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Amusement Rotor Ride Explanation

  1. Feb 18, 2017 #1
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    1. The problem statement, all variables and given/known data

    An amusement park ride consists of a large vertical cylinder that spins about its axis fast enough that any person inside is held up against the wall when the floor drops away (see figure). The coefficient of static friction between person and wall is μs, and the radius of the cylinder is R.

    2. Relevant equations
    1. Show that the maximum period of revolution necessary to keep the person from falling is T = (4π2Rμs/g)1/2
    2. If the rate of revolution of the cylinder is made to be somewhat larger, what happens to the magnitude of each one of the forces acting on the person? (Select all that apply.)
    a. The normal force remains the same.
    b. The gravitational force remains the same and the frictional force increases.
    c. The gravitational force and frictional force remain constant.
    d. The normal force increases.
    3. If the rate of revolution of the cylinder is instead made to be somewhat smaller, what happens to the magnitude of each one of the forces acting on the person? (Select all that apply.)
    a. All the forces decrease.
    b. The gravitational force remains constant.
    c. The gravitational force decreases.
    d. The normal and frictional force decrease.

    3. The attempt at a solution
    I already got the right answers, but mostly with trial and error. In the 2nd question the answer are: c & d; and for the 3rd question the answers are b & d.
    So, now I am confused. How come when the period is larger, the gravitational force remains constant, along with the friction, but when it gets smaller the normal and friction decrease. What I am trying to say is that I see the relationship between normal force and friction, so why doesn't friction increase when normal force increases in the first scenario. I am just trying to get my head around it. Thanks!
     
  2. jcsd
  3. Feb 19, 2017 #2
    here the normal reaction by the wall is due to centrifugal force of the cylinder
    condition for a person to held up inside is μsFcf=mg
     
  4. Feb 19, 2017 #3

    TSny

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    In question 2 the "rate of revolution" is increased. That means the cylinder rotates faster. So, would the period increase or decrease?

    Does the formula ##f_s = \mu_s N## give the static friction force under all circumstances or does it only specify the force of static friction when the object is on the verge of slipping?

    I believe question 2 is assuming that you increase the rate of revolution above the minimum rate necessary to keep the people from slipping. So, after the rate is increased would the people still be on the verge of slipping? Does ##f_s = \mu_s N## give the force of friction in this case?

    In question 3, the rate is decreased somewhat below the rate necessary to keep the people from slipping. So, after the rate is decreased will the people slip? If so, are you still dealing with static friction?

    When thinking about this problem, make sure you have drawn a good free-body diagram.
     
  5. Feb 19, 2017 #4
    I'm trying to think of an example that may help. Let's say you have your picnic basket set on the picnic table outside. And let's say there is a strong wind that keeps blowing the basket off the table. So you decide to add weight inside the basket to try to keep it from blowing off the table. So as weight is added, the normal force from the table acting on the basket increases, which increases the maximum friction between the table and the basket. If enough weight is added, the maximum friction force finally is great enough to counter the force from the wind, preventing the wind from blowing the basket off the table. And if more weight is added to the basket after that, even though the normal force will increase, the friction force that results from that normal force will not increase. The friction force produces an equal and opposite force to the wind as a reaction to the wind force as the wind decreases and increases up to its maximum level per fs = μN.

    The amusement park ride uses the same principle. As the cylinder spins faster and faster, the normal force from the cylinder wall to the back of the human body increases. But once a high enough velocity is attained to produce a friction force equal to the weight of the human body, then the body will not slide down. And as the spinning velocity increases beyond that, the normal force increases, but the friction force does not increase beyond that. The friction force remains equal to the weight of the human body.
     
  6. Feb 19, 2017 #5

    haruspex

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    You have some good responses to this above, but I'll throw in another thought. There are only two vertical forces acting, gravity and friction. At the critical speed, the two just balance. If increasing the speed left gravity the same but increased the frictional force, what would the net vertical force be? What would be the consequence for the customer?
     
  7. Feb 19, 2017 #6

    CWatters

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    Its worth remembering that the friction equation F=uN gives a value for the maximum force that can be applied before slipping occurs. Not the value of a force in its own right.
     
  8. Feb 20, 2017 #7
    Thank you! The basket analogy was really useful. I understood the others only after I read yours, so thank you to everyone too.
     
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