Physics help needed ASAP?

Question

The cosmoclock 21 ferris wheel in japan has a diameter of 100 m. Its name comes from its 60 arms which can function as a second hand (so that it makes one revolution every 60 sec)

a) Find the speed of the passengers when the ferris wheel is rotating at this rate.

b) A passenger weighs 882 N at the weight-guessing booth on the ground. What is the apparent weight at hte highest and at the lowest point in the ferris wheel?

c) What would be the time for one revolution if the passenger's apparent weight at the highest point were zero?

d) What then would be the passenger's apparent weight at the lowest point?

Answer 1

r = 50 m
ω = 1 rpm = π / 30 rad/s
W = 882 N
m = 882 / 9.8 = 90 kg

a) v = r*ω = π * 50 / 30 = 5/3 * π ≈ 5.236 m/s

b)At the top of the ferris wheel, his weight is down. The centripetal acceleration is also pointing down (it always points to the center). However, the seat is holding him up, which is the apparent weight (the normal force)

remember, ac = v² / r

So

-W + N= -m * ac
-882 + N = -90 * (5.236)²/60
N = 840.876 N

At the bottom of the ride, his weight is still down, but the centripetal accelaration is now up! The normal force is still the apparent weight.

-W + N = +m ac
-883 + N = 90 * (5.236)²/60
N = 923.124 N

c) Use the first force equation, set N = 0, and solve for v.
-W = -m v² / r
882 = 90 * v² / 60
v = 24.249 m/s

d) Using the second force equation.....

-W + N = +m v² / 4

-882 + N = 90 * (24.249)²/60

N = 1764 N

Answer 2

a) Since the speed of the wheel is 1 rpm, we must calculate this distance: the circumference = 2pi*r = 100pi = 314.16 m.
The speed of the passengers therefore is 314.16 m per minute or 5.24 m per sec.