How Is the Coefficient of Friction Calculated for an Amusement Park Ride?

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Homework Help Overview

The discussion revolves around calculating the coefficient of friction required to keep individuals from falling in an amusement park ride scenario, where people are spun in a cylinder at a specified angular velocity.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to calculate the coefficient of friction using centripetal force and friction force equations, while others question the setup of the problem, particularly the orientation of the cylinder and the forces acting on the riders.

Discussion Status

Some participants are exploring the implications of the cylinder's orientation and the forces involved, with one participant indicating a misunderstanding regarding the role of friction versus normal force in this context. The discussion is ongoing with clarifications being sought.

Contextual Notes

There is a potential ambiguity regarding whether the cylinder is vertical or horizontal, which affects the interpretation of forces acting on the riders. The original poster's calculations may be based on an assumption that needs further examination.

chicagobears
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Homework Statement


In an amusement park ride, people are spun at 5 radians per sec in a 3 m cylinder. What is the coefficient of friction to prevent people from falling down?


Homework Equations


Centripetal Force= (mv^2)/r
Force of friction=(mu)(mg)
Velocity=(r)(ω)

The Attempt at a Solution


V=(3 m)(5 rad/sec)=15 m/s
Centripetal Force=(M(15^2))/3=75M
Centripetal force=force of friction...
75M=9.8(mu)M
75/9.8=(mu)=7.653?
 
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No need to answer. I get this now...
 
Is it a vertical cylinder of radius 3 m? (A 3m cylinder suggests the diameter is 3.)
With the people above the floor of the cylinder?
If so, the centripetal force is the normal force, not the force of friction.
 
yea that was my mistake...
 

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