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An amusement park ride consists of a round room. Once you are placed against the outer wall, the room begins to spin. After the room attains a critical rotational speed, the floor drops out, but you remain against the wall. The radius of the room is 30m, and the coefficient of friction between you and the wall is u = .02. What is the minimum period of rotation that the room must spin in order for you not to fall? What restriction shold the park enforce regarding the maximum weight of the rider?

Okay, well I know that

a_c = (v^2)/r

v = 2(pi)(r)/(T)

a_c = 4(pi^2)r/(T^2)

F_c = m(v^2)/r

F_c = (u_s)(F_n) = (u_s)(m)(g) = (m)(v^2)/r

v = sqrt ((u_s)(g)(r))

I'm assuming "period of rotation" means centripetal acceleration. In that case,

v = sqrt ((u_s)(g)(r)) = sqrt ((.02)(9.8m/s^2)(30 m)) = sqrt (5.88 m^2/s^2) = 2.425 m/s

a_c = ((2.425 m/s)^2)/(30m) = .196 m/s^2

From this equation, F_c = (u_s)(F_n) = (u_s)(m)(g) = (m)(v^2)/r, we see that the m's cancel out, so in regards to the weight of the rider, it shouldn't matter when he/she rides the ride.

I was just wondering if someone could check my work and see if I got the right answer. If I did not, could someone please give me a lead of what to do next, but not tell me the answer until I get it myself? Thanks