Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Assume a ball is placed on a hill and the time is measured for the ball to reach the ground (y=0). If the time taken for the ball to reach the ground is independent on the position the ball is placed on the hill, determine the equation or set of possible equations for the hill (y as a function of x).

If this is not possible can you provide a proof of why this is not possible?

Also assume gravity is 9.8m/s.

share|cite|improve this question
Assuming that friction is to be ignored, you’re looking for the tautochrone curve. – Brian M. Scott Sep 18 '12 at 7:08
Yes that does seem to answer what I was asking. Thanks! – Mew Sep 18 '12 at 7:10
@Brian: That's what might be called the winter-games version of the problem; but the summer-games version has the same form of solution, just a different travel time; see my answer. – joriki Sep 18 '12 at 10:14
up vote 1 down vote accepted

Brian's solution remains valid if we make a slightly more realistic assumption by ignoring dissipative friction but taking friction to be strong enough to cause the ball to roll rather than slide. In this case the ball's energy is given by

$$ \begin{align} E &=\frac12mv^2+\frac12I\omega^2+mgz \\ &=\frac12(1+\alpha)mv^2+mgz\;, \end{align} $$

where the factor $\alpha=I/(mR^2)$, with $I$ the ball's moment of inertia and $R$ its radius, depends on the mass distribution and is $\frac25$ for a solid ball of constant density.

Thus the rotational energy effectively increases the ball's inertial mass, or equivalently reduces the gravitational acceleration. The solution is still a cycloid; only the time required to reach the ground is now

$$ T=\pi\sqrt\frac{r(1+\alpha)}g $$

(where $r$ is the radius of the cycloid), i.e. the rolling ball takes slightly longer than the sliding ball, by a factor of $\sqrt{7/5}\approx1.18$ in the case of a solid ball of constant density.

Note that this treatment ignores the fact that for a ball with non-zero radius $R$ the ball's centre of mass moves on a different curve than its surface; this approximation is only valid for $r\gg R$.

share|cite|improve this answer
I’m glad that someone did this; the last time I did a physics problem of this kind was in 1965, and I’m just a wee bit rusty! – Brian M. Scott Sep 18 '12 at 18:01

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.