Final answer:
To calculate the minimum coefficient of static friction in a Rotor-ride, apply equations of circular motion and friction. The normal force provided by the ride's walls and equated to the centripetal force for an object in circular motion, helps us find the frictional force needed to equal the gravitational force. This allows us to derive the formula for the minimum coefficient of static friction.
Step-by-step explanation:
To determine the minimum coefficient of static friction that keeps people from slipping down in a "Rotor-ride" at a carnival, we can apply concepts of circular motion and friction. Since the riders are in a vertical cylindrical rotor, the frictional force must match the gravitational force to prevent slipping when the floor drops out.
Force of friction (f) = Normal force (N) * coefficient of static friction (μs)
Centripetal force (Fc) = Normal force (N)
The centripetal force needed to keep the person moving in a circle is provided by the frictional force, which equals the person's weight (mg) in the absence of other vertical forces when the floor drops away.
So, f = mg = N μs and Fc = m(v^2)/r = N where m is the mass of the person, g is the acceleration due to gravity, v is the linear velocity, and r is the radius of the room.
To find v, we use the relation v = 2πr f, where f is the rotational frequency. For this ride, f is 0.44 revolutions per second which can be converted to radians per second by multiplying by 2π. The linear velocity (v) can then be calculated.
Substituting Fc for N in the frictional force equation and rearranging for μs, we obtain:
μs = (v^2)/(rg)
After calculating v, we can plug in the values of r = 5.8 m and g = 9.8 m/s2 to find the minimum coefficient of static friction.