Answer
(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) $a_{tan} = 0.180~m/s^2$
$a_{rad} = 0.754~m/s^2$
$a = 0.775~m/s^2$
Work Step by Step
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.
$a = \sqrt{a_{rad}^2+a_{tan}^2}$
$a = \sqrt{(0.754~m/s^2)^2+(0.180~m/s^2)^2}$
$a = 0.775~m/s^2$
