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Help finding Centripital force angles

  1. Mar 6, 2005 #1
    :confused: I'm stumped on my hw...can anyone help?

    The question is long, but as follows: a merry-go-round has circular platform that is 1m from the central axis at its inner edge and is 5m from the central axis at its outer edge. The ride takes 10s for 1 rotation. A passenger holds himself to the surface with a pair of very stickey shoes and is most comfortable when he orients his body length along the line of the net force on him. Determine the angle his body makes to the vertical A) 1m from the axis, B) 3m from the axis, and c) 5m from the axis. :confused:

    Any ideas...just looking to understand how to set this one up equationwise.
  2. jcsd
  3. Mar 6, 2005 #2
    well, there are two forces action on the passanger, assuming no air resistance. There is a centripital force and a tagential force.

    [tex] F_{centripital} = \frac {mv^2}{r} [/tex]

    [tex] F_{tangential} = mgsin \theta [/tex]

    the tangential force is due to gravity and the centripital one is due to the circular motion. To solve your problem you have to draw some free body diagrams and use vector addition to get the result. Split up your vectors into component form and get your resultant that way. The person will be facing in the angle of this vector. If you need more help, pm me.


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