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physics...centripetal acceleration?
A 43.0 kg child takes a ride on a Ferris wheel that rotates four times each minute and has a diameter of 24.0 m.
(a) What is the centripetal acceleration of the child?
(b) What force does the seat exert on the child at the lowest point of the ride?
Magnitude?
(c) What force does the seat exert on the child at the highest point of the ride?
Magnitude?
(d) What force does the seat exert on the child when the child is halfway between the top and bottom?
Magnitude?
Direction ° (from upward)?
2 Answers
- Ossi GLv 79 years agoFavorite Answer
Hello
frequency = 4 rpm = 4/60 rps
angular velocity ω = 2pi*frequency = 8/60*pi = 2/15*pi
a) acceleration a = ω^2*R = (2pi/15)^2*12 = 2.1055 m/s^2
b) F = mg + ma = 43(9.81 + 2.1055) = 512.37 N
c) F = mg - ma = 43(9.81 - 2.1055) = 331.29 N
d) F = mg = 43*9.81 = 421.83 N
Regards
- 9 years ago
(a) Velocity = (12*2*pi)/(60/4) = 5.0265 m/s
Centripetal acceleration = 2.105 m/s^2
(b) centripetal force = {43*(5.0265)^2}/12 = 90.54 N
Required Force by seat = 90.54+(43*9.81) = 512.37 N upward
(c)Required Force by seat =(43*9.81) - 90.54 = 331.29 N upward
(d)Required Force by seat = 90.54 N
