Answer
The angular speed is $~~\omega = 62.6~rad/s$
Work Step by Step
We can use Equation (11-10) to find the acceleration of the solid cylinder down the incline:
$a = \frac{g~sin~\theta}{1+I/MR^2}$
$a = \frac{g~sin~\theta}{1+\frac{1}{2}MR^2/MR^2}$
$a = \frac{g~sin~\theta}{1+\frac{1}{2}}$
$a = \frac{2}{3}~g~sin~\theta$
$a = \frac{2}{3}~(9.8~m/s^2)~sin~30^{\circ}$
$a = 3.27~m/s^2$
We can find the speed after traveling a distance of $L = 6.0~m$:
$v_f^2 = v_0^2+2aL$
$v_f^2 = 0+2aL$
$v_f = \sqrt{2aL}$
$v_f = \sqrt{(2)(3.27~m/s^2)(6.0~m)}$
$v_f = 6.26~m/s$
We can find the angular speed:
$\omega = \frac{v}{R}$
$\omega = \frac{6.26~m/s}{0.10~m}$
$\omega = 62.6~rad/s$
The angular speed is $~~\omega = 62.6~rad/s$
