twowheelsbg said:
James, i can't imagine your presentation,
please explain which forces and how create 'counter torque exerted by the ground'
OK, sorry I wasn't clearer.
Start with a stationary bike. Let's make it move left to right in the x-direction.
Up will be the y-direction and towards us the z-direction.
As the cyclist begins to accelerate he is peddling to make the wheel turn clockwise which means a torque in the -z direction. (Right hand fist with thumb pointing away from you, fingers roll CW.)
Without friction the wheel would accelerate gaining angular momentum (peeling out) and the bike moves nowhere. Given the friction slowing the rate of wheel acceleration it is exerting a "counter torque" i.e. tending to turn it in a counter-clockwise direction (in this case impeding clockwise acceleration) so that's a torque in the +z direction. (Right hand fist thumb extended toward you fingers curl in CCW direction.)
This torque is due to the force acting by the ground on the tread and this force is also pushing the bike forward.
Finally note that though the torques are almost equal in magnitude and opposite. The tire does have some angular acceleration (not as much if it slid freely), it gains angular momentum as the bike accelerates forward and so the two torques can't quite cancel out... the sprocket torque must be slightly larger in magnitude because there is a net angular acceleration in the clockwise direction.
Figure the angular acceleration times the moment of inertia of the tire and this is the net torque.
Once the bike is up to speed then no more acceleration so the torques exactly cancel. As mentioned they correspond to a force from the ground just enough to overcome wind resistance. (This assumes a frictionless bearing in the wheel... if not then there's a third frictional ("counter") torque acting in the +z direction which must be considered).
