You are standing in the middle of a disk of radius $2\:\mathrm{m}$ rotating at $\pi\:\mathrm{rad s^{-1}}$, and there's a smooth puck at $1.5\:\mathrm{m}$ away from the centre. If the puck was released, what is the relative velocity of the pluck from you? i.e. how would you observe it?

I saw an answer trying to vectorally take away the velocity of the edge of the disk from the velocity of the puck, but I think it's wrong because if you just replace the disk with a larger one, the answer using this approach would be changed, but I think it shouldn't.

I'm sure there is an animation showing this, but I can't find any. Would be much appreciated if anyone could find one.

  • $\begingroup$ I believe you've meant 'puck,' but 'pluck' evokes the image of a defeathered fowl scrabbling about on a slick merry-go-round, which is substantially more engaging as an experimental setup. $\endgroup$
    – Asher
    Commented Nov 14, 2015 at 21:32
  • $\begingroup$ Concerning the question, are you taking friction into account? A frictionless treatment allows a simple mathematical solution, while frictional treatment requires more initial conditions and data and allows, eventually, a synchronous 'orbit' for the puck. $\endgroup$
    – Asher
    Commented Nov 14, 2015 at 21:37
  • $\begingroup$ Ah my bad, it's a smooth puck so no friction $\endgroup$ Commented Nov 14, 2015 at 21:48

1 Answer 1


Consider the diagram below:

It shows you and the puck on the rotating disc. Since as you're in the centre of the disk the puck is rotating around you. The disc is rotating with an angular velocity $\omega=\pi\:\mathrm{rads^{-s}}$.

You and the puck

Pictured is the tangential velocity vector $\vec{v}$ which of course changes every instant as the position vector $\vec{r}$ changes every instant. This change in direction is caused by the centripetal acceleration, in vector notation $\vec{a_c}$:

$$\vec{a_c}=\frac{d \vec{v}}{dt}$$

This acceleration acc. Newton requires a centripetal force $\vec{F_c}$, as:


where $m$ is the mass of the puck and $\vec{F_c}$ is on the same line and points in the same direction as $\vec{a_c}$.

Now imagine that at the instant the diagram represents, we release the puck, that is we remove $\vec{F_c}$ altogether, then acc. Newton's Law the puck will simply continue to move along the line and direction of $\vec{v}$. And without friction or any other forces acting on it... forever so! The linear speed (scalar) will be $v=1.5 \times \pi \:\mathrm{m/s}$ or $4.71\:\mathrm{m/s}$.

Regards the question of relative velocity it doesn't make a great deal of sense because the observer is not translating. One can say that the relative velocity of the puck with regards to the observer, in the horizontal direction it is zero, while in the vertical direction it is $v=1.5 \times \pi \:\mathrm{m/s}$.


OP seems also interested in how long it would take for the puck to leave the disc altogether (without friction). That time is essentially the time to cover the distance between the puck at time of release (call it point $P$, call $O$ the centre of the disc) and the point $E$, that is $|PE|$. With Pythagoras we know:




The time $t$ to cover that distance is:


  • $\begingroup$ A nice answer as always, but did you not skip this question: "What is the relative velocity of the pluck from you?" $\endgroup$ Commented Nov 14, 2015 at 22:01
  • $\begingroup$ @AshishGupta: as the observer is stationary, the concept of relative speed is really a bit surplus to requirement here, don't you think? :) $\endgroup$
    – Gert
    Commented Nov 14, 2015 at 22:05
  • $\begingroup$ Yeah I know. Not really required here but it has been asked by the questioner. Anyway, if you don't mind, can you tell me this: If in this case, the initial velocity of the puck had an x component, how would that be affected over time? $\endgroup$ Commented Nov 14, 2015 at 22:14
  • $\begingroup$ @AshishGupta: same principle: any $\vec{v_x}$ would be conserved at the instant of release. Remember that $x$ and $y$-components are independent of each other. $\endgroup$
    – Gert
    Commented Nov 14, 2015 at 22:18
  • $\begingroup$ Does that mean that we will have to deal with $v_x$ and $v_y$ separately and that they will both be affected by the circular motion over time? (Because after some time, there will be a component of the acceleration which will affect $v_x$ if I'm not wrong.) $\endgroup$ Commented Nov 14, 2015 at 22:21

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