How Does Physics Explain Forces on The Roundup Amusement Park Ride?

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SUMMARY

The discussion focuses on calculating the forces acting on a rider in The Roundup amusement park ride, which features a 16.0 m-diameter rotating ring. The ride completes one rotation every 4.10 seconds, and the rider's mass is 54.0 kg. Key calculations involve determining the force exerted by the ring on the rider at both the top and bottom of the ride, using the equations of motion and centripetal force. The correct angular velocity is derived as 1.53 radians/second, and the necessary equations for force calculations are established.

PREREQUISITES
  • Understanding of centripetal force and gravitational force
  • Familiarity with angular velocity and linear velocity calculations
  • Knowledge of basic physics equations related to circular motion
  • Ability to manipulate equations involving mass, acceleration, and radius
NEXT STEPS
  • Study the derivation of centripetal acceleration formulas
  • Learn how to apply Newton's second law in circular motion contexts
  • Explore the concept of angular momentum in rotating systems
  • Investigate the effects of varying rotation speeds on forces experienced by riders
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Physics students, educators, and amusement park engineers interested in understanding the dynamics of rotating rides and the forces acting on passengers.

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Homework Statement



In an amusement park ride called The Roundup, passengers stand inside a 16.0 m-diameter rotating ring. After the ring has acquired sufficient speed, it tilts into a vertical plane.

Part A: Suppose the ring rotates once every 4.10 s. If a rider's mass is 54.0 kg, with how much force does the ring push on her at the top of the ride?

Part B: Suppose the ring rotates once every 4.10 s. If a rider's mass is 54.0 kg, with how much force does the ring push on her at the bottom of the ride?

Part C: What is the longest rotation period of the wheel that will prevent the riders from falling off at the top?


Homework Equations



Fr=nr+(Fg)r=n+mg=mv^2/r
w=v/r
1 rev=2pi radians
Might be more needed..

The Attempt at a Solution


4.10s/1 rev=1 rev/2pi radians=6.44 rad/s=w
w=v/r=6.44rad/s/8m=51.52m/s
Then do I plug it into the Fr equation? I don't have the acceleration..
Any help or how to think about it would be advantageous.
Thanks.
 
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Those calcs don't look right to me.
You could do w = 1 rev/4.1s = 2*pi/4.1 = 1.53 radians/s
Or just work with v = 2*pi*r/T.
The n+mg=mv^2/r looks good.
 

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