A car traveling 20 m/s rounds an 80 m radius horizontal curve with the tires on the verge of slipping. How fast can this car round a second curve of radius 320 m? (Assume the same coefficient of friction between the car's tires and each road surface.) A) 70 m/s B) 40 m/s C) 30 m/s D) 80 m/s E) 1.6 × 10² m/s

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Quizplus · Accepted Answer

The centripetal force required to keep a car moving in a circle is provided by the frictional force between the tires and the road. This force can be expressed as $F = \mu mg$, where $\mu$ is the coefficient of friction, $m$ is the mass of the car, and $g$ is the acceleration due to gravity. The centripetal force is also given by $F = \frac{mv^2}{r}$, where $v$ is the velocity of the car and $r$ is the radius of the curve. Setting these two expressions for $F$ equal to each other gives $\mu mg = \frac{mv^2}{r}$. The mass of the car cancels out, leading to $\mu g = \frac{v^2}{r}$. Since $\mu g$ is constant (the same coefficient of friction and the same gravitational acceleration), we can say that $v^2 \propto r$, meaning the square of the velocity is directly proportional to the radius of the curve.Given that the car can travel at 20 m/s around an 80 m radius curve, we can find the proportional velocity for a 320 m radius curve by setting up the ratio $\frac{v_1^2}{r_1} = \frac{v_2^2}{r_2}$, where $v_1 = 20$ m/s, $r_1 = 80$ m, and $r_2 = 320$ m. Solving for $v_2$, we get $v_2 = \sqrt{\frac{r_2}{r_1}} \cdot v_1 = \sqrt{\frac{320}{80}} \cdot 20 = \sqrt{4} \cdot 20 = 2 \cdot 20 = 40$ m/s.

A Car Traveling 20 M/s Rounds an 80 M Radius