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A passenger on the ferris wheel normally

weighs 367 N. The ferris wheel has a 14 m radius and is

equipped with a powerful motor. The operator revs it up so that the customers at the top of the wheel feel zero g's (they momentarily

lift slightly of their seats). The acceleration of gravity is 9.8 m/s^2

1) At what angular velocity will this occur?

Answer in units of 1=s.

N = 367 N

R = 14m

g = 9.8 m/s^2

N = 0

Fc = mg = m(V^2/R)

gR = V^2

(9.8)(14) = v^2 = 137.2 = 11.71

(Angular Velocity) W= V/R

11.71 / 14 = .835

2) Assume: The rotating angular velocity is

same as in Part 1.

What weight does the customer feel at the

bottom of the wheel? Answer in units of N.

Fc = N - mg = m(V^2/R)

N = mg + m(V^2/R)

V = WR

N = mg + m [(W^2 x R^2)/(R)]

= mg + mW^2R

= (367)(9.8) + (367) (.84)^2 (14)

= 7221.9728

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# Homework Help: Angular velocity of a ferris wheel

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