Answer
The angular velocity of the tooth at t = 20 seconds is 2210 rpm.
Work Step by Step
We can find the angular velocity of the tooth at t = 20 seconds as;
$\omega = \omega_0+\int_{0}^{20}\alpha(t)~dt$
$\omega = \omega_0+\int_{0}^{20}(20-t)~dt$
$\omega = \omega_0+(20t-\frac{1}{2}t^2)\vert_{0}^{20}$
$\omega = \omega_0+[(20)(20)-\frac{1}{2}(20)^2]$
$\omega = 300~rpm+(200~rad/s)$
$\omega = 300~rpm+(200~rad/s)(\frac{60~s}{1~min})(\frac{1~rev}{2\pi~rad})$
$\omega = 300~rpm+1910~rpm$
$\omega = 2210~rpm$
The angular velocity of the tooth at t = 20 seconds is 2210 rpm.
