In my physics textbook, I have the following worked out example:

A table with a smooth horizontal surface turns at an angular speed $ω$ about its axis. A groove is made on the surface along a radius and a particle is gently placed inside the groove at a distance $a$ from the centre. Find the particle's speed as its distance from the centre becomes $L$.

The author solves it in the frame of the table. I have attempted to do it from the ground frame. I fixed the X-axis along the groove in the initial position, the Y-axis on the same plane, and the Z-axis perpendicular to the plane. Then I considered the case $dt$ time after putting the system into motion. Initially, $v=ωr$, and after a time dt, $v+dv=ω(r+dr)$, thus giving me $dv=ωdr$ (after integrating, I got $v=ω(L-a)$, the answer did not match). My question is: is the use of this formula here valid considering the fact that the particle is not going around in a circle?

  • $\begingroup$ Related - What real force causes outward acceleration in rotation? $\endgroup$
    – Farcher
    Commented Dec 23, 2022 at 10:55
  • $\begingroup$ The answers below I think said everything that needs to be said, but I want to add one small detail about why in circular motion $v=\omega r$ holds. In circular motion with radius $r$ and velocity $v$, it takes the object $t=s/v=2\pi r/v$ for one revulution. In other words, the object goes around the circle with a frequency of $f=1/t = v/2\pi r$. Rearange that to $f 2\pi r = v$ and use $\omega = 2\pi f$, you get $v = \omega r$. You see that this derivation relies on the motion being along a circle. $\endgroup$
    – Samuel
    Commented Dec 23, 2022 at 13:31

2 Answers 2


It's convenient to use a set of polar coordinates with the origin in the center of rotation of the table. Using this set of coordinates, the kinematic quantities of the point mass can be written as:

  • position: $\mathbf{r} = r \mathbf{\hat{r}}$
  • velocity: $\mathbf{v} = \dot{r} \mathbf{\hat{r}} + r \dot{\theta} \boldsymbol{\hat{\theta}}$
  • acceleration: $\mathbf{a} = \left( \ddot{r} -r\dot{\theta}^2\right) \mathbf{\hat{r}} + \left( 2\dot{r}\dot{\theta} + r \ddot{\theta} \right) \boldsymbol{\hat{\theta}} $.

Since the tangential velocity of the mass equal the local tangential velocity of the groove, the following kinematic constraint holds:

$r \dot{\theta} = r \Omega \qquad \rightarrow \qquad \dot{\theta} = \Omega$.

Now we can write the dynamical equation using the second principle of dynamics, with the assumption that there is no friction along the groove in radial direction and the force exchange acts only in azimutal direction (normal to the groove):

$r: m a_r = 0 \qquad \ \ \ \rightarrow \qquad \ddot{r} -r\dot{\theta}^2 = 0$
$r: m a_{\theta} = N_{\theta} \qquad \rightarrow \qquad 2\dot{r}\dot{\theta} + r \ddot{\theta} = \dfrac{N_{\theta}}{m}$ .

Using the kinematic constraint, you can integrate the radial component of the equation to get $r(t)$. Once you know $r(t)$, you can find the normal force $N_{\theta}$ using the tangential component.

Assuming $\Omega$ const, the radial component of the equation of motion gives you

$r(t) = A e^{\Omega t} + B e^{-\Omega t}$,

where the integration constants must be determined using initial conditions. Assuming initial distance $r(0) = a$ and no radial velocity, $\dot{r}(0) = 0$ we get

$0 = \dot{r}(0) = \Omega (A - B) \qquad \rightarrow \qquad A = B$
$a = r(0) = A + B \qquad \quad \ \ \rightarrow \qquad A = B = \dfrac{a}{2}$

so that the solution reads

$r(t) = \dfrac{a}{2} \left( e^{\Omega t} + e^{-\Omega t} \right)$.

So, the mass reaches $r = L$ at time $t_L$ for which

$L = r(t_L) = \dfrac{a}{2} \left( e^{\Omega t_L} + e^{-\Omega t_L} \right)$.

To find $t_L$ you need to solve a second order equation in $e^{\Omega t_L}$, after multiplying the last equation by $e^{\Omega t_L}$ itself. Once you find $t_L$, the velocity reads

$\mathbf{v}(t_L) = \dot{r}(t_L) \mathbf{\hat{r}}(t_L) + r(t_L) \Omega \boldsymbol{\hat{\theta}}(t_L)$

  • $\begingroup$ So, the problem can't be solved (without a lot of trouble) in the coordinate system I've chosen? $\endgroup$ Commented Dec 23, 2022 at 16:18
  • $\begingroup$ Not so many troubles, indeed. But, as a suggestion, you have to start from principles, whenever you have any doubts in Physics, without inventing anything (wrong) from scratch $\endgroup$
    – basics
    Commented Dec 23, 2022 at 16:41

Is the use of this formula here valid considering the fact that the particle is not going around in a circle?

No. $v=r\omega$ is only true if the particle is moving in a circle. But since the particle is not constrained radially, it is not moving in a circle - its path is a spiral, so $v > r\omega$. In vector notation

$\mathbf v = \dot r \mathbf{\hat r} +r \omega \mathbf{\hat \theta}$

and $\dot r \ne 0$ so $|\mathbf v| > r \omega$.


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