Answer
(a) $\alpha = 1.5\times 10^{-4}~rad/s^2$
(b) $a_R = 0.012~m/s^2$
$a_{tan} = 6.2\times 10^{-4}~m/s^2$
Work Step by Step
$\omega = (1.0~rpm)(2\pi \frac{rad}{rev})(\frac{1~m}{60~s}) = \frac{\pi}{30}~rad/s$
(a) $\alpha = \frac{\omega}{t}$
$\alpha = \frac{\frac{\pi}{30}~rad/s}{12\times 60~s} = 1.5\times 10^{-4}~rad/s^2$
(b) At t = 6.0 min, $\omega = \frac{\pi}{60}~rad/s$ because the spacecraft is halfway through the period of acceleration.
We can find the radial acceleration $a_R$ as:
$a_R = \omega^2 r = ( \frac{\pi}{60}~rad/s)^2(4.25~m)$
$a_R = 0.012~m/s^2$
We can find the tangential acceleration $a_{tan}$ as:
$a_{tan} = \alpha ~r = (\frac{\pi}{30\times 12 \times 60} ~rad/s^2)(4.25~m)$
$a_{tan} = 6.2\times 10^{-4}~m/s^2$
