# A boy, a cart, a loop , an icecream-cone

1. Nov 27, 2011

Was sitting 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][1]

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.

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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 account 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.

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

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

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

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Any comments, suggestions, hints soloutions would be much obliged =)

2. Nov 27, 2011

### dextercioby

The time for the boy to slide down the semicircle is found from integrating the conservation of energy formula. It is however an elliptic integral so you'd have to make an approximation somewhere.

3. Nov 27, 2011