Question

In an old-fashioned amusement park ride, passengers stand inside a 5.0-m-diameter hollow steel cylinder with their backs against the wall. The cylinder begins to rotate about a vertical axis. Then the floor on which the passengers are standing suddenly drops away! If all goes well, the passengers will "stick" to the wall and not slide. Clothing has a static coefficient of friction against steel in the range 0.62 to 1.0 and a kinetic coefficient in the range 0.40 to 0.70. A sign next to the entrance says "No children under 30 kg allowed."

Q: What is the minimum angular speed, in rpm, for which the ride is safe?

Answer 1

As the cylinder begins to rotate about a vertical axis, the people feel the cylinder’s vertical, interior wall pushing on their back. This force is the centripetal force. The centripetal force is the force that causes the friction.

Fc = m * v^2/r

F friction = µ * m * v^2/r

The friction force prevents the people from sliding downward. Their weight causes them to slide downward. Since the people do not slide downward, the friction force must equal their weight.

µ * m * v^2/r = m * g

µ * v^2/r = g

v = (r * g ÷ µ)^0.5

As the velocity increases, the people feel like they are “glued to the wall. So the static coefficient of friction should be used.

Radius = 5 ÷ 2 = 2.5 m

range 0.62 to 1.0

To be safe, I would use the lower value.

v = (2.5 * 9.8 ÷ 0.62)^0.5 ≈ 6.29 m/s

OR

For 1.0

v = (2.5 * 9.8 ÷ 1)^0.5 ≈ 4.95 m/s

This is the linear velocity of a person on the outer edge of the circle.

Angular velocity = linear velocity ÷ radius

Angular velocity = 6.29 ÷ 2.5 = 2.51 rad/s

= 2.51/2pi *60 rpm

= 24 rpm