- #1
koodawg
- 6
- 1
I remember this problem from my freshman physics class ~38 years ago. As I recall it was explained to us how a motorcycle rider could actually take a corner faster by putting his inside knee down on the ground. We see this in motorcycle racing where the riders have a knee pad that let's them place their knee on the ground, effectively sliding it along the pavement as they lean into turns..
What I remember and/or presume is that there is (1) the static friction between the tires and the road surface. The static friction would presumably increase/decrease proportionally with weight of the bike & rider. The static friction results in centripetal force keeping the bike in the turn. If rider takes the turn too fast then the inertial force (apparent centrifugal force) overtakes the static friction of the tires/pavement and the bike slides out of the turn.
Then there is (2) the apparent centrifugal force "pushing" the bike towards the outside of the turn along the turn radius.
So the idea is that, if the rider puts his inside knee down on the ground while in the turn, that corresponding amount of weight is removed from the apparent centrifugal force. So there's a little less centrifugal force so that allow the rider to go a little bit faster then when his knee was not sliding on the ground.
I'm ignoring the fact that the static friction would be reduced at the tires as well. But as I recall the way it works out the rider could take the corner faster.
Any insights on how this problem is solved and the result (however small an increase is possible, or not) would be appreciated. Just looking to resolve a friendly discussion.
Thanks
Mike
What I remember and/or presume is that there is (1) the static friction between the tires and the road surface. The static friction would presumably increase/decrease proportionally with weight of the bike & rider. The static friction results in centripetal force keeping the bike in the turn. If rider takes the turn too fast then the inertial force (apparent centrifugal force) overtakes the static friction of the tires/pavement and the bike slides out of the turn.
Then there is (2) the apparent centrifugal force "pushing" the bike towards the outside of the turn along the turn radius.
So the idea is that, if the rider puts his inside knee down on the ground while in the turn, that corresponding amount of weight is removed from the apparent centrifugal force. So there's a little less centrifugal force so that allow the rider to go a little bit faster then when his knee was not sliding on the ground.
I'm ignoring the fact that the static friction would be reduced at the tires as well. But as I recall the way it works out the rider could take the corner faster.
Any insights on how this problem is solved and the result (however small an increase is possible, or not) would be appreciated. Just looking to resolve a friendly discussion.
Thanks
Mike