In a \"rotor-ride\" at a carnival, people are rotated in a cylindrically walled

Question

In a "rotor-ride" at a carnival, people are rotated in a cylindrically walled "room." The room radius is 4.6m, and the rotation frequency is 0.50 revolutions per second when the floor drops out. What is the minimum coefficient of static friction so that the people will not slip down?

The Part I need help on: I have found the circumference (28.9m), the velocity in which the ride is spinning (14.45m/s), and the centripetal acceleration (45.39m/s2). However I am stuck after this point.

Any help is appreciated. This is my first time using this site and therefore I am not entirely sure what other information I should include. If anything else is needed/missing, please ask and I will supply.

Thanks!

Explanation / Answer

rotational acceleration = v^2 / r circumference of wall = 2xPixr=28.9 m this means 28.9 m is travelled for one revolution 1 rev = 28.9 m so velocity = (28.9 m/rev ) x (0.5 rev/sec) = 14.45 m/s acceleration = 14.45^2 / 4.6 m = 45.4 m/s2 Since no mass is given for people, let's just use mass=M downward force from gravity is = Mxg = 9.81xM This means Friction Force = 9.81xM to prevent people from falling coefficient of friction=(Friction Force)/(Force acting normal ) Force acting normal = m x (rotational acceleration) = Mx45.4 minimum coefficient of friction=(9.81xM)/(45.4xM) =0.2161

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects