DynamicsEasy

Centripetal force — road curve

A car of m = 1200 kg takes a curve of radius R = 80 m at speed v = 60 km/h.\nCalculate the necessary centripetal force and the minimum static friction coefficient.

Given data

m = 1200 kgR = 80 mv = 60 km/hg = 9.81 m/s²

Review the theory: Dinamica

Steps

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Convert the speed v = 60 km/h to m/s.
Write your working — the numerical result must appear in the text m/s
What is the centripetal force needed to keep the car on the curve?
What is the weight of the car?
What is the minimum static friction coefficient for the car not to skid on the curve?

Full worked solution

Convert the speed v = 60 km/h to m/s.
$v = 60 \div 3.6$
$v = 60\,\mathrm{km/h} \div 3.6 = \mathbf{16.67\,m/s}$ . Converting km/h to m/s requires dividing by 3.6.
What is the centripetal force needed to keep the car on the curve?
$F_c = \dfrac{mv^2}{R} = \dfrac{1200 \cdot 16.67^2}{80}$
$F_c = mv^2/R = 1200 \times (16.67)^2 / 80 = 1200 \times 277.9 / 80 = \mathbf{4168\,N} \approx \mathbf{4.17\,kN}$ . The centripetal force points radially toward the centre of the curve.
What is the weight of the car?
$P = mg = 1200 \cdot 9.81$
$P = mg = 1200 \times 9.81 = \mathbf{11772\,N}$ . On a horizontal road, the normal force equals the weight: $N = P$ .
What is the minimum static friction coefficient for the car not to skid on the curve?
$\mu_s = \dfrac{F_c}{mg} = \dfrac{4168}{11772}$
$\mu_s \ge F_c/N = 4168/11772 = \mathbf{0.35}$ . Dry asphalt has $\mu \approx 0.7{-}0.9$ , well above the minimum; on ice ( $\mu \approx 0.1$ ) the car would skid.

Result:Centripetal force: 4168 N. Minimum friction coefficient: μ_s ≈ 0.35.

Centripetal force — road curve

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