Q.A car of mass $1200\ \text{kg}$ moves round a level (unbanked) circular curve of radius $50\ \text{m}$ at a constant speed of $54\ \text{km/h}$. Find

Given: mass $m = 1200\ \text{kg}$, radius $r = 50\ \text{m}$, speed $v = 54\ \text{km/h} = 54 \times \dfrac{1000}{3600} = 15\ \text{m/s}$, $g = 9.8\ \text{m/s}^2$.\n\n**(a) Centripetal force:** $F_c = \frac{mv^2}{r} = \frac{1200 \times 15^2}{50} = \frac{1200 \times 225}{50} = 5400\ \text{N}$\n\n**(b) Minimum coefficient of friction.** Since the road is level (unbanked), the entire centripetal force must be supplied by friction between the tyres and the road, up to its maximum value $f_{\max} = \mu_s N = \mu_s mg$. For the car to just barely avoid skidding, $$\mu_s mg = \frac{mv^2}{r} \q …

Convert the speed to m/s, find $F_c = mv^2/r$, then set this equal to the maximum available friction $\mu_s mg$ and solve for $\mu_s = v …

• Forgetting to convert $54\ \text{km/h}$ to $15\ \text{m/s}$ before substituting. …