Question

2. A brav e 90 kg man, named Steve, agrees to give a Gravitron a test ride for the first time. The Gravitron has a diameter of 10 m. It is programed to spin faster and faster until it achieves ution in 15 s. Once it maintains this speed, the floor of the Gravitron lowers itself, s. Once it leaving the riders to hang in midair. raw a picture and an FBD of Steve as if you are looking at him head on at one point in time, while the Gravitron is spinning. (a) D (b) Determine Steve's speed as the Gravitron is spinning. (c) What is the centripetal acceleration of Steve?

Answer 1

Gravitron is doing one revolution in 15 s.

so, w = (1 / 15) rps

w = (1 / 15)*2*pi = 0.4188 rad / s

(b)

Speed of steve,

v = w*r

v = 0.4188 * (10 / 2)

v = 2.094 m/s

(c)

Centripetal acceleration of steve,

a_c = v^2 / r = (2.094)^2 / 5

a_c = 0.877 m/s^2

(d)

Normal force exerted by walls,

N = m*v^ / r = m*a_c

N = 90*0.877

N = 78.95 N

(e)

When steve starts to slip downwards with 4 m/s^2,

gravitational force on steve, Fg = m(g - a)

Fg = 90 (9.8 - 4)

Fg = 522 N