A Ferris wheel 50ft in diameter makes one revolution every 40sec. If the center of the wheel is 30ft above the ground, how long after reaching the low point is a rider 50ft above the ground?

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A Ferris wheel 50ft in diameter makes one revolution every 40sec. If the center of the wheel is 30ft above the ground, how long after reaching the low point is a rider 50ft above the ground?

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Answer 1 · 2016-06-08T01:29:47

I found: $15.91s$ BUT check my maths....

Explanation:

Consider the (confused) diagram:
enter image source here
We can evaluate the angle $alpha$ using trigonometry applied to the orange small triangle with height: $50-30=20"ft"$ and hipotenuse equal to the radius $r=25"ft"$
so:
$20=25sin(alpha)$
$alpha=arcsin(20/25)=53.13^@$
So $50"ft"$ of height corresponds to the total angle: $90^@+53.13^@=143.13^@=2.498"rad"$

Now:

Angular velocity is $omega=(2pi)/T=(2pi)/40=0.157"rad/s"$
To describe $2.498"rad"$ it will take:
$t=2.498/0.157=15.91s$

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A Ferris wheel 50ft in diameter makes one revolution every 40sec. If the center of the wheel is 30ft above the ground, how long after reaching the low point is a rider 50ft above the ground?

1 Answer

Explanation:

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