Answer
(a) The wheel rotates through an angle of $157~rad$ in $1.0~s$
(b) The linear speed of a point on the rim is $47.1~m/s$
(c) The frequency of rotation is $25.0~s^{-1}$
Work Step by Step
(a) We can find the angular speed:
$\omega = \frac{(2.0~rev)(2\pi~rad/rev)}{0.080~s} = 157~rad/s$
We can find the angle through which the wheel rotates in $1.0~s$:
$\theta = \omega ~t = (157~rad/s)(1.0~s) = 157~rad$
The wheel rotates through an angle of $157~rad$ in $1.0~s$
(b) We can find the linear speed of a point on the rim:
$v = \omega~r = (157~rad/s)(0.30~m) = 47.1~m/s$
The linear speed of a point on the rim is $47.1~m/s$
(c) We can find the frequency of rotation:
$f = \frac{\omega}{2\pi} = \frac{157~rad/s}{2\pi} = 25.0~s^{-1}$
The frequency of rotation is $25.0~s^{-1}$
