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/s

2

.

Given g = 9.8 m/s

2

, the coefficient µ =

0.614 of static friction between a person and

the wall, and the radius of the cylinder R =

5.2 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 73 kg person

from falling down.

Relevance

To keep from sliding down, the minimum available friction force must be equal to the weight of the rider.

mg = μN

mg = μ(mω²R)

g = μ(ω²R)

g/μR = ω²

√(g/μR) = ω

√(g/μR) = 2π/T

T = 2π/√(g/μR)

T = 2π/√(9.8/(0.614(5.2))

T = 3.586350...  s

The question numerals are limited to two significant digits by the values of g and R.

T ≈ 3.6 seconds