What Is the Minimum RPM Needed for Safety on This Amusement Park Ride?

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SUMMARY

The minimum angular speed required for safety on an amusement park ride, where passengers are held against the wall of a rotating cylinder, is determined by the balance of forces acting on the passengers. The static coefficient of friction against steel ranges from 0.63 to 1.0, which is crucial for ensuring that passengers do not slide down when the floor drops. The critical velocity can be calculated using the formula SQRT(rg), where 'r' is the radius of the cylinder and 'g' is the acceleration due to gravity. Properly accounting for the normal force and friction in the free body diagram is essential for solving this problem accurately.

PREREQUISITES
  • Understanding of circular motion dynamics
  • Knowledge of static and kinetic friction coefficients
  • Ability to construct and analyze free body diagrams (FBD)
  • Familiarity with Newton's second law (F=ma)
NEXT STEPS
  • Calculate the minimum angular speed using the formula for critical velocity SQRT(rg)
  • Explore the effects of varying the static coefficient of friction on ride safety
  • Investigate the relationship between radius and angular speed in circular motion
  • Review examples of free body diagrams in rotational systems
USEFUL FOR

Physics students, mechanical engineers, amusement park ride designers, and safety analysts interested in the dynamics of rotating systems and frictional forces.

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Homework Statement



In an old-fashioned amusement park ride, passengers stand inside a 5.5m-diameter hollow steel cylinder with their backs against the wall. The cylinder begins to rotate about a vertical axis. Then the floor on which the passengers are standing suddenly drops away! If all goes well, the passengers will "stick" to the wall and not slide. Clothing has a static coefficient of friction against steel in the range 0.63 to 1.0 and a kinetic coefficient in the range 0.40 to 0.70. A sign next to the entrance says "No children under 30 kg allowed." What is the minimum angular speed, in rpm, for which the ride is safe?

Homework Equations



Critical velocity=SQRT(rg)
F=ma

The Attempt at a Solution



I know how to solve these kinds of problems in the absence of friction, but I don't understand what I need to do with these coefficients. Friction points tangentially to the circular motion in the free body diagram, but how does that influence the (Fnet)y?
 
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JoshMP said:

Homework Statement



In an old-fashioned amusement park ride, passengers stand inside a 5.5m-diameter hollow steel cylinder with their backs against the wall. The cylinder begins to rotate about a vertical axis. Then the floor on which the passengers are standing suddenly drops away! If all goes well, the passengers will "stick" to the wall and not slide. Clothing has a static coefficient of friction against steel in the range 0.63 to 1.0 and a kinetic coefficient in the range 0.40 to 0.70. A sign next to the entrance says "No children under 30 kg allowed." What is the minimum angular speed, in rpm, for which the ride is safe?

Homework Equations



Critical velocity=SQRT(rg)
F=ma

The Attempt at a Solution



I know how to solve these kinds of problems in the absence of friction, but I don't understand what I need to do with these coefficients. Friction points tangentially to the circular motion in the free body diagram, but how does that influence the (Fnet)y?

Could you show us your FBD? If the force of gravity would normally pull the people down out of the ride, what force is resisting their movement?
 
berkeman said:
Could you show us your FBD? If the force of gravity would normally pull the people down out of the ride, what force is resisting their movement?

The normal force is what keeps people from falling- when the veolcity decreases such that the normal force is less than 0, the people would fall off the ride.

My FBD has gravity pointing down, normal force pointing down, and friction pointing to the left.
 
JoshMP said:
The normal force is what keeps people from falling- when the veolcity decreases such that the normal force is less than 0, the people would fall off the ride.

My FBD has gravity pointing down, normal force pointing down, and friction pointing to the left.

The forces on the FBD don't sound right. The normal force points toward the center of the circular motion, and is based on the centripetal acceleration. The friction force is what opposes the falling-downward motion of the person, so it has to point _____
 
berkeman said:
The forces on the FBD don't sound right. The normal force points toward the center of the circular motion, and is based on the centripetal acceleration. The friction force is what opposes the falling-downward motion of the person, so it has to point _____

Up? I thought friction points tangential to circular motion?
 
JoshMP said:
Up? I thought friction points tangential to circular motion?

Up is correct in this problem. Now re-draw the forces on your FBD, and write the sum of forces equations in the vertical direction.

I have to bail for a few hours. Keep at it!
 
I'm stuck. I've got an inequality set up, but I can't isolate the velocity. Am I even on the right track?
 
JoshMP said:
I'm stuck. I've got an inequality set up, but I can't isolate the velocity. Am I even on the right track?

Beats me. Show us your FBD and equations.
 
My FBD consists of Gravity pointing down, normal force pointing down, and static friction pointing up. (The net force along the radial axis is equal to Gravity + Normal force - static friction. This net force is equal to the mass multiplied by the radial acceleration, which = v^2/r. Using this equation I solved for the normal force, then used the inequality of n greater than or equal to 0. The inequality looks like this:

(m(v^2/r - g))/(1-mu static) is greater than or equal to 0. I need to solve for v. Any ideas?
 
  • #10
JoshMP said:
My FBD consists of Gravity pointing down, normal force pointing down, and static friction pointing up.

The normal force is always perpendicular to the surface that an object is being pushed against, no matter if it is by gravity or any other force, or to keep it on a circular path.
 

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