A tricky game of catch

Classical Mechanics Level 5

You and a friend are going to ride on a Ferris wheel. For a little extra fun, when your friend is at the very top of the Ferris wheel she's going to drop a tennis ball. How far around the Ferris wheel in radians from your friend should you sit so you can catch the ball?

Details and assumptions

The Ferris wheel radius is $R=20~\mbox{m}$ .
The Ferris wheel goes around at a constant angular velocity $\omega=0.2~\mbox{rad/s}$ .
The acceleration of gravity is $-9.8~\mbox{m/s}^2$ .
Neglect air resistance.

The answer is 2.0145.

2 solutions

Anish Puthuraya
Feb 10, 2014

alt text

From the diagram, it is clear that,
$\displaystyle y = \frac{1}{2} gt^2$

$\displaystyle x = R\omega t$

$\displaystyle\Rightarrow y = \frac{g}{2R^2\omega^2} x^2$

Now, let us the find the point of intersection of this parabolic path of the ball, with the circle (Ferris wheel), whose equation is :
$\displaystyle x^2 + (y-R)^2 = R^2$

Substituting $\displaystyle y = \frac{g}{2R^2\omega^2} x^2$ in the equation of the circle, we get,

$\displaystyle x^2 \approx 120m$

$\Rightarrow \boxed{x = \sqrt{120m}}$

$\displaystyle\Rightarrow \boxed{y = 36.75m}$

Now, to find the angle $\theta$ , we shall take the dot product of $\displaystyle (x\hat{i} + y\hat{j})-R\hat{j}$ and the vertical unit vector $\displaystyle \hat{j}$ .

Thus,
$\displaystyle (x\hat{i}+(y-R)\hat{j})\cdot\hat{j} = \sqrt{x^2+(y-R)^2} \cos\theta$

Putting the values of $\displaystyle x,y,R$ , we get,

$\displaystyle \cos\theta = \frac{-67}{80}$

$\displaystyle \Rightarrow \boxed{\theta = 2.563rad}$

Now, considering the catcher of the ball,
$\displaystyle \theta = \theta_o +\omega t$

We already have $\displaystyle \theta$ , thus, to find $\displaystyle \theta_o$ , we just need to find $\displaystyle t$ , which we will derive from the fact that,
$\displaystyle x = R\omega t$
$\displaystyle\Rightarrow \boxed{t = 2.738 sec}$

Substituting the values of $\displaystyle \theta,\omega,t$ , we get,
$\displaystyle \boxed{\theta_o = 2.015rad}$ is the final answer.

Hmm, I solved it differently, but anyways, nice solution.

Sam Thompson - 7 years, 4 months ago

Post your solution (if you can)...I want to take a look at it.

Anish Puthuraya - 7 years, 4 months ago

My Solution: Set the center of the Ferris wheel as the origin. Obviously, $x=4t$ , and $y=20-4.9t^2$ . Since the equation of the Ferris Wheel is $x^2+y^2=400$ , we can get $t^{2}(24.01t^{2}-180)=0$ , or $t=2.738$ . After plugging this in for the x and y-coordinates, I did some bashing, and I finally found the answer to be 2.0153....

Sam Thompson - 7 years, 4 months ago

Bashing Part: The coordinate where the ball intersects the person is (10.952, -16.735). The angle of this point relative to the top of the Ferris Wheel is 2.5618, which means that the person is at an angular distance of 2.5618-2.738*.2= $\framebox{2.0142}$ from the other person. This solution was more accurate than the last solution, since I used more significant digits for each number.

Sam Thompson - 7 years, 4 months ago

Dave Andika
May 20, 2014

It is clear that the ball will never reach the bottom of the wheel, since it has a positive velocity at the top:

Using the center of the wheel as the coordinate origin, the path of the ball is

1) Yb = -g x²/(2V²) + 20 where V = R w = 20*0.2 = 4 m/s The wheel height is given by

2) Yw = √[20² - x²]

Equating (1) & (2) → x = 10.952 m

Thus, the intersection angle below horizontal Θi = arccos[10.952/20] = 0.9913 rad, and

Total angle Θ = (π/2) + Θi = 2.562 rad

Time for the ball to fall to the intersection point t = √[2y/g] where y = 20(1+sinΘi) = 36.734 m →t = 2.7385 sec

The angle your friend travels through during this time is Θf = w t = 0.2 2.7385 = 0.5477 rad

So, the angle your friend should sit in front of you ß = Θ - Θf = 2.014 rad

A tricky game of catch

The answer is 2.0145.

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