While at the county fair, you decide to ride the Ferris wheel. Having eaten too many candy apples and elephant ears, you find the motion somewhat unpleasant. To take your mind off your stomach, you wonder about the motion of the ride. You estimate the radius of the big wheel to be 17 m , and you use your watch to find that each loop around takes 27 s .

a. What is your speed?
b. What is the magnitude of your acceleration?
c. What is the ratio of your weight at the top of the ride to yourweight while standing on the ground?
d. What is the ratio of your weight at the bottom of the ride toyour weight while standing on the ground?

a) 3.97 m/s

b) 0.93 m/s²

c) 0.905

d) 1.905

Explanation:

Given,

r = 17 m

t = 27 s

Speed is gotten using

V = 2πr/t

V = 2π*17/27

V = 106.828/27

V = 3.97 m/s

Acceleration would be v²/r

a = 3.97²/17

a = 15.7609/17

a = 0.93 m/s²

Because of the circular motion of the wheel, the net force on the rider at the top of the wheel is downwards. Thus,

Σ F(y) = F(n) - mg = -ma(r) so that

F(n) - mg = -ma(r)

F(n) = mg - ma(r)

F(n) = m[g - a(r)]

F(n) = m(9.8 - 0.93)

F(n) = 8.87m

the ratio of apparent weight to true weight at the top is

W(app) / W = F(n) / mg

8.87*m / 9.8*m

W(app) / W = 0.905

and yet again, as a result of the circular motion, the net force at the bottom of the wheel is upwards.

Σ F(y) = F(n) - mg = ma(r)

F(n) = mg + ma(r)

F(n) = m[g + a(r)]

F(n) = m(9.8 + 0.93)

F(n) = 10.73m

The ratio of apparent weight to true weight at the bottom is

W(app) / W = F(n) / mg

10.73*m / 9.8*m

W(app) / W = 1.095

the answer is the index of refraction of the second medium is higher. i just took the test.