Q.Explain why a cyclist bends while negotiating a curved road. Arrive at the expression for the angle of bending for a given velocity.

Step 1. Setting up the system.

Consider a cyclist negotiating a level (unbanked) circular road of radius $r$ at speed $v$. Treat the cyclist and cycle together as one system of mass $m$, with combined center of gravity $C$. The system moves in a circle of radius $r$ about some center $O$ of the curve; let $A$ be the point where the wheels touch the road, and $B$ the foot of the perpendicular from $C$ onto the vertical through $A$, so that $A$, $B$, $C$ form a right triangle with the lean angle $\theta$ (measured from the vertical) at $A$.

Step 2. Why a rotating frame is used.

Because the system as a whole is going around the curve, it is most convenient to analyse it in a frame that co-rotates with the cyclist, in which the cyclist appears momentarily at rest. This frame is non-inertial (it is itself accelerating centripetally), so Newton's laws only apply in it once a pseudo (centrifugal) force of magnitude $\dfrac{mv^2}{r}$ is included explicitly, acting outward through the system's center of gravity $C$.

Step 3. The four forces on the system.

In this rotating frame, four forces act: (i) the weight $mg$, acting vertically downward through $C$; (ii) the normal reaction $N$ from the road, at the contact point $A$; (iii) friction $f$ from the road, also at $A$; and (iv) the centrifugal pseudo-force $\dfrac{mv^2}{r}$, acting horizontally outward through $C$. Since the cyclist appears at rest in this frame, the system is in equilibrium here: both the net force and the net torque must vanish, exactly as in ordinary static equilibrium.

Step 4. Taking torques about the contact point $A$.

Choosing $A$ as the reference point is deliberate: both $N$ and $f$ act exactly at $A$, so neither contributes any torque about it, leaving only $mg$ and the centrifugal force in the torque equation. The weight's torque is $mg\,(AB)$, tending to rotate the system clockwise (taken negative); the centrifugal force's torque is $\dfrac{mv^2}{r}(BC)$, tending to rotate it anticlockwise (taken positive). Setting the net torque to zero:

Step 5. Bringing in the geometry.

From the right triangle $ABC$, with $\theta$ the angle the cyclist leans from the vertical at $A$: the horizontal leg is $AB=AC\sin\theta$ and the vertical leg is $BC=AC\cos\theta$. Substituting these into the torque-balance equation:

Step 6. Solving for $\theta$.

The mass $m$ and the common length $AC$ cancel from both sides, leaving

Step 7. Interpretation.

So the required lean angle grows with the square of the speed and shrinks as the turning radius increases — a sharper, faster turn genuinely demands a more pronounced inward lean, exactly matching everyday cycling experience. Leaning too little leaves a net outward torque (the cyclist tends to fall outward); leaning too much leaves a net inward torque (the cyclist tends to fall inward) — only the angle given by $\tan\theta=v^2/(rg)$ balances the system exactly.

Torque balance in the rotating (non-inertial) reference frame, using the centrifugal pseudo-force, taken about the point of contact

• Forgetting that the centrifugal force is a pseudo-force that appears only in the rotating (cyclist's) frame — analysing the problem in the ground (inertial) frame without introducing it, or without instead correctly using the real centripetal requirement, leads to a missing or wrong torque term.
• Taking torques about the center of gravity $C$ instead of the contact point $A$ — this leaves the unknown reaction force $N$ and friction $f$ both present in the torque equation, since they no longer act through the reference point, so the derivation does not simplify cleanly to $\tan\theta=v^2/(rg)$.
• Swapping which side of the triangle is $\sin\theta$ and which is $\cos\theta$ — since $\theta$ is measured from the VERTICAL, the horizontal leg $AB=AC\sin\theta$ and the vertical leg $BC=AC\cos\theta$; reversing these inverts the final formula.