If a cyclist moving with a speed of 4.9 m/s on a level road can take a sharp circular turn of radius 4 m, then

Question

If a cyclist moving with a speed of 4.9 m/s on a level road can take a sharp circular turn of radius 4 m, then coefficient of friction between the cycle tyres and road is:

Accepted Answer

1. **Principle:** For a cyclist to negotiate a circular turn on a level road without skidding, the necessary centripetal force is provided by the force of static friction between the tyres and the road [Source 72].
2. **Formula:** The limiting condition where the maximum static friction balances the centripetal force is:
   $$ \mu m g = \frac{m v^2}{R} $$
   Rearranging for the coefficient of friction ($\mu$):
   $$ \mu = \frac{v^2}{Rg} $$
3. **Calculation:**
   - Speed ($v$) = $4.9 	ext{ m/s}$
   - Radius ($R$) = $4 	ext{ m}$
   - Acceleration due to gravity ($g$) $\approx 9.8 	ext{ m/s}^2$ [Source 73].
   Substitute the values:
   $$ \mu = \frac{(4.9)^2}{4 	imes 9.8} $$
   $$ \mu = \frac{4.9 	imes 4.9}{4 	imes 2 	imes 4.9} $$
   $$ \mu = \frac{4.9}{8} = 0.6125 $$
   Rounding to two decimal places, $\mu \approx 0.61$.

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