Amusement park ride gravitation problem

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In the amusement park ride scenario, passengers are expected to remain against the wall of a rotating cylinder after the floor drops away, relying on friction to prevent sliding. The static coefficient of friction for clothing against steel is between 0.64 and 1.0, while the kinetic coefficient ranges from 0.40 to 0.70. The calculations attempted involve determining the necessary velocity for passengers to stick to the wall, but an error is identified in assuming the normal force equals the passenger's weight. The correct approach requires using the frictional force equation and recognizing that the normal force is influenced by the centripetal acceleration due to rotation. The discussion seeks clarification on the misapplication of these concepts without requesting a direct answer.
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Homework Statement



In an old-fashioned amusement park ride, passengers stand inside a 5.4-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 0.64 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 kg allowed."


Homework Equations


f(fric)=(us)(n)
Fr=n=mv^2/r
Fz=f=mg
Ft=0
w=v/r


The Attempt at a Solution


f=(0.64)(9.8*30)=188.16N
188.16N=(30)(v)^2/2.7, v=4.1151m/s
4.115m/s/2.7=1.524rad
1.524rad/1s=1rev/2piradians=60sec/1min=143.65rpm
This answer is not correct, however. Any guidance is appreciated. I don't want the answer, just what I am doing incorrectly.
 
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holmeskaei said:

Homework Statement



In an old-fashioned amusement park ride, passengers stand inside a 5.4-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 0.64 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 kg allowed."


Homework Equations


f(fric)=(us)(n)
Fr=n=mv^2/r
Fz=f=mg
Ft=0
w=v/r


The Attempt at a Solution


f=(0.64)(9.8*30)=188.16N use your 3rd relevant equation, f(fric)= mg. You are incorrectly assuming that the Normal force is the person's weight. [/color]
188.16N=(30)(v)^2/2.7, v=4.1151m/s
4.115m/s/2.7=1.524rad
1.524rad/1s=1rev/2piradians=60sec/1min=143.65rpm
This answer is not correct, however. Any guidance is appreciated. I don't want the answer, just what I am doing incorrectly.
Please see above in red[/color]
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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