# Chapter 5 - Dynamics of Uniform Circular Motion - Problems - Page 138: 15

$v_{max}=22m/s$

#### Work Step by Step

During a turn, static friction $f_s$ acts as centripetal force $F_c$ that keeps the vehicle along the curve and not skidding off in another direction. In other words, $$f_s=F_c=\frac{mv^2}{r}$$ Since $f_s$ has a maximum value $f_s^{max}$, we have $$f_s^{max}=\frac{mv^2_{max}}{r}$$ The formula for $f_s^{max}$ is $f_s^{max}=\mu_sF_N$. In these cases, since the vehicle has no vertical acceleration, $F_N=mg$, so $f_s^{max}=\mu_smg$ Therefore, $$\mu_smg=\frac{mv^2_{max}}{r}$$ $$\mu_sg=\frac{v^2_{max}}{r}$$ First, we need to find radius of the curve $r$. From car A, we have $\mu_s=1.1$, $v_{max}=25m/s$ and $g=9.8m/s^2$ $$r=\frac{v^2_{max}}{\mu_sg}=58m$$ Now, for car B, we have $\mu_s=0.85$ and $g=9.8m/s^2$ $$v_{max}=\sqrt{\mu_sgr}=22m/s$$

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