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/s 2 . Given g = 9.8 m/s 2 , the coefficient µ = 0.521 of static friction between a person and the wall, and the radius of the cylinder R = 4.4 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 40 kg person from falling down. Answer in units of s.

Answer 1

Let N be the normal force that forces the person against the wall.

Then u N = m g is the frictional force supporting the person's weight

and N = m g / u

also, N = m v^2 / R is the normal force providing the centripetal acceleration

So, m g / u = m v^2 / R

v^2 = g R / u

since v = 2 pi R T

4 pi^2 R^2 T^2 = g R / u and T^2 = g / (4 u pi^2 R)

T = 1/ (2 pi) (g /(u R))^1/2 = .159 * (9.8 m/s^2 / (.521 * 4.4 m)) ^1/2

T = .68 / s

Do you see any thing wrong here?

T should have units of seconds not 1 / seconds

v should be 2 * pi * R / T where T is the time for 1 revolution

So you need to make that correction in the above formula for v.