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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 min rotational velocity in revolutions per second for which the ride is safe?

User Inshan
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1 Answer

6 votes

Answer:

1.25 rev/s

Step-by-step explanation:

N = mv^2/r (normal force )

f = μN [Frictional force ]

f = mg

μN = mg

μ(mv^2/r) = mg

v = √(rg/μ)

min rotational velocity

v = √(rg/μ) = √(2.5 * 9.8/0.40) = 7.83 rad/sec

= 7.83/2π = 1.25 rev/s

User Antony Shumskikh
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