# Ferris wheel and projectile [closed]

Imagine there's a little girl/boy sitting on one of the seats of a Ferris wheel. She/he throws out a stone and sees that it's experiencing the velocity v. Meanwhile, another observer standing on the ground sees that the stone is thrown with a horizontal velocity, u. What's the connection between u and v? (I hope I explained what's on my mind well enough. )

## closed as off-topic by tpg2114♦, Abhimanyu Pallavi Sudhir, user10851, Brandon Enright, Qmechanic♦Nov 12 '13 at 14:57

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – tpg2114, Abhimanyu Pallavi Sudhir, Brandon Enright, Qmechanic
If this question can be reworded to fit the rules in the help center, please edit the question.

• is there a mixture of rotational, relative and projectile motion? – Sogol Nov 11 '13 at 21:36

This will depend on a few things...

Angular velocity of the Ferris wheel. I will call that "$\omega$"

Position of the Ferris wheel at the moment the rock is thrown. call that: "$\theta$"

And, The radius of the Ferris wheel. call that: "$R$"

So, I will assume we are not talking about any special 3D point of view. This should be solvable on a 2D plane I assume (correctly I hope). The basic equation for a moving object in this case is: $$\vec{r}(t) = {{ \vec{a} *t^2} \over {2}} + \vec{V_o} * t + \vec{r_o}$$

We know the observer on the ground will see a parabolic trajectory based on the initial $X$ and $Y$ velocity components put into the stone. The real question is what the stone thrower will see. We could say $(\omega \approx 0)$ thus the thrower of the stone will see the same as the ground observer. But that would be too easy, so lets say the Ferris wheel is spinning quite fast. Like, say just fast enough that at the peak of the ride 0-Gs are felt by a rider having fun. At this point the rider could simply drop the stone and it would appear to move directly forward from the current position. As the Ferris wheel spins on it will become more evident to the rider the stone is slowing, but this is in the $X$ direction. They both move down together until the rider is more than 90 degrees from the point of drop. At that point the rider starts to see the stone fall and move away as the Ferris wheel slows the rider in the $Y$ direction and speeds the rider in the $X$ direction. So how do we make this into a general case.

Say the peak is at $(\theta = 0)$ and at $(t=0)$

The angular position of the rider is $(\theta = t \omega)$. Using the classic polar coordinate system we can put a vector on the $(X, Y)$ position of the rider. $$\vec{p}(t) = \left[ \begin{matrix} R \cos(t \omega) \\ R \sin(t \omega) \end{matrix} \right]$$

Now to find the relative $(X, Y)$ position of the stone to the rider. We need only take the total position of the stone and subtract the position of the rider. $$Fnal \space Answer = \vec{r}(t) - \vec{p}(t)$$

Some values you may need to put into the equation for " $\vec{r}$(t) " $$\vec{a} = \left[ \begin{matrix} 0\\ -g\end{matrix} \right] \space \space \space \space \vec{V_o} =\left[ \begin{matrix} V_x + R \omega \\ V_y \end{matrix} \right] \space \space \space \space \vec{r_o} =\left[ \begin{matrix} R \\ 2R \end{matrix} \right] \space \space \space \space$$

Hope this helps. If you need more info, make a note of it.

Edit: Just want to let you know the final formula does not simplify much... $$\vec{f}(t) = {{ \vec{a} *t^2} \over {2}} + \vec{V_o} * t + \vec{r_o} - \vec{p}(t)$$ Unless you got some numbers you can plug in...