## University Physics with Modern Physics (14th Edition)

(a) $a_{tan} = 0.180~m/s^2$ $a_{rad} = 0$ $a = 0.180~m/s^2$ (b) $a_{tan} = 0.180~m/s^2$ $a_{rad} = 0.377~m/s^2$ $a = 0.418~m/s^2$ (c) $a_{tan} = 0.180~m/s^2$ $a_{rad} = 0.754~m/s^2$ $a = 0.775~m/s^2$
We can find the tangential acceleration $a_{tan}$ of the wheel. $a_{tan} = \alpha ~r = (0.600~rad/s^2)(0.300~m)$ $a_{tan} = 0.180~m/s^2$ (a) At the start: $a_{tan} = 0.180~m/s^2$ $\omega = 0$ $a_{rad} = \omega^2~r = (0)(0.300~m)$ $a_{rad} = 0$ We can find the resultant acceleration. $a = \sqrt{a_{rad}^2+a_{tan}^2}$ $a = \sqrt{(0)^2+(0.180~m/s^2)^2}$ $a = 0.180~m/s^2$ (b) After the wheel has turned through $60.0^{\circ}$: $a_{tan} = 0.180~m/s^2$ $\omega^2 = 2\alpha~\theta$ $a_{rad} = \omega^2~r$ $a_{rad} = (2)(0.600~rad/s^2)(\frac{\pi}{3}~rad)(0.300~m)$ $a_{rad} = 0.377~m/s^2$ We can find the resultant acceleration. $a = \sqrt{a_{rad}^2+a_{tan}^2}$ $a = \sqrt{(0.377~m/s^2)^2+(0.180~m/s^2)^2}$ $a = 0.418~m/s^2$ (c) After the wheel has turned through $120.0^{\circ}$: $a_{tan} = 0.180~m/s^2$ $\omega^2 = 2\alpha~\theta$ $a_{rad} = \omega^2~r$ $a_{rad} = (2)(0.600~rad/s^2)(\frac{2\pi}{3}~rad)(0.300~m)$ $a_{rad} = 0.754~m/s^2$ We can find the resultant acceleration. $a = \sqrt{a_{rad}^2+a_{tan}^2}$ $a = \sqrt{(0.754~m/s^2)^2+(0.180~m/s^2)^2}$ $a = 0.775~m/s^2$