In a "Rotor-ride" at a carnival, people are rotated in a cylindrically walled "room." (See Fig. 5-35.) The room radius is 5.4 m, and the rotation frequency is 0.8 revolutions per second when the floor drops out.
What is the minimum coefficient of static friction so the people will not slip down?
To determine the minimum coefficient of static friction required for the people to not slip down in the "Rotor-ride," we need to consider the forces acting on the people and analyze the equilibrium condition.
First, let's identify the forces involved. When the floor drops out, the only force acting on the people in the vertical direction will be the weight (mg), where m is the mass of each person and g is the acceleration due to gravity (approximately 9.8 m/s^2).
To prevent slipping, there must be enough static friction between the people and the cylindrical wall to provide the necessary centripetal force. The centripetal force is given by:
Fc = mv^2 / r,
where m is the mass of each person, v is the linear speed of the people, and r is the radius of the cylindrical wall.
In this case, the linear speed is related to the rotation frequency (f) in revolutions per second:
v = 2πrf.
Substituting this expression for v, the centripetal force can be written as:
Fc = m(2πrf)^2 / r.
Now, we can equate the centripetal force to the maximum static friction force (fs) to find the minimum coefficient of static friction (μs) needed:
Fc = fs = μsN,
where N is the normal force exerted by the cylindrical wall on each person, which is equal to their weight (mg).
Substituting N = mg and rearranging, we have:
m(2πrf)^2 / r = μsmg.
Canceling out the mass (m) and simplifying, we get:
(4π^2f^2r^2) / r = μsg.
Canceling out the radius (r), we have:
4π^2f^2r = μsg.
Finally, solving for the coefficient of static friction (μs), we find:
μs = (4π^2f^2r) / g.
Plugging in the given values of f = 0.8 revolutions per second and r = 5.4 m, along with the acceleration due to gravity (g), which is approximately 9.8 m/s^2, we can calculate the minimum coefficient of static friction (μs).
friction=mg
mu*mv^2/r=mg
mu*w^2 r= g
w= 2PI*.8 rad/sec
solve for mu