In the following question:

A puck is placed on a friction less disk that is rotating with angular velocity $\vec \omega$. What is the equation of motion with respect to the rotating frame of the puck?

The contradiction that I can not resolve is that it seems really obvious that the puck will rotate with angular velocity $ - \vec \omega $ in the frame of the disk. But if look at the puck from the frame of reference of the disk, we can see that there is a pseudo centrifugal force acting on the puck. This is the only force that acts on the puck in this frame.

Hence, the puck should accelerate radially outward in this frame...but this is an exact opposite to the expected answer.


2 Answers 2


The centrifugal force is not the only one which acts on the puck in the rotating frame, there is also the Coriolis force. The sum of the two is given by $$ \vec{F} = -m \vec{\omega} \times (\vec{\omega}\times\vec{r})-2m\vec{\omega}\times\vec{v} $$ but in the rotating frame the velocity of the puck is $$ \vec{v}=-\vec{\omega}\times\vec{r} $$ and substituting we find $$ \vec{F} = m \vec{\omega} \times (\vec{\omega}\times\vec{r}) $$ which is equal and opposite to the centrifugal force, and is exactly the force needed for the observed circular motion of constant speed.


I apologize, but the question is not quite clear to me. Let me take a stab nonetheless with my interpretation of it:

I'm assuming the question goes like this: a puck is dropped onto a spinning (frictionless) disc and we'd like to know the position of the puck with respect to an arbitrary point on the disc as a function of time.

Let's define some things (more apologies - not sure how to get Greek letters on here): let r' be the position the puck is dropped to (and orient the reference frame such that is lies on the y-axis)

let r be the position (as a function of time) of the reference point on the disc

let R be the quantity we're looking for: the position of the puck relative to our reference point, r.

Technically, this problem is fairly simple, though your final observation makes it more interesting. R = r'-r. Now we just need to define r, r' with respect to time.

r' is easy - the disc is frictionless, so the puck will not move (according to a 3rd-party observer in the puck's intertial reference frame), so we can define it in cartesian coordiantes as <0,r>

r is less so but still not terribly complicated. it's moving in a circle over time and can be described using basic trig as r' = (I've taken the liberty of assuming that at t=0, our reference point also aligns with the y-axis)

Given those definitions, R=r'-r = <0,r'> - R = <-rsin(wt), r'-rcos(wt)> And, noting that sin(-wt)=-sin(wt) and cos(-wt) = cos(wt), R =

Finally, to address the portion of your question dealing with "there's centrifugal force on the puck, so why doesn't it move?" - The centrifugal force on the puck only exists in the reference frame of the disc. Move back to an intertial reference frame and the puck feels no centrifugal force (and therefore will not fly off into the distance)

It's actually a good thing that the puck sees a centrifugal force in the disc's frame: With respect to the disc, the puck is actually moving in a curved path, and this requires a centrifugal force.

Hope this helps!

  • $\begingroup$ Some equations are not complete. Also notice that MathJax is enabled on this site (write formulas between dollar signs). $\endgroup$
    – Martin
    Commented Mar 19, 2016 at 1:23

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