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 for which the ride is safe?

Answer 1

Answer:

0.407 rev/s

Explanation:

The centripetal force on the passenger equals the normal force acting on the passenger due to the hollow steel cylinder.

So mrω² = N

Also, the frictional force equals the passenger's weight

F = μN = mg

So. μmrω² = mg ⇒ ω = √g/μr.

where ω is the angular frequency,

ω = √(g/μr)

Given that r = radius of cylinder = 5.0/2 = 2.5 m and μ = 0.6 which is the minimum value of the coefficient of static friction given,

ω = √(g/μr) = √(9.8/(0.6 × 2.5)) = √(9.8/1.5) = 2.556 rad/s

ω = 2.556 rad/s ÷ 2π = 0.407 rev/s

Answer 2

Answer:

Minimum rotational frequency for which the ride is safe is 24.402 rpm

Explanation:

For the rotational frequency to be minimum, the frictional force acting on the passengers by the wall should be maximum

$f_{r} = \mu R\\R = mr\omega^{2} \\f_{r} = mg$

$mg =\mu mr \omega^{2} \\g = \mu r \omega^{2} \\g = 9.8 m/s^{2} \\diameter, d= 5.0 m\\radius, r = d/2 = 5/2\\r = 2.5 m\\\mu = 0.6\\9.8 = 0.6 * 2.5 * \omega^{2} \\\omega^{2} =9.8/1.5\\\omega^{2} =6.53\\\omega = \sqrt{6.53} \\\omega = 2.56 rad/sec$

Multiply by 60/2π to convert to rev/min

$\omega = 2.56 * (60/2\pi )\\\omega = 24.402 rpm$