Question

One of the rides found at carnivals is the rotating cylinder. The riders step inside the vertical cylinder and stand with their backs against the curved wall. The cylinder spins very rapidly, and at some angular velocity, the floor is pulled away. The thrill-seekers now hang like flies on the wall. If the radius of the cylinder is 5 m and the coefficient of static friction between the people and the wall is 0.4, what is the maximum period (in s) of rotation of the cylinder for the floor to be removed safely?

Answer 1

The maximum period of rotation of the cylinder is 2.84 seconds

Step-by-step explanation:

Radius of the cylinder, r = 5 m

Coefficient of static friction,
`\mu = 0.4`

The frictional force,
`F = \mu R`

The normal reaction, R is a centripetal force:

`R = (mv^(2) )/(r)`

`F = (\mu mv^(2) )/(r)`

The force due to gravity, F = mg

For equilibrium, the force due to gravity must equal the frictional force

`mg = (\mu mv^(2) )/(r)`

`gr = \mu v^(2)`

`9.8 * 5 = 0.4v^(2) \\v^(2) = (9.8*5)/(0.4) \\v = \sqrt{(9.8*5)/(0.4)}`

v = 11.07 m/s

The angular speed, w = v/r

w = 11.07/5

w = 2.214 rad/s

For a complete revolution through a circle, θ = 2π rad

w = θ/t

t = θ/w

t = 2π/2.214

t = 2.84 seconds