Answer
a. $ 160\pi$ rad/min
b. $6.19$ mi/h
Work Step by Step
Suppose a point moves along a circle of radius $r$
and the ray from the center of the circle to the point
traverses $\theta$ radians in time $t$.
Let $ s=r\theta$ be the distance the point travels in time $t$.
The angular speed of the point is $\omega=\theta/t$.
The linear speed of the point is $v=s/t$.
$ v=r\omega$.
1 revolution = $ 2\pi$ radians.
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A point on the sprocket and a point on the pedal have equal linear speeds.
The linear speed of the pedal,
$ v_{p}=r_{p}\omega =($4 in$)\cdot(40\cdot 2\pi$ rad/min$)=320\pi$ in/min
a.
$\displaystyle \omega=\frac{v_{S}}{r_{s}}=\frac{320\pi\ in/min}{2\ in}=160\pi$ rad/min
b.
The linear speed of a point on the outer rim of the wheel is the speed of the bicycle.
1ft=12 in
1 mi=5280 ft
1 h = 60 min
$ v=r\omega$
$=(13\ in)(\displaystyle \frac{1\ ft}{12\ in})(\frac{1\ mi}{5280\ ft})\cdot(\frac{160\pi\ rad}{1\ min})(\frac{60\ min}{1\ h})$
$\approx 6.19$ mi/h
