A truck with 42-in.-diameter wheels is traveling at 60 mi/h.

"A truck with 42-in.-diameter wheels is traveling at 60 mi/h. ..."

60mph = 63360 inches/minute

Circumference of wheel = 42pi inches

Number of revolutions per minute = 63360/42pi ≈ 480

Number of radians per minute (angular speed) = Number of radians per revolution*Number of revolutions per minute = 2pi*63360/42pi ≈ 3017 

A truck with 42-in.-diameter wheels is traveling at 60 mi/h.

1. Find the angular speed of the wheels in rad/min:

Given: Linear speed V = \(60 \ \frac{mi}{h}\), 

Radius of Circular path \(r = \frac{42}{2}\ in.\)

The Angular Speed  \(\color{red}\mathbf{\omega = \frac{V}{r}}\)
\(\begin{array}{rcll} &=& \dfrac{ 60\ \frac{mi}{h} } {\frac{42}{2}\ in.} \\\\ &=& \dfrac{ 60\ \frac{mi}{h} } {21\ in.} \\\\ &=& \dfrac{ 60\ \frac{mi}{h}\cdot \frac{1\ h}{60\ min.}\cdot \frac{5280\cdot 12\ in. }{1\ mi.} } {21\ in.} \\\\ &=& \dfrac{ \frac{5280\cdot 12\ in. }{min.} } {21\ in.} \\\\ &=& \dfrac{5280\cdot 12 }{21}\cdot \frac{rad}{min.} \\\\ &=& 3017.14285714\ \frac{rad}{min.} \end{array} \)

2. How many revolutions per minute do the wheels make?

T = revolution around in time

Angular Speed is given by \(\color{red}\mathbf{\omega = \frac{2\pi}{T}}\).

\(\begin{array}{rcll} \dfrac{1}{T}&=& \dfrac{\omega} {2\pi} \\\\ &=& \dfrac{3017.14285714\ \frac{rad}{min.} } {2\pi\ rad} \\\\ &=& \dfrac{ 3017.14285714 } {2\pi} \frac{revolutions}{min.} \\\\ &=& 480.193199729\ \frac{revolutions}{min.} \end{array} \)