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missashley
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Homework Statement
Sasha's favorite ride at the fair is the Ferris wheel that has a radius of 7.0m.
1. If the ride takes 20.0s to make one full revolution, what is the linear speed of the wheel?
2. What centripetal force will the ride exert on Sasha's 50.0kg body?
3. In order for Sasha to feel weightless at the top of the ride at what linear speed must the Ferris wheel turn?
4. At this speed, how much will she appear to weigh at the bottom of the Ferris wheel?
use 10 m/s^2 for gravity
Homework Equations
T = 1/f
f = 1/T
speed = (2)(pi)(r)/period
centripetal acceleration = (linear speed)^2 / radius
centripetal force = mass * centripetal acceleration ------- mv^2 / r
angular momentum = mvr
The Attempt at a Solution
1. v= 2pir/t
v = (2)(pi)(7)/(20) = 2.199 m/s
2. centripetal force = mv^2 / r
Fc = (50)(2.199)^2 / 7 = 34.544 N
3. centripetal force i think must equal the same as Sasha's weight
(50)(v^2) / 7 = (50)(10)
50v^2 / 7 = 500
50v^2 = 3500
v^2 = 70
v = 8.366 m/s
4. at 8.366 m/s, she will weigh 418.33 N because
(8.366)(50) = 418.33