Was siting in class thinking about this problem, did some rough sketches of a solution but never really managed to solve it.
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 $$
Best Answer
First, R and r are proportional, by dimensional analysis, so set r to 1. Set g to 1 by choosing the unit of time to be $\sqrt{r/g}$.
The ice-cream height at any time is $2 - {t^2\over 2}$, where time starts when everything is at the top of the loop, the first instant of falling. So the ice-cream gets to the bottom at t=2.
At angle $\theta$ from the top, the velocity of the car is, by conservation of energy, $\sqrt{2(R + 1 - \cos(\theta))}$. The time taken to drop is the integral of the distance over the velocity along the circle, or
$$ {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.