In an old-fashioned amusement park ride, passengers stand inside a 5.4--diameter

Question

In an old-fashioned amusement park ride, passengers stand inside a 5.4--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 allowed.

Explanation / Answer

The frictional force must balance the riders weight

Ff = uFn and the normal force, in this case, is the centripetal force

So, umv^2/r = mg (mass cancels)

The least value for static friction is .62, so

(.62)(v^2)/(2.45) = (9.8)

v = 6.22 m/s

Then apply v = (omega)(r)

Omega = 6.22/2.45 = 2.54 rad/sec

Converting that to rpm (2)(pi) rad per rev and 60 sec per min, gives us 24.3 rpm

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