- #1
capricornpeer
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Homework Statement
Consider a cyclist turning around a circular track with radius r and speed v as shown
in attached fig.
Let W be the weight of the system (cyclist + bicycle) (W = mg)
N = normal force by ground on the system
f = static friction by ground on the system
θ = angle of inclination between cyclist and the vertical
h = height of the centre of mass C of the system above the ground
d = perpendicular distance from the point of contact between the ground and
the tyres to the line of application of the weight
Horizontally:
f = m v2 / r
Vertically , for equilibrium:
N = mg
for rotational equlibrium, taking moment about the centre of mass (C), we get:
N d = f h
My question is : For rotational equilibrium, can't we take moment about the point of contact of cycle with track? As far as I know, for rotational equilibrium, moments can be taken about any point and their summation must be zero.
However, in this case, moment about the point of contact can't be zero as W produces a torque while torques due to N and f are zero. So the condition of rotational equilibrium gives two different conditions for two different points. Where is the fallacy?
Homework Equations
Moment about the bottom most point :
moment = W x d