Question

Ferris wheel has a radius of 10.7m and completes one revolution every 42 seconds.
a. Draw a FBD for a rider on the Ferris wheel.
b. Write a ΣFy equation for the rider at the bottom of the wheel.

c. Write a ΣFy equation for the rider at the top of the wheel.
d. Find the apparent weight of a rider of mass 72 kg at the bottom of the wheel.
e. Find the apparent weight of a rider of mass 72 kg at the top of the wheel.

Answer 1

here Fg = gravitational force and Ft = centripetal force (this is the free body diagram of the situation when rider is at top of the ferris wheel)

ferris wheel completes one revolution in 42 sec

that means it covers distance equal to its circumference in 42 sec

speed = distance / time

speed = 2 * pi * radius / time

speed = 2 * pi * 10.7 / 42

speed = 1.6 m/s

ΣFy at the bottom will be total of all the vetical forces

only two forces are acting vertically at bottom most position weight and centripetal force and they are in same direction

so,

ΣFy = mg + mv^2 / r

ΣFy at top, again two forces are acting weight and centripetal force but they are in opposite direction as centripetal force always act toward radially outward direction

ΣFy = mg - mv^2 / r

apparent weight of a rider at bottom = mg + mv^2 / r

apparent weight of a rider at bottom = 72 * 9.8 + 72 * 1.6^2 / 10.7

apparent weight of a rider at bottom = 722.82 N

apparent weight of a rider at top = mg - mv^2 / r

apparent weight of a rider at top = 72 * 9.8 - 72 * 1.6^2 / 10.7

apparent weight of a rider at top = 688.37 N