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In the problem of computing leaning angle ($\theta$) of a turning bike, the traditional approach is to move to the accelerating reference frame and compute the balanced torques with respect to the ground contact point generated by gravity and centrifugal force. I am ok with this solution and able to derive $$\tan(\theta) = \frac{v^2}{gR},$$ where $v$ is the bike speed, $g$ the acceleration of gravity and $R$ is the radius of the turn.

My question is how to derive the above formula from the stationary inertial reference frame. This post Lean angle of a turning bicycle has a derivation of a more generalised formula taking into account the gyroscope effect. However, we can assume the ideal case that there is no wheel, meaning that the wheel of the bike has zero radius (hence not rotating) so that the angular momentum carried by wheel is zero. In this case the derivation therein seems not valid. Another way to say this is that the leaning angle formula is obviously independent of the internal structure of the bike.

Consider we observe the circular motion in the stationary frame. My naive idea was that the net nonzero torque caused by gravity should be used to change some angular momentum computed from some reference point. Some people argue that taking the ground contact point as the reference point is invalid, but I don't see why.

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  • $\begingroup$ No feedback to the answers given? $\endgroup$
    – KDP
    Commented Mar 8 at 22:28

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For organizing the computation for the stationary frame:

For simplification we can instead think of it as the lean of an ice skater, going around the curve of an ice rink

In order to go around the curve a centripetal force is required. The skater leans into the curve. The skate blades grip the ice. The reaction force exerted by the ice on the skater can be decomposed in a vertical and a horizontal component. The vertical component is equal to the force required to counteract gravity, the horizontal component is providing the required centripetal force. The larger the required centripetal force the larger the corresponding required lean, since the magnitude of the vertical component is a given.

The practical choice of reference point is the midpoint of the circular motion; while going around the curve: the required centripetal force is continuously towards that midpoint. While going around the curve: with uniform angular velocity the angular momentum with respect to the midpoint of the curve is a constant.

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  • $\begingroup$ I am fine with skater going circular motion and force decompositions etc. But again, what if we take the contact point of the reference for computing torque? in inertial frame. $\endgroup$
    – chichi
    Commented Feb 25 at 11:31
  • $\begingroup$ @chichi In the stationary frame the contact point is in non-inertial motion, so for reference of the motion that is not an option. In any context: in order to be able to set up computation the point of reference must be a point that is in inertial motion. (That requirement of reference-point-in-inertial-motion applies equally to using a rotating coordinate system; the equation of motion for the rotating coordinate system contains the angular velocity of the rotating coordinate system wrt the non-rotating coordinate system. The inertial coordinate system is always the underlying reference.) $\endgroup$
    – Cleonis
    Commented Feb 25 at 11:43
  • $\begingroup$ In inertial frame which is stationary, we can still do computation with respect to a stationary contact point, right? We simply pick the stationary ground point O which coincides with the contact point at a moment to do the analysis. $\endgroup$
    – chichi
    Commented Feb 25 at 11:50
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    $\begingroup$ @chichi, you can do that. But the since the bike is accelerating with reference to such a point, you get a changing angular momentum, even if the lean angle remains constant. You can't assume $\Delta L = 0$ as you can in the accelerated frame. $\endgroup$
    – BowlOfRed
    Commented Feb 27 at 4:40
  • $\begingroup$ @BowlOfRed I challenge your 'you can do that'. What chichi proposes is self-contradicting. Chichi proposes to use a stationary frame, and at some point in time the contact point coincides with the zero point of that stationary frame. That's not even an instantaneously co-moving frame (which wouldn't work either.) Chichi can introduce the acceleration as a fictitions acceleration, but the magnitude of that acceleration is obtained from the angular velocity wrt the inertial coordinate system. The only way to obtain a result is to use the inertial coordinate system as the underlying reference. $\endgroup$
    – Cleonis
    Commented Feb 27 at 20:40

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