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Homework Help: Bicycle Leaning

  1. Mar 8, 2015 #1
    Suppose that a bicyclist is moving in a circle of radius R, leaning inwards at an angle a from the verticle, and that the height of the person (h) is less than R but not negligible. I am trying to find the angular velocity the bicyclist must move at.

    I tried doing this in two ways: using torque equilibrium and forces but I got different answers.

    Using forces, I found the total centripetal force acting on the body using an integral. If the person is a rod of mass per length density q, the centripetal force on one piece of the rod of length dr and a distance r from the point of contact is q dr * w^2 (R - r sin a). After evaluating the integral, I got Mw^2 R - 1/2 M h w^2 sin a. This centripetal force came from friction at the point of contact. I used geometry to get friction /Mg = tan a so friction = Mg tan a. Equating this with the expression for the integral gave me w^2 = g/R tan a (1 - h/2R sin a)^-1.

    However using torque, I found the differential expression of the torque to be q dr * w^2 (R - r sin a) r cos a. After evaluating the integral I got Mw^2 h cos a (R/2 - h/3 sin a). In the rotating reference frame, this must balance the torque due to gravity which is Mg h/2 sin a. Equating this gave me
    w^2 = g/R tan a (1 - 2h/3R sin a)^-1.

    I am convinced that the answer from the torque method is correct but I don't know what is wrong with my first method.
  2. jcsd
  3. Mar 8, 2015 #2


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    I moved the thread to the homework section. I think you are supposed to treat the mass of the bicycle rider as a point at height h.


    Torque around what?
  4. Mar 8, 2015 #3
    If you draw a free body diagram of the person, the vector sum of the normal and friction forces should be parallel to the person's body so the two vectors form a right triangle with one of the angles being a. And torque is taken about the point of contact.

    I think my mistake was saying that the friction force was the same as the sum of all the centripetal forces since they act at different points.
  5. Mar 8, 2015 #4


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    That may be true, but it's not clear to me. Taking torque about the point of contact leaves an unbalanced torque from gravity. That's ok, because there will be effectively an angular acceleration about the point of contact with the ground. But what is the relationship between that net torque and w?
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