When on a bicycle you need to lean in to the turn in order to prevent yourself from falling over. Let’s say we’re turning right. The frictional force is then rightwards, there’s a normal force acting upwards and there’s the weight acting downwards to the right of where the wheel and the ground meet. My question is as follows: the normal and frictional forces act at the pivot (point where wheel and ground meet) however the weight acts to the right of it and will thus generate a torque which will lead to rotation about the pivot clockwise. How come this doesn’t actually happen?

  • $\begingroup$ It does $\endgroup$ – Patrick Jan 23 '18 at 12:19
  • $\begingroup$ To expand on the answer by @nopjo: just because the bike/rider system is balanced does not automatically mean that that it is static. A turning bike is constantly accelerating towards the center of the turning circle. $\endgroup$ – MichaelK Jan 23 '18 at 12:50
  • $\begingroup$ @Patrick Interesting video but I don't think it is an answer to the question. $\endgroup$ – paparazzo Jan 23 '18 at 15:19
  • $\begingroup$ @Paparazzi the torque op mentions indeed leads to rotation, turning the wheel due to precession. This is shown in the video. Doesn't it answer the question? $\endgroup$ – Patrick Jan 23 '18 at 15:48
  • $\begingroup$ @Patrick My opinion and don't care to debate $\endgroup$ – paparazzo Jan 23 '18 at 15:55

You are expecting the bicycle to turn over from the point of view of the bicycle itself (fixed pivot point). That is, you are adopting the (accelerating) rotating reference frame. In that case, you need to include the virtual centrifugal force, acting to the left, which counteracts the weight.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.