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Torque with constant circular velocity

  1. Mar 20, 2014 #1
    1. The problem statement, all variables and given/known data

    a passenger with height 175cmis driving on a city bus. The center of mass is at h=110cm above the middle point of the shoe which are 30 cm long. The passenger is standing in the direction of the ride.
    shoe size = 30 cm

    the bus is driving in a circle which has the radius of 20m with a constant speed of 20 km/h.For what angle relative to the vertical axis should the passenger lean so that he/she would not tip over.

    2. Relevant equations

    3. The attempt at a solution
    So here is what I tried to do
    fc=m*(20/3,6)2/20.................. I got Fc=1,543m
    and here is something I don't understand. Since the centripetal force is acting toward the center of the circle, why is it incorrect to just use the Fx component of m*g
    I tried doing it like this
    Fc=1,543m Fx=m*g*sinα......since the body is not moving Torque=0
    sinα=(1.543*m)/(m*g) the masses cancel out

    which is incorrect. The right answer is 8.9° and the end equation should look like this
    Can someone please explain or help me get to this equation tanα=v2/(r*g) [/b]
    Thank you for reading
    Last edited: Mar 20, 2014
  2. jcsd
  3. Mar 20, 2014 #2

    Simon Bridge

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    Homework Helper

    Are you given the mass of the passenger?
    What are the units for Fc?

    ... you mean the horizontal component of the weight?
    Cartesian coordinates don't make a lot of sense here.

    ... but the body is moving - in fact, for the term "centripetal force" to make sense, it must be accelerating: i.e. moving in a circle. So it must be acted on by an unbalanced force.

    If I read you correctly, you tried to arrange to have zero horizontal force.

    In the inertial frame, the net force on the body must point horizontally towards the center of the turn and sufficient to maintain circular motion. There should be zero net torque as well because the body has zero angular acceleration - it is rotating because it is constantly oriented along the axis of the turning bus.
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