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

[chicagobears]

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?

 

[chicagobears]

No need to answer. I get this now...

 

[Delphi51]

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.

 

[chicagobears]

yea that was my mistake...

 

[asteriatic]

I would first clarify the units used in the problem. Radians per second is not a unit of speed, so I would ask for clarification on the units of the velocity. Additionally, I would also clarify if the 3 m cylinder is the radius or the diameter of the cylinder.

Assuming that the velocity is actually 5 m/s and the cylinder has a radius of 3 m, the coefficient of friction can be calculated using the equations provided. However, it is important to note that the coefficient of friction is dependent on the materials involved and the surface conditions, so it may not be a fixed value. It would also be important to consider the safety measures in place on the amusement park ride to prevent people from falling down, as the coefficient of friction alone may not be enough.