Question

A curve in a speed track has a radius of 1000 ft and a rated speed of 120 mi/h. (From Sample Prob. 12.7 is the definition of rated speed: The rated speed of a banked highway curve is the speed at which a car should travel to have no lateral friction force exerted on its wheels.) Knowing that a racing car starts skidding on the curve when traveling at a speed of 160 mi/h, determine

Answer 1

Answer:

$tan \theta = \frac{(176ft/s)^2}{1000 ft 32.2 ft/s^2}= 0.962$

$\theta = tan^{-1} (0.962) = 43.89$

Explanation:

If the question is: Determine the banking angle θ

We have the forces involved on the figure attached.

For this case we know that the weight is given by:

$W = mg$

And for this case the centripetal acceleration would be given by:

$a=\frac{v^2}{r}$

If we analyze the sum of forces on x and y we have:

$\sum F_x = m a_x$

$F + W sin \theta = ma cos theta$

And if we solve for the force we got:

$F = ma cos \theta - mg sin \theta = \frac{mv^2}{r} cos \theta - mg sin \theta$

$\sum F_y = m a_y$

$N - W cos \theta = ma sin \theta$

If we solve for the normal force we got:

$N =W cos \theta + ma sin \theta = \frac{mv^2}{r} sin \theta + mg cos \theta$

In order to find the banking angle we use the fact that F =0

$0 = \frac{mv^2}{r} cos \theta - mg sin \theta$

$tan \theta= \frac{v^2}{rg}$

The velocity on this case is 120 mi/h if we convert this into ft/ s we got:

$120 mi/h * \frac{5280 ft}{1mi} *\frac{1hr}{3600 s}= 176 ft/s$

And then we have this:

$tan \theta = \frac{(176ft/s)^2}{1000 ft 32.2 ft/s^2}= 0.962$

$\theta = tan^{-1} (0.962) = 43.89$