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?

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

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

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

momentum of soft ball three times less than baskey ball

it is a negative acceleration