Question

question 8
1 pts
a truck makes a turn around a circular stretch of highway with some radius
\. the truck can travel at a maximum speed of \v\ without veering off of the road. it takes some centripetal force provided by friction equal to \f\ to keep the truck on the road. if the radius is decreased to
/3\ one third of the original radius size, how large of a centripetal force will be needed to keep it on the road?

Explanation:

Step1: Write the formula for centripetal force

The centripetal force formula is \(F = \frac{mv^{2}}{r}\), where \(m\) is the mass of the truck, \(v\) is the speed, and \(r\) is the radius of the circular path.

Step2: Analyze the initial and new - radius cases

Initially, \(F=\frac{mv^{2}}{r}\). When the radius is \(r'=\frac{r}{3}\), let the new centripetal force be \(F'\). Then \(F'=\frac{mv^{2}}{r'}\).
Substitute \(r' = \frac{r}{3}\) into the formula for \(F'\): \(F'=\frac{mv^{2}}{\frac{r}{3}}\).
Using the rule of dividing by a fraction (\(\frac{a}{\frac{b}{c}}=\frac{ac}{b}\)), we get \(F' = 3\times\frac{mv^{2}}{r}\).
Since \(F=\frac{mv^{2}}{r}\), then \(F' = 3F\).

Answer:

The centripetal force needed is \(3F\).