Uniform Circular Motion — Question 5

Question 5

Friction on a bank changes direction across the design speed

Take $g=10$ m/s$^2$. A banked turn has horizontal radius 30 m, $\tan\theta=0.5$ and $\mu_s=0.25$. The road rises outward. Treat the vehicle as a particle; exclude rollover. Let signed friction f be positive down the bank, toward the center.

1. Write vertical and inward force equations for steady motion.

2. Derive the minimum and maximum allowed speeds.

3. Find the speed at which friction vanishes and explain its direction near each endpoint.

Question 5 - Solution

Friction on a bank changes direction across the design speed

(a) Keep signed friction in both components.

Resolve forces vertically and horizontally inward, retaining the signed friction: $$\boxed{N\cos\theta-f\sin\theta=mg},\qquad \boxed{N\sin\theta+f\cos\theta=\frac{mv^2}{r}}.$$ The contact constraint is $|f|\leq\mu_sN$, where positive $f$ points down the bank.

(b) Use both limiting directions.

Let $q=f/N\in[-.25,.25]$. Dividing the component equations gives $$\frac{v^2}{rg}=\frac{\tan\theta+q}{1-q\tan\theta},\qquad \frac{d}{dq}\frac{\tan\theta+q}{1-q\tan\theta}=\frac{1+\tan^2\theta}{(1-q\tan\theta)^2}>0.$$ The extrema therefore occur at the two friction limits: $$v_{\min}^2=300\frac{.5-.25}{1+.125}=\frac{200}{3},\qquad v_{\max}^2=300\frac{.5+.25}{1-.125}=\frac{1800}{7}.$$ $$\boxed{8.165\leq v\leq16.036\ \mathrm{m/s}}.$$

(c) Locate the frictionless speed.

At zero friction, $$\boxed{v_0=\sqrt{rg\tan\theta}=\sqrt{150}=12.247\ \mathrm{m/s}}.$$ The low-speed endpoint has $q=-\mu_s$: friction points up the bank. The high-speed endpoint has $q=+\mu_s$: friction points down the bank. Throughout the interval, $$N=\frac{mg}{\cos\theta-q\sin\theta}>0,$$ since $1-q\tan\theta\geq.875>0$.