Question

In an old-fashioned amusement park ride, passengers stand inside a 3.0-m-tall, 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 0.60 to 1.0 and a kinetic coefficient in the range 0.40 to 0.70. What is the minimum rotational frequency, in rpm, for which the ride is safe?

Answer 1

Answer:

$\omega = 2.56 rad/s$

Explanation:

As the cylinder rotates the centripetal force on all the passengers is due to normal force due to the wall

So here we can say

$N = m\omega^2 R$

now when floor is removed all the passengers are safe because here friction force on the passenger is counter balanced by the weight of the passengers

so we can say

$F_f = mg$

$\mu_s F_n = mg$

$\mu_s (m\omega^2 R) = mg$

$\mu_s \omega^2 R = g$

$\omega = \sqrt{\frac{g}{\mu_s R}}$

for minimum rotational speed we have

$\omega = \sqrt{\frac{9.8}{0.60(2.5)}$

$\omega = 2.56 rad/s$