A passenger with mass 85 kg rides in a Ferris wheel. The seats travel in a circle of radius 35 m. The Ferris wheel rotates at constant speed and makes one complete revolution every 25 s. Calculate the magnitude and direction of the net force exerted on the passenger by the seat when she is
(a) one-quarter revolution past her lowest point and
(b) one quarter revolution past her highest point
Thank you in advance! :)
(a) one-quarter revolution past her lowest point and
(b) one quarter revolution past her highest point
Thank you in advance! :)
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first let's use the data to find the speed of the Ferris wheel:
v = dist/time = 2 pi r/25s = 2*pi*35m/25s = 8.8m/s
in both cases, she is level with the center of the Ferris wheel
in the vertical direction, her weight is balanced by the vertical component of the normal force, so we have
Nv = mg
in the horizontal direction, the horizontal component of the normal force exerts a forceof mv^2/r
when she is at the level of the center of the wheel, these forces are perpendicular so the total magnitude of force is
force = sqrt[(mg)^2 +(mv^2/r)^2]
substituting numbers I get 854.9N
the direction is given by tan(theta) = Nv/Nh = mg / mv^2/r = g r/v^2 = 4.43
theta = 77.3 deg up from the horizontal pointing toward the center of the circle
v = dist/time = 2 pi r/25s = 2*pi*35m/25s = 8.8m/s
in both cases, she is level with the center of the Ferris wheel
in the vertical direction, her weight is balanced by the vertical component of the normal force, so we have
Nv = mg
in the horizontal direction, the horizontal component of the normal force exerts a forceof mv^2/r
when she is at the level of the center of the wheel, these forces are perpendicular so the total magnitude of force is
force = sqrt[(mg)^2 +(mv^2/r)^2]
substituting numbers I get 854.9N
the direction is given by tan(theta) = Nv/Nh = mg / mv^2/r = g r/v^2 = 4.43
theta = 77.3 deg up from the horizontal pointing toward the center of the circle