# Amusement Park Centripetal Force Question.

• HangingOnByAPulley
In summary: I don't remember if it was rotating or not. My primary concern was my shirt rising over my face and knocking my glasses off.
HangingOnByAPulley
Homework Statement
The amusement park ride shown in Figure 10 is a
large, rapidly spinning cylindrical room with a radius
of 3.0 m. The riders stand up against the wall, and the
room starts to spin. Once the room is spinning fast
enough, the riders stick to the wall. Then the floor
slowly lowers, but the riders do not slide down the
wall. Assume the coefficient of friction between
the wall and the riders is 0.40.

(b) Calculate the minimum speed of the rider
required to keep the person stuck to the wall
when lowering the floor.
Relevant Equations
Fc = mv^2/r

Fs = μsFn

I am not given the mass at all in any portion of the question. Fs = Fc because the static friction is the thing that keeps the rider stuck to the wall

My answer came out to about 3.4 m/s for the minimum speed that keeps the rider stuck to the wall, however, the textbook answer is 8.6 m/s and I am not entirely sure why. Am I supposed to factor in the fact that the floor is being lowered? I am not given the acceleration at which the floor is being lowered.

I isolated for V by doing Fs = Fc
μsmg = mv^2/r
v= √μs(g)(r)

v came out to 3.26 m/s. Where did I go wrong?

Where is Figure 10? You got the directions of the forces mixed up. Which force prevents the person from sliding? Which force provides the centripetal acceleration?

Forgot to include it, my bad.

HangingOnByAPulley said:
View attachment 289712

Forgot to include it, my bad.

kuruman said:
Where is Figure 10? You got the directions of the forces mixed up. Which force prevents the person from sliding? Which force provides the centripetal acceleration?
I guess the force that prevents the person from sliding would be the frictional force and the force that provides the centripetal acceleration would be the normal force of the wall pushing back on the person? Is that correct?

bob012345
Yes. Now write F = ma for the horizontal and the vertical direction (two equations).

Ahh, now I see, thank you for the help, I got the correct answer now.

bob012345 and kuruman
HangingOnByAPulley said:
Ahh, now I see, thank you for the help, I got the correct answer now.
Glad to be of assistance. I experienced this ride several decades ago and it didn't quite work as advertised in this physics problem. Yes, the wall surface was some kind of rubbery material with an expected high coefficient of static friction, however, the coefficient of static friction between my shirt and my body was lower. I started sliding down with my shirt being pulled over my head. Thankfully, I was able to stop the sliding using my hands to push against the wall until the ride was over.

bob012345
kuruman said:
Glad to be of assistance. I experienced this ride several decades ago and it didn't quite work as advertised in this physics problem. Yes, the wall surface was some kind of rubbery material with an expected high coefficient of static friction, however, the coefficient of static friction between my shirt and my body was lower. I started sliding down with my shirt being pulled over my head. Thankfully, I was able to stop the sliding using my hands to push against the wall until the ride was over.
I was wondering if the floor went completely away or was just lowered yet still rotating?

bob012345 said:
I was wondering if the floor went completely away or was just lowered yet still rotating?
It was lowered, about 2 meters below the feet of the riders, enough to give one the feeling of being suspended. Not going completely away is probably required by their insurance company and also makes good sense.

bob012345
kuruman said:
It was lowered, about 2 meters below the feet of the riders, enough to give one the feeling of being suspended. Not going completely away is probably required by their insurance company and also makes good sense.
So if you slipped to the floor was it rotating with the cylinder or fixed? Seems if it was also rotating you would not have suffered much damage if you slipped down but if not it could have injured you.

bob012345 said:
So if you slipped to the floor was it rotating with the cylinder or fixed? Seems if it was also rotating you would not have suffered much damage if you slipped down but if not it could have injured you.
I don't remember if it was rotating or not. My primary concern was my shirt rising over my face and knocking my glasses off.

## 1. What is centripetal force?

Centripetal force is the force that keeps an object moving in a circular path. It is directed towards the center of the circle and is necessary for an object to maintain its circular motion.

## 2. How is centripetal force related to amusement park rides?

Amusement park rides, such as roller coasters and spinning rides, use centripetal force to keep riders moving in a circular motion. This force is created by the ride's design and is what allows riders to experience thrilling twists, turns, and loops.

## 3. How does centripetal force affect the human body?

Centripetal force can have a significant impact on the human body, especially at high speeds. It can cause feelings of weightlessness, dizziness, and even nausea. This is why it is important for amusement park rides to be carefully designed and regulated to ensure the safety of riders.

## 4. How does the speed of a ride affect the centripetal force experienced by riders?

The speed of a ride directly affects the amount of centripetal force experienced by riders. The faster the ride, the greater the centripetal force. This is why rides that are designed for more intense thrills, such as roller coasters, often have higher speeds compared to slower, more gentle rides.

## 5. How do engineers calculate and design for centripetal force in amusement park rides?

Engineers use mathematical equations, such as Newton's second law of motion and the centripetal force formula, to calculate and design for the necessary amount of centripetal force in amusement park rides. They also take into account factors such as the ride's speed, radius, and the weight and distribution of riders to ensure a safe and enjoyable experience.

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