Question

A certain light truck can go around a flat curve having a radius of 150 m with a maximum speed of 32.5 m/s. With what maximum speed can it go around a curve having a radius of 66.0 m

Answer 1

Answer:

Approximately $21.6\; {\rm m\cdot s^{-1}}$, assuming that the coefficient of kinetic friction between the tires and the ground stays the same.

Explanation:

Let $\mu_{k}$ denote the coefficient of kinetic friction between the tires and the ground.

Let $m$ denote the mass of the truck. Let $g$ denote the gravitational field strength. The weight of this truck would be $m\, g$. Since the curve is flat, the magnitude of the normal force $F_\text{normal}$ between the truck and the ground would also be $m\, g\!$.

Let $v = 32.5\; {\rm m\cdot s^{-1}}$ denote the speed at which the truck goes around the curve with a radius of $150\; {\rm m}$. The (horizontal) centripetal acceleration of the truck would be:

$\displaystyle a = \frac{v^{2}}{r}$.

The horizontal net force on the truck (of mass $m$) would need to be exactly:

$\displaystyle F_\text{net} = m\, a = \frac{m\, v^{2}}{r}$.

The friction between the tires of the truck and the ground is the only horizontal force on the truck. Thus, this friction would need to supply the entirety of this horizontal net force:

$\displaystyle F_\text{friction} = F_\text{net} = \frac{m\, v^{2}}{r}$.

Notice that increasing $v$ would require an increase in the kinetic friction between the truck, $F_\text{friction}$. However, since $F_\text{friction}\!$ is at most $\mu_{k}\, F_\text{normal}$ (which is equal to $\mu_{k}\, m\, g$):

$\begin{aligned}\frac{m\, v^{2}}{r} &= F_\text{friction} \\ & \le \mu_{k}\, F_\text{normal} \\ &= \mu_{k}\, m\, g\end{aligned}$.

Therefore, when the radius of the curve is $r^{\prime}$, the maximum velocity of this truck would be:

$\begin{aligned}v^{\prime} &= \sqrt{\mu_{k}\, g\, r^{\prime}}\end{aligned}$.

Likewise, it is possible to express $\mu_{k}$ in terms of $r$, $g$, and max speed $v$:

$\displaystyle \mu_{k} = \frac{v^{2}}{r\, g}$.

Substitute this expression for $\mu_{k}$ back into the expression for $v^{\prime}$:

$\begin{aligned}v^{\prime} &= \sqrt{\mu_{k}\, g\, r^{\prime}} \\ &= \sqrt{\frac{v^{2}}{r\, g}\, g\, r^{\prime}} \\ &= \sqrt{v^{2}\, \frac{r^{\prime}}{r}}\end{aligned}$.

Given that $v = 32.5\; {\rm m\cdot s^{-1}}$, $r = 150\; {\rm m}$, and $r^{\prime} = 66.0\; {\rm m}$:

$\begin{aligned}v^{\prime} &= \sqrt{v^{2}\, \frac{r^{\prime}}{r}} \\ &= \sqrt{(32.5\; {\rm m\cdot s^{-1}})^{2} \times \frac{66.0\; {\rm m}}{150\; {\rm m}}} \\ &\approx 21.6\; {\rm m\cdot s^{-1}}\end{aligned}$.