Question

At Magic Mountain there is a ride in which people stand up against the inside wall of a large cylinder of radius 3 m. The cylinder is then spun with a velocity of 15 m/s at which point the floor drops away leaving the riders suspended against the wall. What is the minimum coefficient of static friction between the rider's clothing and the wall in order to keep the rider from sliding down?

Answer 1

0.1308

Step-by-step explanation:

To keep the rider from sliding down, then the friction force
`F_f` must at least be equal to gravity force
`F_p`

`F_f = F_p`

`\mu N = mg`

where μ is the coefficient, N is the normal force acted by the rotating cylinder, m is the mass of a person and g = 9.81 m/s2 is the gravitational acceleration.

According to Newton's 3rd and 2nd laws, the normal force would be equal to the centripetal force
`F_c`, which is the product of centripetal acceleration
`a_c` and object mass m

`N = F_c = a_cm`

Therefore

`\mu a_cm = mg`

`\mu a_c = g`

The centripetal acceleration is the ratio of velocity squared and the radius of rotation

`a_c = v^2/r = 15^2 / 3 = 75 m/s^2`

Therefore

`\mu = g/a_c = 9.81 / 75 = 0.1308`