Physics C - Rotational motion/forces

AI Thread Summary
The discussion revolves around a physics problem involving a rotating amusement park ride where a rider remains against the wall as the floor lowers. The upward force preventing the rider from falling is primarily provided by static friction, which must counteract the gravitational force acting on the rider. Calculations indicate that the normal force acting on a 50 kg rider is 833 N, while the frictional force is 500 N, sufficient to keep the rider in equilibrium. For a 100 kg rider, the increased weight would likely exceed the maximum static friction force, suggesting that this rider would slide down the wall. Understanding the balance of forces and the role of friction is crucial in analyzing the scenario.
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Homework Statement



An amusement park ride consists of a rotating cylinder with rough canvas walls. The rider enters the cylinder and stands on the floor as the cylinder starts spinning, and as it spins, the floor lowers but the rider stays against the wall and does not slide down with it. The mass of the rider is 50 kg, the radius of the cylinder is 5 meters, the rotational speed is 2 radians/second, and the coefficient of static friction is 0.6.

a) Calculate the upward force that allows the rider from falling when the floor lowers, and state what provides that force.

b) At the same rotational speed, would a person of 100 kg slide down the wall? Explain.

Homework Equations



F=ma
Fcentripetal = mv2/r

The Attempt at a Solution



I drew a free-body diagram and got 500 N for weight, which would mean 500 N for friction since they are in opposite directions and the rider is in equilibrium. The normal force is 833 N.

For part a) I think the answer is the frictional force, but I don't know the exact reason.

Part b) I would assume that it won't affect it. Am I right?
 
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