In a "Rotor-ride" at a carnival, people are rotated in a cylindrically walled "room." (See Fig. 5-35.) The room radius is 4.6 m, and the rotation frequency is 0.7 revolutions per second when the floor drops out.

What is the minimum coefficient of static friction so that the people will not slip down?

Change frequency to period T=1/f. Find the velocity v=2(pi)r.

Find radial acceleration a=v^2/r.
Then divide gravity by the radial acceleration that you found. g/a

The force holding them up is mu*fn= mu*mv^2/r= mu*m*w^2 r

set that equal to the force of weight...

m g= mu*m w^2 r
solve for mu.

change .7 rev/sec to rad/sec by multiplying by 2PI.

Help me anyone please

rotor ride a cylinder of radius 3m is rotation at a linear speed of 15 m/s. What minimum coefficient of static friction is required to keep the participant from slipping?

To determine the minimum coefficient of static friction required to prevent people from slipping down in the "Rotor-ride," we need to analyze the forces involved.

1. Start by drawing a free-body diagram for a person in the ride cylindrical wall.

Here's a breakdown of the forces acting on the person:
- The normal force (N) acting perpendicular to the wall, pushing the person into the wall.
- The gravitational force (mg) acting vertically downward, pulling the person downward.
- The frictional force (f) acting parallel to the wall, opposing the slipping motion.

2. Consider the forces in the vertical direction.

The gravitational force mg acts downward, while the normal force N acts upward. Since the person is not moving up or down, these two forces are equal: N = mg.

3. Find the force exerted by the wall in the horizontal direction.

Since the floor drops out, there's no vertical forces in the horizontal direction. Therefore, the frictional force f provides the centripetal force required to keep the person moving in a circular path.

The centripetal force (Fc) is given by Fc = mv^2/r, where m is the mass, v is the velocity, and r is the radius of the ride. In this case, v is the linear velocity on the cylindrical wall, but the angular velocity is given.

Now, convert the angular velocity to linear velocity using v = ωr, where ω is the angular velocity.

Given:
- Radius (r) = 4.6 m
- Angular velocity (ω) = 0.7 revolutions per second

Calculate the linear velocity (v) using the formula v = ωr.

4. Calculate the frictional force.

The frictional force (f) is equal to the centripetal force (Fc) required to stay on the wall without slipping. So, Fc = f.

5. The maximum static friction is μsN. This friction force must equal the centripetal force: f = μsN = μsmg.

We can substitute Fc for f in the equation: Fc = μsmg.

6. Rearrange the equation and solve for μs:

μs = Fc / mg.

Substitute the values of Fc and mg obtained in the previous steps.

μs = (mv^2/r) / mg.

Plug in the values for m, v, r, and g to find μs.

7. Calculate the minimum coefficient of static friction, μs.

Use the given values:
- Mass (m): Assume an average mass, e.g., 70 kg.
- Linear velocity (v) calculated in step 3.
- Radius (r) given as 4.6 m.
- Acceleration due to gravity (g) is approximately 9.8 m/s^2.

Calculate μs = (m * v^2) / (r * g).

By following these steps and plugging in the appropriate values, you can find the minimum coefficient of static friction required to prevent people from slipping down in the "Rotor-ride."