In a "Rotor-ride" at a carnival, people are rotated in a cylindrically walled "room." The radius of the room is 4.0 m, The coefficient of static friction is μ=0.13.
What is the minimum rotation frequency such that the people will not slide down the wall when the floor drops out?
To find the minimum rotation frequency at which people will not slide down the wall, we need to consider the forces acting on them.
Let's start by looking at the forces acting on a person inside the rotating room. There are two main forces to consider:
1. Gravitational force (mg): This force pulls the person downward, trying to make them slide down the wall.
2. Centripetal force (F_c): This force acts towards the center of the circular motion and prevents the person from sliding outwards.
At the minimum rotation frequency, the centripetal force will be equal to the maximum static friction force, preventing the person from sliding down the wall. Therefore, we can set up an equation to find the relationship between these forces.
The centripetal force is given by:
F_c = m⋅ω²⋅r
Where m is the mass of the person, ω is the angular velocity, and r is the radius of the rotating room.
The maximum static friction force is given by:
F_friction = μ⋅N
Where μ is the coefficient of static friction and N is the normal force (equal to the gravitational force in this case).
Since the person is not sliding up or down the wall, the gravitational force is balanced by the normal force:
mg = N
Combining these equations, we can set up an equation to find the minimum rotation frequency:
F_c = F_friction
m⋅ω²⋅r = μ⋅N
m⋅ω²⋅r = μ⋅mg
ω² = (μ⋅g)/r
ω = √[(μ⋅g)/r]
Now we can substitute the values given in the problem to find the minimum rotation frequency. Plugging in the values:
μ = 0.13
g = 9.8 m/s²
r = 4.0 m
ω = √[(0.13⋅9.8)/4.0]
ω = √[1.274]
ω ≈ 1.13 rad/s
Therefore, the minimum rotation frequency at which people will not slide down the wall is approximately 1.13 radians per second.