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Circular Motion of a cyclist

  1. May 7, 2005 #1
    In order to round a bend of radius 20m on a rough and flat level ground. A cyclist must tilt at an angle of 20 degrees to the vertical. What is the spped of the cyclist?

    Can anyone help me with this qn pls...?
     
  2. jcsd
  3. May 7, 2005 #2

    dextercioby

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    HINT:Draw a FBD and then u'll figure out that the tangential (to the trajectory) component of gravity is in fact a centripetal force.

    Daniel.
     
  4. May 7, 2005 #3

    OlderDan

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    I don't think this hint says quite what it was intended to say. There is no component of gravity tangential to the trajectory. But there is a centripetal force that comes from the frictional force acting on the tires of the bike, and there is a normal force countering the weight of the cyclist and bike.

    From here, there are two ways to look at the problem. One is in terms of the fictitious or apparent centrifugal force acting on the cyclist and bike in their frame of reference that is opposite the centripetal force. The sum of the weight plus the centrifugal force must act along the line of the bike inclined 20 degress from the vertical or else the bike would topple over.

    Another way to look at the problem is to convince yourself that the force from the ground (friction plus normal force) must act on a line through the center of the bike. One way to do that is to look at the comparable problem of a bike on a banked frictionless track where the only forces are the normal force and the gravitational force. The bike must be perpendicular to the track or else it would slip. Gravity must act downward, and the normal force in this case must act through the center of the bike, with a vertical component equal to the weight and a horizontal component that is the centripetal force. These are equivalent to the forces that must act on a flat track with friction. The net force from the ground must act through the center of the bike.
     
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