Imagine a 2D circle has a point mass traveling along its circumference with constant speed. The only force experienced is centripetal. However, if we take the 2D plane the circle sits on and rotate it such that it appears 1 dimensional to us, the circle becomes a line. The mass traveling along this circle becomes a mass traveling back and forth along a line, which means this mass must have linear acceleration. How can I make sense of this?

  • 1
    $\begingroup$ The tangential speed is constant. The velocity in the x (or y) direction is sinusoidal assuming the circle is in the x-y plane. $\endgroup$
    – Bill Watts
    May 30 at 22:33
  • $\begingroup$ Try projecting the force vector onto the same plane as you just projected the displacement vector. $\endgroup$
    – g s
    May 31 at 2:03
  • $\begingroup$ You can't just ignore one component of position/velocity/acceleration and only apply the laws of motion to the other component only and expect the laws to make sense. $\endgroup$ May 31 at 15:29
  • $\begingroup$ That's basically how you get a linear harmonic oscillator. $\endgroup$ May 31 at 20:45

2 Answers 2


It makes sense because the acceleration is entirely centripetal, and when you rotate the plane so that the circular motion appears to be linear, you are merely observing the acceleration that indeed exists along that one axis. The particle's velocity, which is a vector maintains constant magnitude but only changes direction, but this tilting of the plane enables you to observe only a single component of this vector, and if a vector has constant magnitude it does not necessarily imply each of its components are constant separately!

To put it more precisely, if we have a particle moving on a circle of radius $R$ with angular speed $\omega$ the usual parametrization of the position vector is:

$$\vec{r} = R\left(\cos(\omega t) \hat{x} + \sin(\omega t) \hat{y}\right)$$

We get the velocity and acceleration vectors by differentiating wrt $t$:

$$\dot{\vec{r}} = \omega R\left(-\sin(\omega t)\hat{x} + \cos(\omega t)\hat{y} \right) $$ $$\ddot{\vec{r}} = -\omega^2 R\left(\cos(\omega t)\hat{x} +\sin(\omega t)\hat{y} \right) $$

Now you can easily see that the acceleration vector is always perpendicular to the velocity vector by noting that:

$$\dot{\vec{r}} \cdot \ddot{\vec{r}} = 0$$

Hence the magnitude of the velocity $\left| \dot{\vec{r}} \right|$ remains constant and the velocity vector only changes its direction due to the centripetal acceleration. You can also see it by directly calculating the time derivative of the squared speed $\frac{d}{dt}\left(\left| \dot{\vec{r}} \right|^2\right) = \frac{d}{dt}\left(\dot{\vec{r}} \cdot \dot{\vec{r}}\right)$, and verifying that it vanishes for the same reason.

The same is not true for each separate component of the velocity $\dot{\vec{r}}$, namely, $\dot{\vec{r}}\cdot \hat{x}$ and $\dot{\vec{r}} \cdot \hat{y}$ clearly oscillate with a linear acceleration given by their respective acceleration components $\ddot{\vec{r}}\cdot \hat{x}$ and $\ddot{\vec{r}} \cdot \hat{y}$. These are the linear accelerations you observe when you rotate the circle to look like a line, indeed they are very much there and they don't contradict the fact that as a vector in $2$ dimensions the velocity of the particle doesn't change its magnitude.


In polar coordinates:

$$ \dot r = 0$$ $$ \dot \theta = v/R $$

Integrate that to get:

$$ r(t) = +C =R $$

$$ \theta(t) = \omega t +\phi$$

with $\omega = v/R$.

Project onto an axis:

$$ x(t) = r(t)\cos{\theta(t)} = R\cos{(\omega t+\phi)} $$


$$ \ddot x(t) = -\omega^2R\cos{(\omega t+\phi)} =-\omega^2 x(t)$$


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