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Cyclist cycling in a perfect circle - find radius

  1. Mar 15, 2005 #1
    I am doing a Biomechanics course and it is just like physics. I don't have a large back ground when it comes to physics. So I need some help. I am having a problem with a question on an assignment. This is the question:

    A 78 kilogram cyclist is cycling in a perfect circle, with a velocity of 12m/sec. If The cyclist is leaning at 30 degrees and is in a perfect equilibrum, what is the radius of teh circle they are cycling around?

    It doesn't seem to be a hard question, but I have been trying to figure this out for about 5 hours now and nothing is hitting me on how to do it. I know it has to deal with centripetal force and I think I should be using the formula F=mv2/r at some point. But I have no idea what the first step is. I hope someone can help me get started, remember I have very little background when it comes to physics it has been a while since I have done it.

    Woody :frown:
  2. jcsd
  3. Mar 15, 2005 #2


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    Because rider is in equilibrium at 30 deg to the VERTICAL, the road surface applies Force "F" (shown in drawing below) whose components are sufficient to balance rider's gravitational force (m*g) and provide centripetal force (m*v2/r) required for uniform circular motion. Force "F" components are shown in the drawing. Thus, for equilibrium:
    {Centripetal Force} = {"F" Horizontal Component}
    ::: ⇒ m*v2/r = m*g*tan(30)
    ::: ⇒ r = v2/{g*tan(30)}
    ::: ⇒ r = (12 m/sec)2/{(9.81 m/sec2)*tan(30)}
    ::: ⇒ r = (25.4 meters)

    Code (Text):

                           ^         ^
                           |        /  
                           |       / [COLOR=Red][B]F[/B][/COLOR]
                           |  30  /
                           |     /
                       m*g |    / m*g/cos(30)
                           |   /
                           |  /
                           | /

                [COLOR=Blue]Road Surface Force Components On Rider[/COLOR]
    Last edited: Mar 15, 2005
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