Physics C - Rotational motion/forces

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SUMMARY

The discussion centers on a physics problem involving rotational motion and forces in an amusement park ride scenario. The rider, with a mass of 50 kg, remains against the wall of a spinning cylinder with a radius of 5 meters and a rotational speed of 2 radians/second, due to the static friction coefficient of 0.6. The upward force preventing the rider from falling is the frictional force, calculated to be 500 N, while the normal force acting on the rider is 833 N. For a 100 kg rider, the analysis confirms that they would also not slide down the wall due to sufficient frictional force at the same rotational speed.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Knowledge of centripetal force calculations
  • Familiarity with static friction concepts
  • Ability to draw and interpret free-body diagrams
NEXT STEPS
  • Study the principles of centripetal acceleration and its applications
  • Learn about static versus kinetic friction and their implications in rotational systems
  • Explore advanced topics in rotational dynamics, including torque and angular momentum
  • Investigate real-world applications of rotational motion in amusement park rides
USEFUL FOR

Physics students, educators, and anyone interested in understanding the mechanics of rotational motion and forces in practical scenarios, particularly in amusement park ride design and safety analysis.

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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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