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Homework Help: What is the centripetal acceleration of the child?

  1. May 20, 2005 #1
    Hey guys,
    I'm new here and my name's Tracy. I just found out about this site and i think its fabulous.

    Anyway i had a question and did'nt exactly know where to post it so if this is the wrong forum to post it i'm sorry.

    Could somebody help me out and solve this question for me.

    Q. A child weighing 40kg takes a ride on a Ferris wheel that rotates four times each minute and has a diameter of 18 m.

    a) What is the centripetal acceleration of the child?

    b) What is the magnitude and direction of the force that the seat exerts on the child at the lowest point of the ride?

    c) What force does the seat exert on the child at the highest point of the ride?

    d) What force does the seat exert on the child when he is midway between the top and the bottom?

    That's the question i need solved. Also please show your working.

    Thanks a lot! And hi again to everyone!
  2. jcsd
  3. May 20, 2005 #2
    Why dont you show us your working so we can show you where you went wrong?
    The first thing you need to do is find out how fast shes going, so find angular velocity, and linear velocity.
  4. May 20, 2005 #3
    I can't really show you my working since i have'nt even tried the problem because i don't know where to start. Thats why i need help.

    Simple problems like such always confuse me :frown:.

    I have a test tommorow and while studying i stumbled across this problem. Help would be greatly appreciated.
  5. May 20, 2005 #4
    Ok tell me,
    what the centripetal force is and what it does
    the equation for centripetal force
    tell me what each variable means, and what its value is in this problem.
  6. May 20, 2005 #5
    EDIT: Ok. Maybe my hint was too revealing.

    Take it away, Whozum.
    Last edited: May 20, 2005
  7. May 21, 2005 #6
    The centripetal force is that force pulling an object towards the center of a circular path as the object goes around the circle. An object can travel in a circle only if there is a centripetal force on it.

    The equation i'm not sure of.
  8. May 21, 2005 #7
    This is correct but sounds better if you rephrase it: An object's net force is centripetal if the consequent motion is circular.

    Does F = mv^2/r ring any bells?
  9. May 21, 2005 #8

    So once i find the centripetal force using that formula i can calculate the centripetal acceleration using a=v²/r ?

    Also in the first equation what would be the value of 'v' ?
  10. May 21, 2005 #9
    V is the linear velocity of the object in motion. In the problem, they dont give you this directly, however you can figure it out with what theyve given you.

    The bolded material gives you a time, and a rotation rate. Can you figure out the linear velocity?

    Hint: Figure out the distance the child travels per rotation, then how many rotations per minute, then rotations per second.
  11. May 21, 2005 #10
    Thanks for the help Whozum!

    What about the rest of the question :confused:.

    If you don't mind can you eplain the answer too.
  12. May 21, 2005 #11
    have you figured out the velocity in terms of the period yet?
  13. May 21, 2005 #12
    You can use [itex]v = r\omega[/itex].
  14. May 21, 2005 #13
    I'll post you my solution to #1:

    Step One: Write down all that you know:
    Child's weight = m = 40kg
    Ferris Wheel Diameter = d = 18m
    Ferris Wheel Radius = r = 9m
    Rotation rate = [itex] \omega [/itex] = ?
    Centripetal Acceleration = a = ?
    Centripetal Force = F = ?

    Step Two: Identify what you need.
    Ultimately we want the centripetal force, which is just the mass times the centripetal acceleration. The rotation rate needs to be solved for from the given information.

    Step Three: Identify your equations
    [tex] \omega = \frac{\Delta \theta}{\Delta t}[/tex]

    [tex] v = r \omega[/tex]

    [tex] a = \frac{v^2}{r}[/tex]

    [tex] F = ma = m \frac{v^2}{r} [/tex]

    Step Four: Solve your equations for your unknowns:

    They are all solved for their unknowns.

    Step Five: Plug in your numbers:

    [tex] \omega = \frac{\Delta \theta}{\Delta t} = \frac{2\pi}{15s} = \frac{0.42 rad}{1 sec} [/tex]

    [tex] v = r\omega = (9)(0.42) = 3.77m/s [/tex]

    [tex] a = \frac{v^2}{r} = \frac{(3.77)^2}{(9)} = 1.58 m/s^2 [/tex]

    [tex] F = m \frac{v^2}{r} = ma = (40kg)(1.58m/s^2) = 63.2N [/tex]

    Step Six: Answer the question:

    The child experiences a centripetal acceleration of 1.58m/s^2
    The child experiences a centripetal force of magnitude 63.2N

    Can you follow the same procedure for the other two, and fully answer #2?
    Last edited: May 21, 2005
  15. Sep 29, 2010 #14
    would it be the centripidal force + the gravitational force for when the kid is at the lowest point of the ride?
  16. Sep 29, 2010 #15
    You need to use the formula ac = v^2 / r
    where a sub c = the centripetal acceleration
    v^2 = the velocity squared of the object in question
    and r is ofcourse the radius of a circle around which an object is moving
    Try it out and see what you get!
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