Uniform Circular Motion — Question 3

Question 3

A banked road can turn a vehicle without friction

Take $g=10$ m/s$^2$. A frictionless road is banked upward away from a circular turn center. Its angle satisfies $\tan\theta=0.5$ and its horizontal turn radius is 50 m. A 1000 kg vehicle is modeled as a particle.

1. Find the speed that permits a steady horizontal circle.

2. Find the normal force.

3. Explain why setting N equal to mg is wrong and whether the same bank supports every speed without friction.

Question 3 - Solution

A banked road can turn a vehicle without friction

(a) Use one force for two required components.

The normal force has both vertical and inward components: $$N\cos\theta=mg,\qquad N\sin\theta=\frac{mv^2}{r}.$$ Dividing eliminates $N$ and $m$: $$\boxed{v=\sqrt{rg\tan\theta}=\sqrt{50(10)(.5)}=5\sqrt{10}=15.811\ \mathrm{m/s}}.$$

(b) Use vertical balance.

From $\tan\theta=.5$, $\cos\theta=2/\sqrt5$. Thus $$\boxed{N=\frac{mg}{\cos\theta}=\frac{1000(10)}{2/\sqrt5}=5000\sqrt5=11180.34\ \mathrm N}.$$

(c) Distinguish a component from the full force.

The component balances require $$N_y=N\cos\theta=mg,\qquad N_r=N\sin\theta=\frac{mv^2}{r}.$$ Therefore $N>mg$ for this nonzero bank angle. At fixed $r$ and $\theta$, only $v^2=rg\tan\theta$ satisfies both equations without friction; other speeds require a different force balance or path.