Consider an point particle moving on a frictionless semicircular hill (curve). The particle's initial kinetic energy is equal to the potential energy on the top of the hill, i.e it has the necessary energy to climb the hill.

Will it reach the top of the hill in infinite or finite time?

In my proof it needs infinite time and this is quite non-intuitive because, though the particle has the necessary energy to climb the hill, it needs an infinite time. Also if we reverse the time, when the particle is in equilibrium on the top of the hill, it will never go down, therefore this process is time irreversible.

  • $\begingroup$ Would things be different if the hill was a cone? (or simply a ramp)? $\endgroup$
    – Georg
    Aug 5 '11 at 11:24
  • $\begingroup$ Any smooth curve with zero slope at the top will do. I pick semicircle just for simplicity. $\endgroup$
    – Andreas K.
    Aug 5 '11 at 11:30
  • $\begingroup$ Can you show the proof? $\endgroup$ Aug 5 '11 at 11:35
  • $\begingroup$ Is the profile of the hill concave or convex? $\endgroup$ Aug 5 '11 at 12:25
  • $\begingroup$ @ Raskolnikov, it's convex, i.e $y=\sqrt{1-x^2}$. I will write my proof which is familiar to your. But I think that your proof have some mistakes. $\endgroup$
    – Andreas K.
    Aug 5 '11 at 13:07

EDIT: OK, I misinterpreted ANKU's question and had a bowl shape in mind whereas he had an upturned bowl in mind. This changes the energy equation to

$$ mgR = \frac{1}{2}mR^2\dot{\theta}^2 + m g R (\sin \theta) \; .$$

I measure the angle from the horizontal here. After similar manipulations as below, I get

$$ T = \sqrt{\frac{R}{2g}} \int_0^{\pi/2} \frac{d\theta}{\sqrt{1-\sin\theta}}$$

which is indeed a divergent integral. However, this does not mean that the motion is irreversible. The reverse motion starts from the top of the hill, but the top of the hill is an equilibrium point, albeit an unstable one, which implies that it takes an infinite amount of time to roll down the hill.

From conservation of energy, you can write down the following formula:

$$ mgR = \frac{1}{2}mR^2\dot{\theta}^2 + m g R (1-\cos \theta) \; .$$

The left hand side represents the potential energy at the top of your semi-circular hill, the right hand side the total energy at any point of the trajectory. (Angle measured from the vertical.)

Rearranging, you can write this as

$$ \frac{2 g \cos \theta}{R} = \dot{\theta}^2$$

or after some additional work and intergrating

$$ T = \sqrt{\frac{R}{2g}} \int_0^{\pi/2} \frac{d\theta}{\sqrt{\cos\theta}}$$

A quick check with wolframalpha gives a finite number for the right integral. Therefore the time it takes for the ball to roll up the hill is finite.

  • $\begingroup$ Where did you get $$mgR=\tfrac12mR^2\dot{\theta}^2+mgR(1-\cos(\theta))$$ from? The right-hand side of this equation is zero for a particle resting at the top of the hill, and so it should be. Also, it should be $\cos(\theta)-1$ rather than $1-\cos(\theta)$, because the potential energy is highest on top of the hill. $\endgroup$ Aug 5 '11 at 12:46
  • 2
    $\begingroup$ Yeap, it's $ mgR = \frac{1}{2}mR^2\dot{\theta}^2 + m g R \cos \theta $. So you get an divergent integral as I did in my proof. Therefore the time is infinite. $\endgroup$
    – Andreas K.
    Aug 5 '11 at 13:12
  • $\begingroup$ It's easier to get insight by using an approximation. Starting from ANKU's corrected version of the conservation of energy equation, use a Taylor series to approximate $\cos\theta$ for small $\theta$. The result is $\dot{\theta}=\pm k\theta$, where $k=\sqrt{g/R}$. The solution is $\theta=Ae^{\pm kt}$. The only way to have $\theta=0$ at any time is to have $A=0$, so that $\theta=0$ at all $t$. Basically this is a simple harmonic oscillator with a negative "spring constant," i.e., an unstable equilibrium rather than a stable one. $\endgroup$
    – user4552
    Aug 5 '11 at 14:57
  • $\begingroup$ To be harmonic, the hill has to be very flat, ie not a semisphere. (analog to limiting a normal pendulum to some degrees swing to be harmonic) $\endgroup$
    – Georg
    Aug 5 '11 at 15:11
  • $\begingroup$ @Georg: It's not harmonic. It's similar to harmonic, but with a negative "spring constant." The equations of motion for a pendulum near the bottom of its arc and the equations of motion near the top are of exactly the same form, the only difference being the sign of the "spring constant." Note that in ANKU's corrected version of the equations (3 posts up), $\theta=0$ is the top of the circle, not the bottom. (Raskolnikov's mistake may have caused some confusion.) $\endgroup$
    – user4552
    Aug 5 '11 at 16:38

Interesting question: The time it takes to get at the top is naturaly infinite. It has sense, since the particle arrives at rest, it should move slower and slower as it get closer to the top. The main part of the time that the particle needs for to reach the top is spent in the last portion of its trayectory. The process is already time reversivle; the point is that the both exmples you mention doesnt correspond to the same proces. In the former the particle never reaches the top. I suggest to solve the problem in the case when the particle has initialy a little more energy than the potential at the top, say MgR+epsilon; then take the limit epsilon to zero. Probably it might clarify the aspects of the problem. Hope I have been of help.


The correct answer is that the time needed to reach the top of the hill is both infinite and finite. Which one depends only on the shape of the hill within some small neighborhood of the top, - wherein the potential energy (i.e. the shape of the hill in this neighborhood) can be expressed as: $$U(x)=(\text{total energy of the system}) - mgf(x)$$

where $f(x) \approx x^a$ for some small neighborhood of the top of the hill.

It takes infinite time if $a\leq1$ or $a\geq2$

and it takes finite time if $1 < a < 2$

you can reach this through simple cons.of energy with newtonian mechanics. start with $E=\frac{m\left(\frac{\rm dx}{\rm dt}\right)^2}{2}+U(x)$ solve for dt and play with the integrand to arrive at constraints for a.


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