A car of mass 1200 kg travels around a flat curve of radius 50 m at a steady speed. If the static friction coefficient is μ_s = 0.3, what is the maximum speed to avoid slipping?

A car of mass 1200 kg travels around a flat curve of radius 50 m at a steady speed. If the static friction coefficient is μ_s = 0.3, what is the maximum speed to avoid slipping?

• A. sqrt(g R)
• B. sqrt(μ_s g R)
• C. sqrt(μ_s g / R)
• D. sqrt(μ_s g R)

Answer: (D)

The key idea is that static friction provides the centripetal force needed to keep the car moving in a circle on a flat road. The maximum static friction is μ_s times the normal force, and on a flat surface the normal force is mg. For circular motion, the required centripetal force is m v^2 / R. Setting the maximum friction equal to the needed centripetal force gives m v^2 / R = μ_s m g. The mass cancels, leaving v^2 / R = μ_s g, so v = sqrt(μ_s g R). With μ_s = 0.3, g ≈ 9.8 m/s^2, and R = 50 m, v_max ≈ sqrt(0.3 × 9.8 × 50) ≈ sqrt(147) ≈ 12 m/s. So the expression sqrt(μ_s g R) is the correct form for the maximum speed to avoid slipping.

The key idea is that static friction provides the centripetal force needed to keep the car moving in a circle on a flat road. The maximum static friction is μ_s times the normal force, and on a flat surface the normal force is mg. For circular motion, the required centripetal force is m v^2 / R. Setting the maximum friction equal to the needed centripetal force gives m v^2 / R = μ_s m g. The mass cancels, leaving v^2 / R = μ_s g, so v = sqrt(μ_s g R).

With μ_s = 0.3, g ≈ 9.8 m/s^2, and R = 50 m, v_max ≈ sqrt(0.3 × 9.8 × 50) ≈ sqrt(147) ≈ 12 m/s. So the expression sqrt(μ_s g R) is the correct form for the maximum speed to avoid slipping.