Question

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 15 m, and you use your watch to find that each loop around takes 23 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?

Answer 1

Step-by-step explanation:

From the given information:

radius = 15 m

Time T = 23 s

a) Speed (v) =
`(2 \pi r)/(T)`

`v = (2* \pi * 15)/(23)`

v = 4.10 m/s

b) The magnitude of the acceleration is:

`a = (v^2)/(r) \\ \\ a = ((4.10)^2)/(15)`

a = 1.12 m/s²

c) True weight = mg

Apparent weight = normal force

From the top;

the normal force = upward direction,

weight is downward as well as the acceleration.

true weight - normal force = ma

apparent weight =mg - ma

`(apparent \ weight)/(true \ weight) = ((mg - ma))/((mg))`

`=1- (1.12)/(9.8)`

= 0.886 m/s²

d)

From the bottom;

acceleration is upward, so:

apparent weight - true weight = ma

apparent weight = true weight + ma

`(apparent \ weight )/(true \ weight) =( mg + ma)/(mg)`

`(apparent \ weight )/(true \ weight) =1+ (a)/(g) \\ \\ = 1 + (1.12)/(9.8)`

`= 1 + (1.12)/(9.8) \\ \\`

= 1.114 m/s²