Q.A cyclist speeding at $18\ \text{km/h}$ on a level road takes a sharp circular turn of radius $3\ \text{m}$ without reducing the speed. The co-efficient of static friction between the tyres and the road is $0.1$. Will the cyclist slip while taking the turn?

The key idea is that the cyclist will slip if the required centripetal force exceeds the maximum static friction available.

1. Convert speed to SI units:

   $v = 18\ \text{km/h} = 18 \times \frac{5}{18} = 5\ \text{m/s}$.

2. The required centripetal force is $F_c = \frac{mv^2}{r}$. The maximum static friction is $f_{\text{max}} = \mu_s mg$.

3. For no slip, we need $\frac{mv^2}{r} \le \mu_s mg$, i.e. $\frac{v^2}{r} \le \mu_s g$. …

The key here is understanding static friction limit. When a cyclist turns, friction provides the centripetal force needed to change direction. But friction can only supply so much force — once the demand exceeds that maximum, the tyres lose grip and the cyclist slips outward.

The maximum static friction is $\mu_s N$, and on a level road the normal force $N$ equals $mg$. So the maximum centripetal force friction can provide is $\mu_s mg$. The actual centripetal force required to stay on the circular path is $\frac{mv^2}{r}$. If $\frac{mv^2}{r} > \mu_s mg$, slipping occurs.

Let’s check.

1. Convert speed to SI units.

   $18\ \text{km/h} = 18 \times \frac{1000}{3600} = 5\ \text{m/s}$.

   This is a common conversion — multiply km/h by $\frac{5}{18}$ to get m/s.

2. Write the condition for no slipping.

   The cyclist will not slip if the required centripetal force is less than or equal to the maximum static friction:

Notice the mass $m$ cancels out — the result does not depend on the cyclist’s weight.

So the condition simplifies to:

3. Plug in the numbers.

   $v = 5\ \text{m/s}$, $r = 3\ \text{m}$, $\mu_s = 0.1$, $g = 9.8\ \text{m/s}^2$ (or $10\ \text{m/s}^2$ for quick checks — we’ll use $9.8$ for accuracy).

   Left side: $\frac{v^2}{r} = \frac{5^2}{3} = \frac{25}{3} \approx 8.33\ \text{m/s}^2$.

   Right side: $\mu_s g = 0.1 \times 9.8 = 0.98\ \text{m/s}^2$.

4. Compare. …

Concept: Comparing the Required Centripetal Force to the Maximum Static Friction Available

Step 1: Convert speed to SI units

Step 2: Write the "no slip" condition

The required centripetal force must not exceed the maximum static friction:

(the mass cancels, so the result is independent of the cyclist's weight).

Step 3: Compute both sides

$\frac{v^2}{r} = \frac{5^2}{3} = \frac{25}{3} \approx 8.33\ \text{m s}^{-2}$ …