(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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 v^{2}/ 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?

2. Relevant equations

Moment about the bottom most point :

moment = W x d

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

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# Condition of rotational equilibrium in circular motion

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