Uniform Circular Motion — Question 2

Question 2

A level turn imposes a speed bound through friction

Take $g=10$ m/s$^2$. A vehicle modeled as a particle follows a level circular road of radius 40 m at constant speed. Static tire-road friction is the only horizontal force. Its coefficient is 0.40; ignore aerodynamic forces and rollover.

1. Derive the maximum speed without sliding.

2. Find the fraction of the available friction used at 10 m/s.

3. If the coefficient drops to 0.10, find the new maximum speed and the minimum radius at 10 m/s.

Question 2 - Solution

A level turn imposes a speed bound through friction

(a) Apply the radial demand to the friction bound.

Vertical balance gives $N=mg$. The static-friction bound must cover the radial demand: $$\frac{mv^2}{r}\leq\mu_sN=\mu_smg.$$ $$\boxed{v_{\max}=\sqrt{\mu_sgr}=\sqrt{.40(10)(40)}=4\sqrt{10}=12.649\ \mathrm{m/s}}.$$ Mass cancels in this particle model.

(b) Compare required and available magnitudes.

At 10 m/s, compare the required force with its capacity: $$\boxed{\frac{f_{\mathrm{actual}}}{f_{s,\max}}=\frac{mv^2/r}{\mu_smg}=\frac{100}{160}=.625=62.5\%}.$$ The actual static friction is below its maximum.

(c) Recompute the capacity.

With $\mu_s=.10$, $$\boxed{v_{\max}=\sqrt{.10(10)(40)}=2\sqrt{10}=6.325\ \mathrm{m/s}}.$$ Maintaining 10 m/s instead requires $$\boxed{r\geq\frac{v^2}{\mu_sg}=\frac{100}{.10(10)}=100\ \mathrm m}.$$