Amusement Park Centripetal Force Around A Loop

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SUMMARY

The forum discussion centers on calculating the forces acting on a rider in The Roundup amusement park ride, which features an 18.0 m-diameter rotating ring that completes a rotation every 4.90 seconds. Key calculations involve determining the normal force exerted by the ring on a rider with a mass of 58.0 kg at both the top and bottom of the ride. The discussion emphasizes the need to consider both gravitational force and centripetal acceleration when analyzing forces at different points in the ride. Additionally, the longest rotation period that prevents riders from falling off at the top is a critical aspect of the analysis.

PREREQUISITES
  • Centripetal force and acceleration concepts
  • Basic physics of forces and motion
  • Understanding of gravitational force
  • Knowledge of rotational motion equations
NEXT STEPS
  • Calculate the centripetal acceleration using the formula a = v²/R
  • Determine the normal force at the top and bottom of the ride using F = m(g ± a)
  • Explore the concept of rotational dynamics in amusement park rides
  • Investigate the effects of varying rotation periods on rider safety
USEFUL FOR

Physics students, amusement park ride designers, and engineers interested in the dynamics of rotational motion and safety considerations in ride design.

GoSS190
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In an amusement park ride called The Roundup, passengers stand inside a 18.0 m-diameter rotating ring. After the ring has acquired sufficient speed, it tilts into a vertical plane. Suppose the ring rotates once every 4.90 s. and the rider's mass is 58.0 kg.

A.) With how much force does the ring push on her at the top of the ride?

B.) With how much force does the ring push on her at the bottom of the ride?

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

If anyone could help me with these questions that would be great. Thanks



I tried this using the equations to find v = (2pi(r)) / T

but then i realized that the velocity is different at the top than at the bottom. I am stumped as to how to find the velocity then find the force.

I think the equation for force at the top is (m(vtop)2) / R

Can anyone help me out with this though please
 
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GoSS190 said:
In an amusement park ride called The Roundup, passengers stand inside a 18.0 m-diameter rotating ring. After the ring has acquired sufficient speed, it tilts into a vertical plane. Suppose the ring rotates once every 4.90 s. and the rider's mass is 58.0 kg.

A.) With how much force does the ring push on her at the top of the ride?

B.) With how much force does the ring push on her at the bottom of the ride?

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

If anyone could help me with these questions that would be great. Thanks

I tried this using the equations to find v = (2pi(r)) / T

but then i realized that the velocity is different at the top than at the bottom. I am stumped as to how to find the velocity then find the force.

I think the equation for force at the top is (m(vtop)2) / R

Can anyone help me out with this though please

No. That's not what's going on.

Draw a force diagram. You have 2 forces acting on the 58 kg person.

There's weight - which always acts down and there is the radial acceleration from the rotation of the ride. So there are two forces. What is the difference between looking at the forces between the top and the bottom then?
 
The force at the top would be adding to each other while at the bottom they would be opposing. But how do u find the normal force?
 

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