# Pseudo force in rotating frames

A bug of mass $m$ crawls out along a radial scratch of a phonographic disc rotating at $\omega$ angular velocity. It travels with constant velocity $v$ with respect to the disc. What are the forces (and their directions) which act on the bug?

I have tried thinking in this way: If I look from the point of view of the bug, that is the non-inertial reference frame, forces acting on the bug are friction, centrifugal force and the Coriolis force. Since the bug crawls outward radially, the friction should act radially inward and the centrifugal force acts radially outward. I am totally confused regarding the direction of the Coriolis force.

Does it act on the bug perpendicular to the radial direction?

If so, then while considering the effective force on the bug (from non-inertial point of view), should I break it into horizontal and vertical components? Or can they be written in one equation?

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I think it would be nice to derive the fictitious forces at least once and then use the definitions later on if you remember.

In this case in a rotating frame of reference the force is given as:

$$\frac{\vec F}{m} = \left( \frac{\operatorname{d}}{\operatorname{d}t} + \vec \omega \times \right) \left( \frac{\operatorname{d}}{\operatorname{d}t} + \vec \omega \times \right) \vec x \\ \vec{F} = m \left( \frac{\operatorname{d}}{\operatorname{d}t} + \vec \omega \times \right) \left( \vec v + \vec \omega \times \vec x \right) \\ = m \vec a + m \vec \omega \times \vec v + m \vec \omega \times \vec v + m\vec \omega \times \vec \omega \times \vec x \\ = m \vec a + m \vec \omega \times \vec \omega \times \vec x + 2 m \vec \omega \times \vec v$$

Where $\vec a$ is the acceleration in the non rotating frame of reference and $\vec v$ is velocity in the non rotating frame, whereas $\vec F$ is the experienced force in the rotating frame.

From the expression above you can distinguish the centrifugal force term and the Coriolis force term.

Just to ease our interpretation, let's use the vector tripple product formula to reexpress the centrifugal for term:

$$\vec \omega \times \vec \omega \times \vec x = \vec \omega \cdot \left( \vec \omega \cdot \vec x \right) - \vec x \cdot \left| \vec \omega \right|^2$$

Now the motion of the bug is radial and on the disc, which means that in the centrifugal force term will become:

$$\vec x \cdot \left| \vec \omega \right|^2$$

This means that the centrifugal force indeed acts towards the 'outside' of the disc and the Coriolis force will be acting perpendicular to the direction of motion. The friction force will be acting antiparallel to the direction of motion.

Hope that this long and mathematically rigorous way, will let you picture everything better.

Note: The final expression of the fictitious forces can be expressed in the following way as well:

$$m \vec a = F - m \vec \omega \times \vec \omega \times \vec x - 2 m \vec \omega \times \vec v$$

This can be interpreted that we are always interested in the acceleration in the non-rotating frame $\left(\vec a \right)$ and that we need to subtract from the force $\left( \vec F \right)$ all the fictitious forces we know about.

So this is not inconsistent with the Coriolis force definition given by joshphysics:

$$m \vec a = \vec F + \vec F_{centrifugal} + \vec F_{Coriolis}$$

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I like your answer except that you have assumed $m=1$ which is not always the case. Force is not always numerically equal to acceleration. –  ja72 Mar 11 '13 at 6:13
Yes, that was my fault, I corrected the mistake. I was very much conserned to distinguish the different frames, so that this typo skipped through my eyes. –  gns-ank Mar 11 '13 at 15:07
I have one clarification. The Coriolis force acts perpendicular to the radial direction and the centrifugal force acts radially outward. I can find a resultant of these two mutually perpendicular forces which is equal(only magnitude) to the friction for the bug. Also the friction does not act radially inward but at an angle with the radius(I got this idea because the bug will feel that it is travelling a circular path-non inertial point of view). Am I correct? –  Pallavi Roy Mar 11 '13 at 16:06
Well I do not think so, because the surface which causes friction force will move together with the bug, so as far as the bug is concerned, the friction will be only due to his velocity, which is radial in his frame. –  gns-ank Mar 11 '13 at 17:12
@gns-ank, I edited the answer and added $m$ factors in the expressions for force (the ones you forgot to edit). –  ja72 Mar 12 '13 at 13:19
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Note that the Coriolis force is given by $$\vec F_\mathrm{cor} = -2m\,\vec\omega\times \vec v$$ where here $\vec v$ is the velocity as measured in the rotating frame. Given that $\vec\omega$ points vertically upward and $\vec v$ points radially outward, you can determine the direction of the Coriolis force.

You will find that the net fictitious force on the bug (excluding normal force) points in the plane of the rotating disk with a radial component, and a tangential component.

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I was wondering, why you would not recommend deriving the fictitious forces in rotating frames? –  gns-ank Mar 10 '13 at 20:08
@gns-ank I removed that recommendation; I meant to say that going through such a derivation is unnecessary for solving this problem. –  joshphysics Mar 10 '13 at 21:13
Ah, I see. I completely agree. This simple problem does not require it. –  gns-ank Mar 10 '13 at 23:15

Yep. Coriolis force would be perpendicular. And to solve this problem, I would suggest:

that you make a co-ordinate system with center of disc as origin, with it's x and y components. Try to write x and y as functions of time, in a frame fixed to the co-orodinate system. You'll get an expression with trigonometric as well as linear functions. This is displacement of bug in vector form. Differentiate it twice to get acceleration. There'll be 2 parts. One of them will be acceleration due to centrifugal force and one due to coriolis force.

Do it and you'll find out.

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Consider an instantaneous coordinate system on the bug with its local $x$-axis along the radial direction and $y$-axis along the tangential direction. If the radial distance of the bug is $x$ and the disk is rotating with constant $\vec{\omega} = \hat{k} \Omega$, and accelerating with $\dot{\vec{\omega}}=\hat{k} \dot{\Omega}=0$.
The position vector is: $$\vec{r} = \hat{i} x$$ The velocity vector is its time derivative $$\dot{\vec{r}} = \dot{\hat{i}} x + \hat{i} \dot{x} = (\hat{k} \Omega) \times \hat{i} x + \hat{i} \dot{x} = (\Omega\,x)\,\hat{j} + v\, \hat{i}$$ with $v$ the radial velocity of the bug, and $\dot{v}=0$.
The acceleration vector is its time derivative $$\ddot{\vec{r}} = \frac{\rm d}{{\rm d}t} (\Omega\,x)\,\hat{j} + (\Omega\,x)\,\dot{\hat{j}} +\dot{v}\, \hat{i} + v\,\dot{\hat{i}} \\ = (\Omega\,v)\hat{j} + (\Omega\,x) (\Omega \hat{k} \times \hat{j}) + v (\Omega \hat{k} \times \hat{i}) \\ = (\text{-}x\, \Omega^2) \hat{i} + 2(\Omega\,v) \hat{j}$$
So now the force vector is found by $\vec{F} = m \ddot{\vec{r}}$
The radial force is thus $m\,(\text{-}x\, \Omega^2)$ and the tangential force $m 2(\Omega\,v)$.