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Centripetal Motion

  1. Jun 30, 2007 #1
    1. The problem statement, all variables and given/known data

    A 33kg child stands at the rim of a merry-go-round of radius 3.66m, rotating with an angular speed of 1.2 rad/s. Find the minimum force between his feet and the floor of the carousel that is required to keep him in the circular path.

    2. Relevant equations

    I drew out the force diagram and found that N = mg. Is this it? Or should I use the equation for centripetal motion, which is mv^2/r.

    3. The attempt at a solution

    Thanks for your help!
  2. jcsd
  3. Jun 30, 2007 #2
    It should be the centripetal force is equal to the force of friction since you want the minimum force to keep him on the merry-go-round.

    So Fc = Ff, where Fc is the centripetal force and Ff is the force due to friction.

    Since you already have the mass and the radius you need to convert angular velocity to good ol' velocity in m/s. To do this you need to find the frequency using angular velocity.

    w = 2*pi*f, f is frequency and w is angular velocity.

    Now that you have f, take it's reciprical to find the period of motion (T). So,

    f = 1/T.

    Now that you have T you can find the circular velocity.

    V = ( 2*pi*r ) / T, where r is the radius and V is the velocity.

    Now you have mass, radius, and velocity. Just solve for Fc.

    Fc = mv^2 / r.
  4. Jun 30, 2007 #3
    Thank you so much!
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