Question

A car of mass M = 1000 kg traveling at 50.0 km/hour enters a banked turn covered with ice. The road is banked at an angle θ , and there is no friction between the road and the car's tires as shown in (Figure 1). Use g = 9.80 m/s2 throughout this problem.What is the radius r of the turn if θ = 20.0 ∘ (assuming the car continues in uniform circular motion around the turn)?

Answer 1

The radius of the curved road at the given condition is 54.1 m.

The given parameters:

• mass of the car, m = 1000 kg
• speed of the car, v = 50 km/h = 13.89 m/s
• banking angle, θ = 20⁰

The normal force on the car due to banking curve is calculated as follows;

$Fcos(\theta) = mg$

The horizontal force on the car due to the banking curve is calculated as follows;

$Fsin(\theta) = \frac{mv^2}{r}$

Divide the second equation by the first;

$\frac{Fsin(\theta)}{Fcos(\theta) } = \frac{mv^2}{rmg} \\\\tan(\theta) = \frac{v^2}{rg} \\\\r = \frac{v^2}{g \times tan(\theta)} \\\\r = \frac{13.89^2}{9.8 \times tan(20)} \\\\r = 54.1 \ m$

Thus, the radius of the curved road at the given condition is 54.1 m.

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