Q.While driving along an unbanked circular road, a two-wheeler rider has to lean with the vertical. Why is it so? With what angle does the rider have to lean? Derive the relevant expression. Why is such a leaning not necessary for a four wheeler?

The rider leans to create a torque, about the contact point, that balances the torque due to the centrifugal force at the (elevated) centre of mass; the required angle is $\theta=\tan^{-1}(v^2/rg)$, the same expression as the banking-angle formula. A four-wheeler's wide wheelbase balances this via unequal normal reactions instead, so it needn't lean.

On an UNBANKED (flat) circular road, a two-wheeler taking a turn of radius r at speed v needs static friction $f_s=\frac{mv^2}{r}$ from the road to supply the centripetal force -- but this friction acts at GROUND level, at the single narrow contact patch of the tyres, while the vehicle's weight mg (and, working in the rotating frame, the centrifugal pseudo-force $\frac{mv^2}{r}$) effectively acts at the centre of mass, which is ELEVATED above the ground. Taken about the point of contact, friction (acting through the contact point itself) produces no torque, but the centrifugal force at the elevated centre of mass DOES produce a toppling torque, tending to rotate the vehicle outward and over -- and with nothing else to counter it, an upright two-wheeler taking a fast enough flat turn would simply topple outward.

The rider prevents this by LEANING the whole vehicle (and themselves) INWARD, towards the centre of the curve, by an angle $\theta$ from the vertical. Leaning shifts the line of action of the (now effectively combined) weight-and-centrifugal-force system so that, in the ROTATING frame attached to the vehicle, the net TORQUE about the point of contact becomes zero for the correct angle. Working in this frame, the two 'apparent' forces at the centre of mass are weight mg (vertically down) and the centrifugal force $\frac{mv^2}{r}$ (horizontally outward); for rotational equilibrium about the contact point, the torque of mg (moment arm = the horizontal offset of the centre of mass from the contact point, once leaned) must exactly balance the torque of the centrifugal force (moment arm = the vertical height of the centre of mass above the contact point). Geometrically, leaning at angle $\theta$ from the vertical makes this horizontal offset equal to (height)$\times\tan\theta$, so balancing the two torques (each force times its own moment arm) gives $mg\cdot(\text{height})\tan\theta = \frac{mv^2}{r}\cdot(\text{height}) \quad\Rightarrow\quad \tan\theta=\frac{v^2}{rg} \quad\Rightarrow\quad \theta=\tan^{-1}\left(\frac{v^2}{rg}\right)$ -- the identical mathematical expression as Eq. (1.2), the banking-angle formula, even though this time it is the RIDER'S body angle (not the road's tilt) providing the balance.

A four-wheeler does NOT need to lean because it does not rely on a single narrow contact line for stability -- it has FOUR separate contact points spread across a comparatively wide track/wheelbase. Rather than the whole vehicle body tilting, the toppling torque due to the centrifugal force is instead balanced by a REDISTRIBUTION of the normal reaction among the four tyres: the OUTER tyres (farther from the centre of the turn) experience an INCREASED normal reaction, and the INNER tyres a correspondingly DECREASED one, the resulting unequal-N torque taking the place of the leaning-angle torque a two-wheeler must generate through its own tilt. Only if the speed becomes large enough that the inner tyres' normal reaction would need to go negative does a genuine four-wheeler actually begin to topple (the inner wheels lifting off the ground).

Work in the rotating frame attached to the vehicle and balance the torque of the centrifugal (pseudo) force at the elevated centre of mass against the torque of weight, about the point of ground contact, to find the required lean angle; contrast this with a four-wheeler's wider base, which balances the same toppling torque via unequal normal reactions instead.

Trying to explain the leaning purely through 'centripetal force needs a horizontal component of weight', which is the banked-ROAD argument, rather than the correct TORQUE-balance argument needed here, since the road itself is flat and unbanked in this question.