A ferris wheel 30 meters high turns at a uniform rate of 2.2 revolutions/min. How far above the bottom of the Ferris wheel is a passenger when his or her height is changing at a rate of 2 m/second?
This is my last one and it is due today!
A ferris wheel 30 meters high turns at a uniform rate of 2.2 revolutions/min. How far above the bottom of the Ferris wheel is a passenger when his or her height is changing at a rate of 2 m/second?
This is my last one and it is due today!
First, $\displaystyle \frac{22}{10}\frac{rev}{min}*2*\pi*15\frac{m}{rev} *\frac{1 min}{60 sec}\;=\;\frac{11}{10}\pi\frac{m}{sec}$
At the top, the the height is changing (dv/dt) at 0 m/sec with the horizontal displacement changing (dh/dt) at 11$\displaystyle \pi$/10 m/sec. Vamp 1/4 turn to the far right. At this point, the situation is reversed, dh/dt = 0 m/sec and dv/dt = 11$\displaystyle \pi$/10 m/sec. This should suggest to you some ambiguity in the problem statement. Does it mean +2 m/sec or does it mean 2 m/sec in either direction? I think this is unclear. I would find all four solutions and decide later if the problem statement didn't mean some of them.
In any case, since we're talking about a circle, you should see that:
$\displaystyle \left(\frac{dv}{dt}\right)^{2}+\left(\frac{dh}{dt} \right)^{2}\;=\;\left(\frac{11}{10}\pi\frac{m}{sec }\right)^{2}$
and
$\displaystyle \frac{dv}{dt}\;=\;\frac{11}{10}\pi\frac{m}{sec}*co s(\alpha)$
Where $\displaystyle \alpha$ is the angular location of the passenger with reference to the center of the wheel, $\displaystyle \pi$/2 being the vertical position and 0 being the full forward position.
Simply solving, then,
$\displaystyle 2 m/sec\;=\;\frac{11}{10}\pi\frac{m}{sec}*cos(\alpha)$
and
$\displaystyle -2 m/sec\;=\;\frac{11}{10}\pi\frac{m}{sec}*cos(\alpha)$
Produces 0.954, 2.188, 4.095, and 5.329. It's up to you to decide which are applicable, if any.
Clear as mud?