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

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The discussion focuses on calculating the coefficient of friction necessary to keep riders from falling in a 3-meter cylinder spinning at 5 radians per second. The user initially calculated the velocity as 15 m/s and derived the centripetal force as 75M, equating it to the force of friction. However, it was clarified that in this scenario, the centripetal force acts as the normal force, not the frictional force. The user acknowledges the mistake and indicates a better understanding of the problem. The conversation highlights the importance of correctly identifying forces in physics problems.
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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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