Barrel of Fun Carnival Ride, and the relationship between Friction and ω

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SUMMARY

The "Barrel of Fun" carnival ride operates on the principles of centripetal acceleration and static friction. When the angular speed is doubled from ω1 to ω2 = 2ω1, the frictional force F2 required to keep a person against the wall increases due to the relationship between angular speed and centripetal force. The static friction force must counteract the gravitational force acting on the rider, which remains constant. Therefore, the frictional force F2 can be calculated using the formula F2 = m * (r * ω2²), where m is the mass of the rider and r is the radius of the barrel.

PREREQUISITES
  • Centripetal acceleration concepts
  • Static friction and its relationship with normal force
  • Angular motion equations
  • Basic physics of forces and motion
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  • Study the relationship between angular speed and centripetal force in circular motion
  • Learn about the equations governing static friction, particularly FF = μFN
  • Explore the effects of varying angular speeds on forces in rotating systems
  • Investigate real-world applications of centripetal acceleration in amusement park rides
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Physics students, mechanical engineers, and anyone interested in the dynamics of rotating systems and amusement ride design.

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"Barrel of Fun" Carnival Ride, and the relationship between Friction and ω

Homework Statement



A ”Barrel of Fun” consists of a large vertical cylinder that spins about its axis fast enough so that any person inside will be held against the wall.

Assume: At an angular speed ω1, an upward frictional force F1 holds a person against the wall without slipping.

What is the friction force F2, if the angular speed is doubled;i.e., ω2 = 2ω1.

Homework Equations



This is what has me stumped. I have no idea which equations I should bee looking at!

F=ma
FF= μFN

But all of these are accounting for linear motion, not circular. The rider will be suspended by a static Frictional Force that must counteract the pull of gravity. I have no idea what the relationship is between Force of Friction and Angular Speed.

The Attempt at a Solution

 

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The person is accelerating centripetally (inward toward the center of the barrel, in the horizontal direction), but, as you noted, the person is not accelerating in the vertical direction, so the frictional force is just equal to the person's weight. Remember that this static friction force (static since there is no slipping) is less than or equal to uN. With this information, what is the static friction force when the angular speed is doubled?
 

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