- #1
Jzhang27143
- 38
- 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.
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.