Explain why a cyclist bends while negotiating a curve road?

Explanation

Let us consider a cyclist negotiating a circular level road (not hanked) of radius r with a speed. The cycle and the cyclist are considered as one system with mass m. The center gravity of the system is C and it goes in a circle of radius r with center at O. Let us choose the line OC as X-axis and the vertical line through O as Z-axis as shown in Figure.
The system as a frame is rotating about Z-axis. The system is at rest in this rotating frame. To solve problems in rotating frame of reference, we have to apply a centrifugal force (pseudo force) or the system which will be `(mv^(2))/(r)`. This force will act through the center of gravity. The forces acting on the system are, (i) gravitational force (mg). (ii) normal force (N), (iii) frictional force (f) and (iv) centrifugal force `((mv^(2))/(r))`. As the system is in equilibrium in the rotational frame of reference, the net extemal force and net external torque must be zero. Let us consider all torques about the point A in Figure. For rotational equilibrium, `vec tau_("net")=0`
The torque due to the gravitational force about point A is (mg AB) which causes a clockwise turn that is taken as negative. The torque due to the centripetal force is BC which causes an centriipetal force is `((mv^(2))/(r)BC)` which causs an anti-clockwise turn that is taken as positive.
`-mg AB+(mv^(2))/(r) BC=0`
`mg AB=(mv^(2))/(r) BC`
From `DeltaABC`,
`AB=Ac sin theta BC =AC cos theta`
`mg AC sin theta=(mv^(2))/(r) AC cos theta`
`"tan" theta=(v^(2))/(rg)`
`thata =tan^(-1) ((v^(2))/(rg))`
While negotiating a circular level road of radius r at velocity v, a cyclist has to bend by an angle from vertical given by the above expression to stay in equilibrium (ie, to avoid a fall).