In a popular amusement park ride, a rotating cylinder of radius R = 3.90 m is set in rotation at an angular speed of 6.50 rad/s, as in the figure shown below. The floor then drops away, leaving the riders suspended against the wall in a vertical position. What minimum coefficient of friction between a rider's clothing and the wall is needed to keep the rider from slipping? Hint: Recall that the magnitude of the maximum force of static friction is equal to ?sn, where n is the normal force

Since the riders do not move away or towards the center of the cilinder, the centripetal and centrifugal forces must be the same. In this case, this force will act as the normal force of the riders:

Fc = N = m r ω^2

Now we look for the minimum coefficient of static friction between the clothes of the riders and the cilinder to keep the riders from slipping, this condition can be expressed as:

Ff = Nμ_s = mg

Replacing the expression for N:

m r ω^2 μ_s = mg

Solving for μ_s

μ_s = (mg)/( m r ω^2 ) = (g / r ω^2 )

μ_s = 9.81 / (3.9 * 6.5^2) = 0.059

okpalawalter8 okpalawalter8 · Answer 2 · 2022-04-19T12:37:54+02:00

The minimum coefficient of friction is mathematically given as

μs = 0.059

What minimum coefficient of friction between a rider's clothing and the wall is needed to keep the rider from slipping?

Question Parameter(s):

In a popular amusement park ride, a rotating cylinder of radius R = 3.90 m is set in rotation at an angular speed of 6.50 rad/s,

Generally, the equation for the centrifugal forces is mathematically given as

Fc = N

Fc= m r ω^2

Where

Ff = Nμ_s

Ff= mg

In conclusion

μs = (mg)/( m r ω^2 )

μs= (g / r ω^2 )

μs = 9.81 / (3.9 * 6.5^2)

μs = 0.059