Explain why a cyclist bends while negotiating a curve road?

Explanation

Let us consider a cyclist negotiating a circular level road (not bankde) of radius r with a speed v. The cycle and the cylcist are considered as one system with mass m. The centre of gravity of the system is C and it goes in a circle of radius r with centre at O. Let us choose the line OC as X-axis and the vertical line through O as Z-axis as shown in the figure.

the system as a frame is rotating about z-axis. The system is at rest in this rotating frame. The forces acting on the system are, (i) gravitational force (mg), (ii) normal force (N),(ii) frictional force (f) and (iv) centrifugal force `((mv^(2))/r)`. As the system is in equilibrium in the rotaional frame of reference, the net external force and net external torque must be zero. Let us consider all torques about the point A in th figure.

Conditions for rotational equilibrium are
`ve(tau)_("net")=0`
(i) The torque due to the gravitational force about point a is (mg AB) which causes a clockwise turn that is taken as negative.
(ii) The torque due to the centripetal force is `((mv^(2))/r BC)` which causes an anticlockwise 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` and `BC =AC cos theta`
`mg AC sin theta =(mv^(2))/r AC cos theta`
`tan theta=(v^(2))/(rg)`
`tan theta=(v^(2))/(rg)`
`theta=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 `theta` from vertical `theta=tan^(-1)((v^(2))/(rg))` to stay in equilibrium (i.e, to avoid a fall). So the cyclist bends while negotiating a curve road.