1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Centripital Force problems

  1. Oct 24, 2007 #1
    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 a person inside is stuck to the wall and does not slide down when the floor drops away. The acceleration of gravity is 9.8 m/s^2. Given g=9.8 m/s^2, the coefficient mu=.447 of static friction between a person and teh wall, and the radius of the cylinder R=4.8 m. For simplicity, neglect the persons depth and assume he or she is just a physical point on the wall. The persons speed is v=2(pi)r/T, where T is the rotation period of the cylinder (the time to complete a full circle) Find the maximum rotation period T of the cylinder which would prevent a 74 kg person from falling down. Answer in units of s.


    2. Relevant equations
    A=v^2/r
    v=2(pi)r/T


    3. The attempt at a solution

    i set sum of the forces in the y direction equal to Fn=Fg, since the person is not moving in the y direction
    thus able to find Fn, i plugged that into the formula Fs= (mu)s(Fn), and got 324.1644. Then i did the sum of the forces in the x directoin, which i had as Fs=mv^2/r, thus i found v^2, and plugged v into the v formula to get T. I think i am missing centripital force in my x direction, but i dont know how to represent it, and get a number for it. Please respond with explanation, and answer will be also greatly appreciated.
     
  2. jcsd
  3. Oct 24, 2007 #2

    Kurdt

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I'm not sure what your misunderstanding is. From what you've written as your attempt it sems that you initially done the right thing. You needed to equate the weight of the person with the maximum force needed to overcome static friction. Form that you can work out what the normal force must be. This normal force can be equated to the centripetal force to find the minimum (assuming the question has a typo) T.
     
  4. Oct 24, 2007 #3
    no actually the question says maximum, so does it change then?
     
  5. Oct 24, 2007 #4
    so are you saying the Fc (centripetal force) is equal to 74 kg (the weight of the person)?

    then this would change the whole equation i think
     
  6. Oct 24, 2007 #5

    Kurdt

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    No I'm saying its equal to the normal force on the person from the wall.

    P.s. yes it should be maximum I've just gone mad at the end of a long day.
     
  7. Oct 24, 2007 #6
    yea i had that, but im not sure...Is an answer of about 6 seconds wrong for the T, because im thinking that it is too high
     
  8. Oct 24, 2007 #7

    Kurdt

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I get approximately half your answer.
     
  9. Oct 24, 2007 #8
    Fs=324.1644
    v= 4.585507
    T=6.577089398

    those are the exact answers i got
     
  10. Oct 24, 2007 #9

    Kurdt

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Your Fs is wrong. remember that the static friction has to hold the weight of the person. Therefore Fs is just their weight. You only need this to solve for the normal force. In fact if you work algebraically throughout you see that the mass is not needed at all. So to clarify:

    [tex] F_s = \mu_s N = mg[/tex]

    You will need to solve for N and set that equal to the centripetal force.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Centripital Force problems
  1. Force of ladies push (Replies: 1)

Loading...