Question

While at the county fair, you decide to ride the Ferris wheel.Having eaten too many candy apples and elephant ears, you find themotion somewhat unpleasant. To take your mind off your stomach, youwonder about the motion of the ride. You estimate the radius of thebig wheel to be 15 , and you use your watch tofind that each loop around takes 25 .

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

a) The speed is 3.77 m/s

b) The acceleration is 0.948 m/s²

c) The ratio of your weight at the top to your weight standing on the ground is 0.903

d) The ratio of your weight at the bottom of the ride to your weight while standing on the ground is 1.097

Step-by-step explanation:

a) The speed of the wheel is equal to:

`v=rw`

Where

r = radius of the circle = 15 m

w = angular velocity

`w=(2\pi )/(T)`

`v=(2\pi r)/(T) =(2\pi *15)/(25) =3.77m/s`

b) The acceleration is:

`a=(v^(2) )/(r) =(3.77^(2) )/(15) =0.948m/s^(2)`

c) Applying the Newton´s second law:

`W_(true) -N=ma\\W_(true)-N=(mv^(2) )/(r) \\N=W_(app) =W_(true)-(mv^(2) )/(r)`

Dividing the expression by Wtrue:

`(W_(app))/(W_(true) ) =1-(mv^(2) )/(rW_(true)) \\(W_(app))/(W_(true) )=1-(v^(2) )/(rg) =1-(3.77^(2) )/(15*9.8) =0.903`

d) Applying the Newton´s second law:

`N-W_(true)=ma\\N-W_(true)=(mv^(2) )/(r) \\N=W_(app) =W_(true)+(mv^(2) )/(r)`

Dividing the expression by Wtrue:

`(W_(app))/(W_(true) ) =1+(mv^(2) )/(rW_(true)) \\(W_(app))/(W_(true) )=1+(v^(2) )/(rg) =1+(3.77^(2) )/(15*9.8) =1.097`

Answer 2

a. 0.0628 m

b. 0.000263 m/s2

c.99.997%

d.100.003%

Step-by-step explanation:

Suppose the radius is 15 m and each loop takes 25 minutes (25*60 = 1500 seconds) to complete. The circumference of the Ferris wheel would be

C = 2πR = 2π*15 = 94.25 m

The your speed is the length of each loop per unit of time to complete each loop

v = C/t = 94.25 / 1500 = 0.0628 m/s

b. Since the circular motion is constant, the only acceleration from the motion is the centripetal acceleration, which can be calculated as the following:

`a_c = v^2/r = 0.0628^2/15 = 0.000263 m/s^2`

c. Let g = 9.8 m/s^2. At top of the ride, the centripetal acceleration would be pushing you up, while gravity pushes you down, so the weight you experience (compare to standing on ground would be):

`(g-a_c)/(g) = (9.8 - 0.000263)/(9.8) = 0.99997 = 99.997 \%`

d. At the bottom of the ride, both centripetal acceleration and gravity would be pushing you downward, making you feel like

`(g+a_c)/(g) = (9.8 + 0.000263)/(9.8) = 1.00003 = 100.003 \%`