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Questions about centripetal force

  1. Mar 20, 2009 #1
    1. Mr. B takes his grade 12 physics class to the local amusement park at Hopedale Mall to conduct some experiments on friction and centripetal force in the gravitron. The Gravitron is an amusement park ride shaped like a vertical cylinder with a radius of 7m. Riders stand with tier backs against the inside wall of the cylinder, facing the center. As the cylinder begins to rotate faster and faster, the riders become pinned to the wall. The floor then dissapears and the riders are left suspended against the wall as the ride continues to rotate at top speed.

    a) what must be exerting force on the riders to give them the necessary centripetal force directed towards the center of the cylinder? what type of force is this?
    b) What must be true about the vertical forces on the riders, given that they are moving in a horizontal circle?
    c) what force is holding the riders up against gravity so that they do not fall?
    d) Given the coeffience of friction between a typical rider and the wall is aprroximately 0.4, what is the minimum freequency of rotation required so that the riders do not fall? Express yur answer in rpm.

    2. Relevant equations
    for part d i think we use this equation
    Fc= mass*4*radius^2*freequency^2


    3. The attempt at a solution
    a) i think the answer for this is the gravity of the earth thats exerting force. This is Fg force which is equal to Fn
    b) they should b equal to each other positive
    c) Ff
    d) i didnn know how to solve this at all


    Plz try to help me i m kinda stuck i get the idea but i donnohow to explain it
     
  2. jcsd
  3. Mar 20, 2009 #2

    danago

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    Gold Member

    Gravity acts downwards, so how can it possibly be directed towards the center of the circle of motion as a centripetal force?
     
  4. Mar 20, 2009 #3
    oh so the force thats towards the centre is the force of friction?
     
  5. Mar 21, 2009 #4

    rl.bhat

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    Homework Helper

    No. Frictional force acts in the opposite direction to the gravitational force.
    And the normal force is provided by the centrifugal force due to rotation of the cylinder.
     
  6. Mar 21, 2009 #5
    Alright whats up sheenu32, i have feeling i am your class, so let me attempt question for you.
    a) I believe you are right about the normal force, but the normal is not the force of gravity, it is the force exerted by wall on the passenger.
    b)they cancel out
    c)the force of friction of the wall
    d) well this is how i attempted it:
    Ff=Fn*.4
    Fn=Ff/.4
    Fn=Fc
    Ff=mg(opposite force of gravity)
    mg/.4=m4pi^2rf^2
    m cancel
    g/.4=4*pi^2(7)f^2
    24.5=4*pi^2(7)f^2
    f=.30 Hz
    convert .30 Hz to rpm
    and tada you get your answer
    I think that's right lol
    someone correct me if they see something wrong
    PLEASE WE HAVE BIG TEST ON MONDAY!
     
  7. Mar 22, 2009 #6
    ooo thnx but we need to give more explanations for it though
     
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