Circular Motion Dynamics and Banking of Roads

A particle moving in a circle, even at constant speed, is constantly accelerating because its velocity direction changes. By Newton's second law, this acceleration must be caused by a net force directed toward the centre — the centripetal force.

Concept

For uniform circular motion of radius $r$ with speed $v$, the centripetal acceleration is

$a_c = \frac{v^2}{r} = \omega^2 r$

directed toward the centre. The centripetal force needed is

$F_c = \frac{mv^2}{r} = m\omega^2 r$

This force is not a new kind of force — it is whatever real force happens to point toward the centre: tension in a string for a stone on a string, gravity for a satellite, friction for a car on a flat curve, the horizontal component of normal force for a car on a banked curve, electrostatic attraction for an electron around a nucleus, and so on.

Banking of roads. Curves on highways are tilted inward so that the horizontal component of the normal force provides centripetal force, reducing reliance on friction.

For a frictionless banked road at angle $\theta$, the safe speed is

$v = \sqrt{rg \tan\theta}$

With friction (coefficient $\mu$), the maximum safe speed is

$v_{\max} = \sqrt{\frac{rg(\mu + \tan\theta)}{1 - \mu\tan\theta}}$

Derivation

Frictionless banked curve. Let the road be inclined at angle $\theta$ to the horizontal. The forces on a car of mass $m$ are weight $mg$ down and normal force $N$ perpendicular to the road surface.

Resolve $N$ into:

• vertical: $N\cos\theta$ balancing gravity → $N\cos\theta = mg$.
• horizontal: $N\sin\theta$ providing centripetal force → $N\sin\theta = \frac{mv^2}{r}$.

Dividing,

$\tan\theta = \frac{v^2}{rg}$

$\boxed{v = \sqrt{rg\tan\theta}}$

With friction (maximum speed). At maximum speed, friction acts down the slope (preventing the car from skidding outward). Resolving:

• Perpendicular to road: $N = mg\cos\theta + \frac{mv^2}{r}\sin\theta$.
• Parallel to road (toward centre — wait, easier to use horizontal/vertical):
• Vertical: $N\cos\theta - \mu N\sin\theta = mg$.
• Horizontal: $N\sin\theta + \mu N\cos\theta = \frac{mv^2}{r}$.

Dividing,

$\frac{v^2}{rg} = \frac{\sin\theta + \mu\cos\theta}{\cos\theta - \mu\sin\theta} = \frac{\mu + \tan\theta}{1 - \mu\tan\theta}$

$\boxed{v_{\max} = \sqrt{\frac{rg(\mu + \tan\theta)}{1 - \mu\tan\theta}}}$

Worked Example

A curve on a highway has radius 200 m and is banked at angle $\theta$ such that no friction is needed for a car travelling at 20 m/s. (a) Find $\theta$. (b) If the road has $\mu = 0.2$, find the maximum safe speed. Take $g = 10 \, \text{m/s}^2$.

Solution:

(a) $\tan\theta = v^2/(rg) = 400/(200 \times 10) = 0.2$. So $\theta \approx 11.3°$.

(b) With $\mu = 0.2$:

$v_{\max}^2 = \frac{rg(\mu + \tan\theta)}{1 - \mu\tan\theta} = \frac{200 \cdot 10 (0.2 + 0.2)}{1 - 0.04} = \frac{800}{0.96} \approx 833 \, \text{m}^2/\text{s}^2$

$v_{\max} \approx 28.9 \, \text{m/s}$

Common Confusions

• "Centrifugal force pushes you outward." In an inertial frame, there is no centrifugal force; you feel a tendency to continue in a straight line (Newton's first law). In the rotating frame, centrifugal force is a pseudo force.
• Centripetal force is not a new force. It's a requirement, met by existing forces (tension, friction, gravity, normal).
• No work done by centripetal force. Because it is always perpendicular to velocity, work $\int \vec{F}\cdot d\vec{r}$ is zero. Hence speed is constant in uniform circular motion.
• Banking reduces friction need, not eliminates it. Real roads use both banking and friction.

Key Takeaways

• Uniform circular motion has centripetal acceleration $v^2/r$ directed toward the centre.
• Centripetal force comes from some real force in the situation (tension, friction, normal).
• For a frictionless banked road: $v = \sqrt{rg\tan\theta}$.
• With friction: $v_{\max} = \sqrt{rg(\mu+\tan\theta)/(1-\mu\tan\theta)}$.
• Centripetal force does no work.