Centripetal Force of an amusement park ride problem

In summary, the conversation discusses a scenario of an amusement park ride consisting of a rotating vertical cylinder with a rider inside. The rider has a mass of 50 kilograms and the cylinder has a radius of 5 meters, with an angular velocity of 2 radians per second. The coefficient of static friction between the rider and the wall of the cylinder is 0.6. The conversation also includes calculations for the centripetal force on the rider when the floor is dropped down and the upward force that keeps the rider from falling. It is also mentioned that friction plays a role in preventing the rider from sliding down the wall and that gravity pulls the rider towards the earth but is counteracted by the friction and the wall of the cylinder.
  • #1
jcmtnez
1
0

Homework Statement


An amusement park ride consits of a rotating vertical cylinder with rough canvas walls. After the rider has entered and the cylinder is rotating sufficiently fast, the floor is dropped down, yet the rider does not slide down, The rider has a mass of 50 kilograms, the radius R of the cylinder is 5 meters, the angular velocity of the cylinder when rotating is 2 radians per second, and the coefficient of static friction between the rider and the wall of the cylinder is 0.6.

(b) Calculate the centripetal force on the rider when the cylinder and state what provides that force.

(c) Calculate the upward force that keeps the rider from falling when the floor is dropped down and state what provides that force.

(d) At the same rotational speed would a rider of twice the mass slide down the wall? Explain you answer.

Homework Equations



F=m*a

a=R*w^2

The Attempt at a Solution



I could derive the centripetal force by applying Newton's second law and the formula for centripetal aceleration. I got 1000 N but i am not sure of my answer. And I don't understand very well what force is involved in pulling up the person when the floor is dropped down and how is friction involved.
 
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  • #2
jcmtnez said:
And I don't understand very well what force is involved in pulling up the person when the floor is dropped down and how is friction involved.


It is not pulling up the person, but instead is acting against W, and is equals to W.
-It's frictional force.
-the person tends to go in a straight line, but the wall prevents him from doing so, by constantly applying a centripetal force on him
-gravity tends to pull him towards the earth, but since he is sticking to the wall, and there is friction between him and the wall, so he doesn't slide.
 
  • #3


I would first clarify the problem and its assumptions. It is important to note that the problem assumes the rider is stuck to the wall due to friction, and the floor is dropped down without any external force acting on the rider. I would also clarify that the rider is in circular motion, with the wall of the cylinder providing the centripetal force.

To calculate the centripetal force on the rider, I would use the formula F=ma, where m is the mass of the rider and a is the centripetal acceleration. Using the equation a=R*w^2, where R is the radius and w is the angular velocity, I would get a value of 1000 N for the centripetal force. This force is provided by the friction between the rider and the rough canvas walls of the cylinder.

When the floor is dropped down, the rider would experience an upward force that prevents them from falling. This force is provided by the normal force exerted by the wall of the cylinder on the rider. The normal force is equal in magnitude to the weight of the rider, which in this case is 490 N (50 kg * 9.8 m/s^2). This force is balanced by the downward force of gravity on the rider.

To answer part (d), I would use the same equations and assumptions. Since the mass of the rider has doubled, the centripetal force on the rider would also double, resulting in a force of 2000 N. However, this does not necessarily mean that the rider will slide down the wall. The coefficient of friction remains the same, and as long as the force of friction is greater than or equal to the force of gravity, the rider will not slide down the wall. Therefore, the rider of twice the mass would not slide down the wall, as long as the coefficient of friction remains the same.
 

1. What is centripetal force?

Centripetal force is the force that keeps an object moving in a circular path. It acts towards the center of the circle and is necessary to maintain the object's curved motion.

2. How does centripetal force apply to amusement park rides?

In amusement park rides, centripetal force is responsible for keeping riders in their seats and preventing them from flying off the ride. The ride's structure and design create a centripetal force that keeps the riders moving in a circular path.

3. How is the centripetal force of an amusement park ride calculated?

The centripetal force of an amusement park ride can be calculated using the formula Fc = mv^2/r, where Fc is the centripetal force, m is the mass of the rider, v is the velocity, and r is the radius of the circular motion.

4. What factors affect the centripetal force of an amusement park ride?

The centripetal force of an amusement park ride is affected by the speed of the ride, the mass of the riders, and the radius of the circular motion. Other factors such as air resistance and friction also play a role in determining the centripetal force.

5. How does the centripetal force of an amusement park ride impact rider safety?

The centripetal force of an amusement park ride is crucial for ensuring rider safety. If the force is too weak, riders may not stay securely in their seats, risking injury or even falling off the ride. On the other hand, if the force is too strong, it can cause discomfort or injury to riders. Therefore, amusement park ride designers must carefully calculate and consider the centripetal force to ensure rider safety and enjoyment.

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