Calculating Centrifugal Force in a "Rotor-ride

In summary, the minimum coefficient of static friction needed in the "Rotor-ride" at a carnival is dependent on the centripetal acceleration, which can be calculated using the formula a=(4π^2(r))/T^2. The centripetal force required to keep a person from slipping can be found using Newton's 2nd law, and the mass of the person may cancel out when solving for the minimum friction force needed.
  • #1
JC Polly
2
0
In a "Rotor-ride" at a carnival, people are rotated in a cylindrically walled room (think of a ride like the gravitron). The room radius is 4.6m and the rotation frequency is .4 revolutions per second when the floor drops out. What is the minimum coefficient of static friction so that the people will not slip down?

I understand that i need to solve for centripetal acceleration and solve for a normal force and fictional force, but I'm having trouble finding the process to get to that solution. Any help would be greatly appreciated.
 
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  • #2
Hi JC Polly, welcome to PF! :smile:

What is the centripetal acceleration using the formula you know for it? What is the equation for the centripetal force in that direction that results from that cenrtripetal acceleration, using Newton's 2nd law? What then is the min friction force required to keep the person from falling?
 
  • #3
I found the centripetal acceleration to be 45.4 m/s^2 a=(4π^2(r))/T^2. The equation I have for centripetal force is F=mass*centripetal acceleration, but my problem is the problem didn't give a mass for a human so that's where I'm stuck at. I know that some problems we did in class the masses of the the object canceled out through manipulating the equations but that was only in examples when the object was moving in the vertical direction.
 
  • #4
Well leave the centripetal force with the m in it, and maybe you will find it cancels out!
 
  • #5


I would approach this problem by first understanding the principles of centripetal force and friction. Centripetal force is the force that keeps an object moving in a circular path, and it is equal to the mass of the object times its centripetal acceleration. Friction is the force that opposes motion between two surfaces in contact.

To solve for the minimum coefficient of static friction, we need to consider the forces acting on the person in the "Rotor-ride." At the point when the floor drops out, the person is experiencing two forces: the centripetal force due to the rotation of the ride, and the force of gravity pulling them down. The centripetal force is given by the equation Fc = mv^2/r, where m is the mass of the person, v is the tangential velocity, and r is the radius of the ride.

To prevent the person from slipping down, the force of friction between their feet and the ride's floor must be equal to or greater than the force of gravity pulling them down. This can be represented by the equation Ff ≥ mg, where Ff is the force of friction and mg is the force of gravity.

We can rearrange these equations to solve for the coefficient of static friction, which is given by the equation μ = Ff/Fn, where μ is the coefficient of friction, Ff is the force of friction, and Fn is the normal force. The normal force is the force perpendicular to the surface that supports the person's weight.

To find the minimum coefficient of static friction, we can set the two equations equal to each other and solve for μ. This will give us the minimum value of μ needed to keep the person from slipping down. The final equation will be μ ≥ v^2/rg.

In this case, we have the values for v (tangential velocity) and r (radius), and we can assume the value for g (acceleration due to gravity) to be 9.8 m/s^2. Plugging in these values, we can calculate the minimum coefficient of static friction needed to be 0.16.

In conclusion, the minimum coefficient of static friction for a person to not slip down in a "Rotor-ride" with a radius of 4.6m and a rotation frequency of .4 revolutions per second is 0.16. It is important to note that this is the minimum value and a higher coefficient of friction
 

1. What is centrifugal force and how does it affect a "Rotor-ride"?

Centrifugal force is a type of apparent force that acts outward on an object that is moving in a circular path. In a "Rotor-ride", this force is responsible for pushing riders against the wall of the rotating cylinder, giving them the sensation of being stuck to the wall.

2. How is centrifugal force calculated in a "Rotor-ride"?

The formula for calculating centrifugal force in a "Rotor-ride" is Fc = mv²/r, where Fc is the centrifugal force, m is the mass of the rider, v is the velocity, and r is the radius of the circular path. It is important to note that the velocity and radius must be in meters and meters per second, respectively.

3. What factors affect the strength of centrifugal force in a "Rotor-ride"?

The strength of centrifugal force in a "Rotor-ride" is affected by the mass and velocity of the riders, as well as the radius of the circular path. The faster the ride rotates and the larger the radius, the stronger the centrifugal force will be. Additionally, heavier riders will experience a stronger force compared to lighter riders.

4. How does centrifugal force impact rider safety in a "Rotor-ride"?

When a "Rotor-ride" is operating within safe and recommended parameters, the centrifugal force experienced by riders is not considered dangerous. However, if the ride is not properly maintained or operated at excessive speeds, the force could potentially cause discomfort or injury to riders.

5. Are there any real-world applications for understanding centrifugal force in a "Rotor-ride"?

Yes, understanding centrifugal force in a "Rotor-ride" can be applied to various real-world situations, such as designing amusement park rides, calculating forces on astronauts in space, and understanding the dynamics of objects in circular motion. It is a fundamental concept in physics and has practical applications in many fields.

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