Question

An amusement park ride consists of a large vertical cylinder that spins about its axis fast enough that a person inside is stuck to the wall and does not slide down when the floor drops away. The acceleration of gravity is 9.8 m/s2.

Given g = 9.8 m/s2, the coefficient μ = 0.569 of static friction between a person and the wall, and the radius of the cylinder R = 5.4 m. For simplicity, neglect the person’s depth and assume he or she is just a physical point on the wall. The person’s speed is

v = (2πR/T)

where T is the rotation period of the cylinder (the time to complete a full circle).

Find the maximum rotation period T of the cylinder which would prevent a 43 kg person from falling down.

Answer in units of s.

Answer 1

Given μ=0.569, R=5.4m, m=43kg, g=9.8m/s²
Person won't slide when frictional force ≥ weight, as given by:
F = μmv²/R = mg

v=√(g*R/μ)

Now solve for min. velocity, then equate to (given) cylinder speed:
v = √(gR/μ) = 2πR/T
Then solve for max. period T:
T = 2π√(μR/g)

T = 2π√(0.569*5.4/9.8)

T=3.518 S