Circular Motion Question - fully algebraic

Question

I'm sort of struggling with this problem, I've explained my reasoning so far below.

This is the problem: A bend in the road is a horizontal circular arc of radius $\ r \ $. The surface of the bend is banked at an angle $\ a \ $ to the horizontal. When a vehicle is driven round the bend there is no tendency to slip. Show that the speed of the vehicle is ${ \sqrt{r g\tan a }}.$

So I know that ${ v = \sqrt{ar }}$, so there is where my answer will come from, but how do I find ${a}$? I did this topic a week ago so I think it might involve some horizontal and vertical resolving, but I can't quite remember how to complete this. I'd really appreciate some help. Thanks

Answer 1

• A stationary vehicle on a horizontally flat road experiences two forces: the downward force of gravity and the upward normal force. We expect these forces to cancel, leading to no net up/down acceleration.

• You may have solved "rolling down a slope" problems before. On a slanted road, gravity still points downward, but the normal force always points perpendicular to the slanted ground, and so it is no longer vertical.

  

• As such, the normal force has some nonzero component pointing toward the center of the circular bend in the road. Every centripetal force comes from somewhere—from the force of tension in a string, from the force of gravity, etc. For us, the centripetal acceleration will come from the component of the normal force that points in the centripetal direction.

• You can divide the normal force into its vertical and horizontal components. Trigonometry will tell you that the horizontal component of the normal force is $N\cdot \sin{\theta}$, where $N$ is the entire magnitude of the normal force and $\theta$ is the angle of the slope with respect to the horizontal (as shown in the figure).

• Similarly, if the normal force is supposed to perfect cancel gravity ("no tendency to slip") so the car makes the turn without sliding up or down the slope, trigonometry tells you that the vertical component of the normal force ($N\cos\theta$) is equal in magnitude to the downward force due to gravity $mg$.
• Combining these two equations, we have that when gravity is canceled out by the normal force, the horizontal component of the normal force is equal to: $$N\cdot \sin{\theta} = \frac{mg}{\cos{\theta}}\sin \theta = mg\tan{\theta}$$
• Of course, using $F=ma$, this means that the acceleration in the horizontal direction is just $$a_N = g\tan{\theta}$$
• Centripetal acceleration, as you point out, follows the rule $a_c = {v^2/r}$. If the horizontal normal acceleration is going to constitute a kind of centripetal acceleration, then $$a_N = g\tan{\theta} = v^2/r$$
• We can solve for $v$, getting the desired result: $$v = \sqrt{r\cdot g \cdot \tan{\theta}}$$ This is the speed required in order to make the turn.