Q.Determine the angle to be made with the vertical by a two wheeler rider while turning on a horizontal track. (Hint: For both this and the four-wheeler case above, find the torque that balances the torque due to centrifugal force and torque due to static friction force.)

Reason it out from torque/force balance:

For a two-wheeler there is no wheel-pair to share a stabilising couple — the rider must lean so that the resultant of the forces passes through the line of contact. In the rider's frame, three effects act about the contact point: the weight mg (down, through the centre of mass), the centrifugal force $mv^2/r$ (outward, at height h along the lean line), and friction at the contact. Balancing the toppling torque of the centrifugal force against the restoring torque of the weight about the contact point: with lean angle θ from the vertical, the moment arms are $h\cos\theta$ for the centrifugal force and $h\sin\theta$ for the weight, giving $\frac{mv^2}{r}\,h\cos\theta = mg\,h\sin\theta \quad\Rightarrow\quad \tan\theta = \frac{v^2}{rg}$ So the faster or tighter the turn, the deeper the lean: $\theta=\tan^{-1}(v^2/rg)$ — precisely the same expression as a banked road's ideal banking angle (Eq. 1.2), which is why a leaning rider and a banked track are two faces of one requirement.

Work in the inertial frame: friction $f=mv^2/r$ (toward centre) and normal reaction N = mg act at the contact; for the resultant of N and f to pass along the rider's body line (no net torque about the centre of mass), $\tan\theta = f/N = v^2/rg$ — the identical answer without any pseudo force.

Measuring θ from the horizontal instead of the vertical; balancing forces instead of TORQUES (the lean condition is fundamentally a torque balance about the wheel contact line).