Was siting in class thinking about this problem, did some rough sketches of a solution but never really managed to solve it.

Ice cream cone and a loop-de-loop

Assume a boy starts at the top of a circle with radius R as described in the picture. It is a snowy day and the path can be considered without friction. The boy enters a loop with radius r at the bottom of the hill. At the top of the loop the boy loses his icecream cone in such a way that it starts faling. The initial velocity of the icream is 0 m/s straight down.

The problem is to find R expressed by r, such that the boy reaches the icrecream just as he reaches the bottom of the loop.

The problem boiled down to finding out how much time the boy uses getting from the top of the loop to the bottom. Any help, solutions or inputs would be great.

My attempt, I know that this is most likely 90% wrong

By using conservation of mechanical energy. The speed at the bottom of the hill equals

$$ v_b^2 = Rmg $$

And the velocity at the top of the loop equals

$$ v_t^2 = 2g\left( R - 2r \right) $$

Vi know that the aceleration is constant and equals $ g $ (Here is where I think I make my mistake, forgot to acount for the angular velocity)

$$ s = \dfrac{v_1 - v_0}{2} t $$

We use this equation to find out how long it takes the boy to get from the top, to the bottom of the loop.

$$ \large t \, = \, \dfrac{2s}{v_1 - v_0} \, = \, \dfrac{2\left( \dfrac{2\pi r}{2}\right)}{\sqrt{2gR} - \sqrt{2g(R - 2r)}} $$

Now we figure out how long it takes the icream to fall the distance of the diameter or $ 2r $ .

$$ s = v_0 + \dfrac{1}{2}gt^2 $$

$$ t = \sqrt{\dfrac{2s}{g}} \, = \, \sqrt{\dfrac{4r}{g}} \, = \, 2 \sqrt{\dfrac{r}{g}} $$

By setting these two equations equal each other, and solving for $$ R $ , we obtain that

$$ R = \dfrac{16+8 \pi^2+\pi^4) r}{8 \pi^2} \cdot r \approx 2.43 r $$

  • 1
    $\begingroup$ The assumption that the ice cream will have a initial velocity "straight down" is erroneous. It has a non-zero horizontal velocity just before it starts falling and has experienced no horizontal force to change that situation. (Unless the ice-cream is held $r$ above the tracks, in which case the boy can't catch both drop it and catch it without making some kind of active move of the cone.) $\endgroup$ Nov 28, 2011 at 1:20
  • $\begingroup$ @dmckee: yeah, that bothered me as well, but it seems to be a given condition. I can see why the authors might want to put that in to create a simpler, if unrealistic, problem. (I guess it could happen if the boy jerks the ice cream cone backwards relative to himself at exactly the right velocity just as he passes the top of the loop...) $\endgroup$
    – David Z
    Nov 28, 2011 at 2:36
  • $\begingroup$ David Zaslavsky thats the plan, jerking the icecreamcone backwards at the exact correct velocity. Sorry for my bad english, someone might want to edit, if they feel like it =) ofcourse the make the problem even more unrealistic, we ignore friction and consider the boy as a dot particle. And loops in rollercoasters are not perfect circles, and so on ^^ $\endgroup$ Nov 28, 2011 at 6:10

3 Answers 3


First, R and r are proportional, by dimensional analysis, so set r to 1. Set g to 1 by choosing the unit of time to be $\sqrt{r/g}$.

The ice-cream height at any time is $2 - {t^2\over 2}$, where time starts when everything is at the top of the loop, the first instant of falling. So the ice-cream gets to the bottom at t=2.

At angle $\theta$ from the top, the velocity of the car is, by conservation of energy, $\sqrt{2(R + 1 - \cos(\theta))}$. The time taken to drop is the integral of the distance over the velocity along the circle, or

$$ {1\over \sqrt{2}} \int_0^\pi {d\theta \over \sqrt{R - 1 - \cos(\theta)}} $$

And setting this time equal to 2 gives a transcendental equation for H=R-1 whose solution is the answer (and extend the domain of integration to the full circle).

$$ \int_0^{2\pi} {d\theta \over \sqrt{H - \cos(\theta)}} = 4\sqrt{2} $$

expand the denominator in powers of $\cos(\theta)$ using the power series for square-root below (a simple Taylor series at x=0):

$$ (1-x)^{-{1\over 2}} = \sum_{N=0}^{\infty} {(2N)!!\over 2^N} {x^N \over N!}$$

Where the (2N)!! is a weird notation product of odd numbers less than 2N, which is given by:

$$ 2N!! = 1\cdot 3\cdot 5 ... \cdot (2N-1) = {(2N)!\over 2^N N!}$$

and use this useful identity for integer N's:

$$ \int_0^{2\pi} (\cos(\theta))^{2N} d\theta = 2\pi {2N \choose N} {1\over 2^{2N}}$$ $$ \int_0^{2\pi} (\cos(\theta))^{2N+1} d\theta = 0 $$

Which is derived by writing cosine as a sum of complex exponentials and multiplying out the product, and noting that only the middle term for even powers survives. This gives a good expansion for the function of R on the left.

The function is monotonic decreasing from infinity at $H=1$ to 0 at $H=\infty$, so there is a unique solution. The resulting expansion is

$$ {1\over \sqrt{H}} \sum_{N=0}^{\infty} { (4N)!!\over (2^N N!)^2 } {1\over H^{2N}} = {2\sqrt{2}\over \pi} $$

And the solution is (by hand) very close to H=1.50, meaning that R is close to 2.50. You can invert the power series, or just solve this in a second with a computer.


I know that this is most likely 90% wrong

Well there's your problem ;-)

Seriously though:

I know that the aceleration is constant and equals g (Here is where I think I make my mistake, forgot to acount for the angular velocity)

Yes, that's incorrect. Assuming that you're talking about the acceleration of the roller coaster cart, it's not constant. For one thing, acceleration is a vector, so it has a direction as well as a magnitude, and even if the magnitude is constant, it doesn't mean the acceleration is constant. But even more importantly: the magnitude of the acceleration is $g$ only for an object in free fall. The roller coaster cart is not in free fall because it has a track pushing on it. So its acceleration changes in both magnitude and direction from point to point on the loop.

Of course, figuring out the acceleration as a function of time is a pretty complex problem in its own right. Perhaps there's some trick you can use that will allow you to avoid doing that.

  • $\begingroup$ Someone told me in order to "solve" this problem I to integrate the conservation of energy formula, but this would give me an elliptic integral. Though I had some problem setting this up $\endgroup$ Nov 28, 2011 at 6:13

This is what I get:

Using conservation of energy, velocity from the top of the loop as a function of $\theta$ is $\sqrt{2gr(1 - \cos(\theta))}$. Distance traveled is arc length, which is $rd\theta$. Therefore, I get time as integral of distance over speed, which is: $\int_0^\pi {rd\theta \over \sqrt{2gr(1 - \cos(\theta))}}$

I'm not sure how to integrate this expression, but I think you could probably take it from there.

  • $\begingroup$ This is assuming $v_t = 0$ which is not correct. To add a non-zero $v_t$ just add it to the denominator of the integral: $\int_0^\pi {rd\theta \over (\sqrt{2gr(1 - \cos(\theta))} + v_t)}$ (It does make the integral a lot more complicated than I know how to handle.) $\endgroup$
    – Joebevo
    Mar 11, 2012 at 8:22

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