$\left( \frac{d \vec{v}_{i} }{dt} \right)_{i} = \frac{d}{dt}(\vec{v}_{r} + \vec{\omega} \times \vec{r})_{r} + \vec{\omega} \times (\vec{v}_{r} + \vec{\omega} \times \vec{r})$

Expanding this out we get

$\left( \frac{d \vec{v}_{i} }{dt} \right)_{i} = \left( \frac{d \vec{v}_{r} }{dt} \right)_{r} + \frac{d}{dt}(\vec{\omega} \times \vec{r}_{r}) + \vec{\omega} \times \vec{v}_{r} + \vec{\omega} \times (\vec{\omega} \times \vec{r}_{r})$

$\vec{a}_{i} = \vec{a}_{r} + 2\vec{\omega} \times \vec{v}_{r} + \vec{\omega} \times (\vec{\omega} \times \vec{r}_{r})$
Multiplying the acceleration by the mass $m$ to get a force

$m\vec{a}_{i} = m\vec{a}_{r} + 2m\vec{\omega} \times \vec{v}_{r} + m\vec{\omega} \times (\vec{\omega} \times \vec{r}_{r})$

So, writing the force in the rotating frame in terms of the force in the inertial frame, we have

$\boxed{ m\vec{a}_{r} = m\vec{a}_{i} - 2m\vec{\omega} \times \vec{v}_{r} - m\vec{\omega} \times (\vec{\omega} \times \vec{r}_{r}) }$

So,

$\boxed{\vec{F}_{r} = \vec{F}_{i} - 2m\vec{\omega} \times \vec{v}_{r} - m\vec{\omega} \times (\vec{\omega} \times \vec{r}_{r}) }$

Notice that there are two extra terms (Term A and Term B) in the equation on the right, I have highlighted them below.

If we compare the force in a rotating frame to an inertial frame, two extra terms (Term A and Term B) arise. Term A is the Coriolis force, Term B is the centrifugal force

Term A is what we call the Coriolis force, which depends on the velocity in the rotating frame $v_{r}$. It is the force which causes water going down a plughole to rotate about the hole and to move anti-clockwise in the northern hemisphere and clockwise in the southern hemisphere. It is also the force which determines the direction of rotation of low pressure systems in the atmosphere. I will discuss the coriolis force more in a future blog.

Term B is the centrifugal force, the force we were aiming to derive in this blogpost. The strength of the centrifugal force depends on the position of the object in the rotating frame – $r_{r}$.

What is the direction of the centrifugal force

The direction of $(\vec{\omega} \times \vec{r}_{r})$ can be found using the right-hand rule for the vector product, which I blogged about here. Remembering that the direction of $\vec{r}_{r}$\$ is radially outwards from the centre of the Earth, and the direction of $\vec{\omega}$ is the direction of the Earth’s axis (pointing north), then the direction of $\vec{\omega} \times \vec{r}_{r}$ is towards the east (right if looking at the Earth with the North pole up).

We now need to take the vector produce of $\vec{\omega}$ with a vector in this eastwards direction, and again using the right-hand rule gives us that the direction of $(\vec{\omega} \times \vec{r}_{r})$ is outwards (not radially from the centre of the Earth, but at right angles to the axis of the Earth). But, notice the centrifugal force has a minus sign in front of it, so the direction of the centrifugal force is outwards, away and at right angles to the Earth’s axis.

The direction of the centrifugal force is away from the axis of rotation, as shown in this diagram

This means that it acts to reduce the force of gravity which keeps us on the Earth’s surface. It also depends on the angle between where you are and the Earth’s axis, so is greatest at the equator and goes to zero at the pole. It means that you will weight slightly less than if the Earth were not rotating, but the effect is quite small and you would not notice such a difference going from the pole to the equator.

What is the strength of the centrifugal acceleration due to Earth’s rotation?

Let us calculate the centrifugal force at the Earth’s equator, where it is at its greatest.

At the equator, we can write that the centrifugal acceleration has a value of

$\omega^{2} r \text{ as } \theta = 90^{\circ}$

We can calculate $\omega$ for the Earth by remembering that it takes 24 hours to rotate once, and $\omega$ is related to the period $T$ of rotation via

$\omega = \frac{2 \pi }{ T}$

We need to convert the period $T$ to seconds, so $T = 24 \times 60 \times 60 = 86400 \; s$. This gives that

$\omega = 7.272 \times 10^{-5} \text{ rad/s }$

If we take the Earth’s radius to be 6,378.1 km (this is the radius at the equator), then we have that

$\omega^{2} r = 0.0337 \text{ m/s/s}$

Compare this to the acceleration due to gravity which pulls us towards the Earth’s surface, which is 9.81 m/s/s and we can see that the centrifugal force at its greatest is only $0.34 \%$ of the acceleration due to gravity. Tiny.

It is, however, noticeable when you are on a roundabout, and is used on fairground rides where you spin inside a drum and the floor moves away leaving you pinned to the wall of the drum. The force you feel pushing against this wall is the centrifugal force, and it is very real for you in that rotating frame!

So, there we have it, centrifugal force does exist in a rotating frame of reference, but does not exist from the perspective of someone in an inertial frame of reference.

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