Problem P49: A Ferris wheel is a vertical, circular amusement ride with radius 10 m. Riders sit on seats that swivel to remain horizontal. The Ferris wheel rotates at a constant rate, going around once in 10.5 s. Consider a rider whose mass is 56 kg. a) At the bottom of the ride, what is the rate of change of the rider's momentum? Answer: <0, 200, 0> kg m/s/s b) At the bottom of the ride, what is the vector gravitational force exerted by the Earth on the rider? Answer: <0, -549, 0> N c) At the bottom of the ride, what is the vector force exerted by the seat on the rider? Answer: <0, 749, 0>
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Given: Radius, R = 10 m Time period, T = 10.5 s Using the formula: Speed = 2πR / T Speed = 2 * π * 10 / 10.5 Speed ≈ 5.98 m/s ** Show more…
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