their reasoning was that when the bike lean towards horizontal the reaction acting on the bike by surface will act in an angle through the center of gravity of the slanted bike(not perpendicular to the surface). therefore the horizontal component of that angled reaction will provide the centripetal force even though friction is not present. they also said that if the friction is present this component will be added to the friction and their resultant will act as centripetal force. i find this hard to believe.i think it is impossible to turn on a frictionless horizontal surface. can someone please help me find the correct answer?
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1$\begingroup$ Sounds like a great way to fall over… $\endgroup$– Jon CusterCommented Sep 10, 2023 at 20:12
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2$\begingroup$ Ask the proponent of this idea to give a practical demonstration at your local ice rink. Stand well clear if they attempt this ... $\endgroup$– gandalf61Commented Sep 10, 2023 at 20:26
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2 Answers
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You are correct. The only forces acting on the bike are gravity (straight down) and normal force (straight up). The net force on the bike will therefore have no horizontal component, so the bike will not turn.
Though our biker is already disappointed at their inability to turn, their day is about to get a whole lot worse. The slant of the biker causes a nonzero torque acting to increase the slant, which will cause the biker to quickly fall over.
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The only way you can turn on a frictionless surface is if there is some horizontal reaction force on the wheels acting towards the center of circular motion. With no friction you would need a banked roadway to provide that force.
To determine the max speed on a banked roadway with no friction, plug in zero for the coefficient of friction and a very small value for $\theta$ for the banking angle in the following link.
http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/carbank.html
Hope this helps.
