In a "Rotor-ride" at a carnival, people pay money to be rotated in a vertical cylindrically walled "room."
If the room radius is 4.8 m, and the rotation frequency is 0.6 revolutions per second when the floor drops out, what is the minimum coefficient of static friction so that the people will not slip down?
People describe this ride by saying they were being "pressed against the wall." Is this true? Is there really an outward force pressing them against the wall?
_ yes
_ no
If so, what is its source? If not, what is the proper description of their situation (besides "scary")? (Hint: First draw the free-body diagram for a person.)
I will be happy to critique your thinking or work on this. Please don't post under multiple names.
To determine the minimum coefficient of static friction required for the people to not slip down in the "Rotor-ride," we need to analyze the forces acting on a person inside the rotating room.
First, let's draw a free-body diagram for a person. Since the person is in circular motion, there will be a net inward force acting towards the center of the circle called the centripetal force. In this case, this force is provided by the friction between the person and the wall of the room.
The centripetal force can be calculated using the formula:
Fc = m * ac
Where:
Fc is the centripetal force
m is the mass of the person
ac is the centripetal acceleration
The centripetal acceleration (ac) can be calculated using the formula:
ac = R * ω^2
Where:
R is the radius of the room
ω is the angular velocity
Given that the radius of the room (R) is 4.8 m and the rotation frequency (f) is 0.6 revolutions per second, we can convert the frequency to angular velocity using the formula:
ω = 2πf
Calculating the values:
ω = 2π * 0.6 = 1.2π rad/s
ac = 4.8 * (1.2π)^2
Now, we can calculate the centripetal force (Fc) by multiplying the mass of the person with ac.
At this point, we need to assume a mass for the person to continue the calculation since the question doesn't provide this information. Let's assume a mass of 70 kg.
Now, let's calculate the centripetal force (Fc) for this mass.
Fc = 70 * ac
To determine the minimum coefficient of static friction (μs), we need to consider two opposing forces acting on the person. The centripetal force (Fc) directed inward and the weight (mg) directed downward.
The maximum static friction force (Fs) that can exist without the person slipping can be calculated using the formula:
Fs ≤ μs * N
Where:
μs is the coefficient of static friction
N is the normal force
In this case, the normal force (N) is equal to the weight of the person (mg).
To avoid slipping down the wall, the static friction force (Fs) must be equal to or greater than the centripetal force (Fc). Therefore, we can set Fs equal to Fc and solve for the coefficient of static friction (μs).
Based on the forces involved, we can conclude that the people inside the rotating room are being "pressed against the wall." The outward force pressing them against the wall is the friction force between their bodies and the wall, which provides the centripetal force necessary for the circular motion.
To summarize:
1. Draw the free-body diagram for a person inside the rotating room.
2. Calculate the centripetal force (Fc) using the formula Fc = m * ac.
3. Calculate the centripetal acceleration (ac) using the formula ac = R * ω^2.
4. Convert the rotation frequency (f) to angular velocity (ω) using the formula ω = 2πf.
5. Assum a mass for the person and calculate Fc using the given radius (R) and calculated angular velocity (ω).
6. Set Fs ≤ Fc and solve for the coefficient of static friction (μs) using the formula Fs = μs * N, where N is the normal force equal to the weight of the person (mg).
Please note that the numerical value of the coefficient of static friction cannot be determined without knowing the mass of the person.