Answer
The angular speed of the flywheel is
$$
\omega=(75 \mathrm{rev} / \mathrm{min})(2 \pi \mathrm{rad} / \mathrm{rev})(1 \min / 60 \mathrm{s})=7.85 \mathrm{rad} / \mathrm{s}
$$
With $r=0.50 \mathrm{m},$ the radial (or centripetal) acceleration is given by Eq. $10-23 :$
$$
a_{r}=\omega^{2} r=(7.85 \mathrm{rad} / \mathrm{s})^{2}(0.50 \mathrm{m}) \approx 31 \mathrm{m} / \mathrm{s}^{2}
$$
which is much bigger than $a_{t}$ . Consequently, the magnitude of the acceleration is
$$
|\vec{a}|=\sqrt{a_{r}^{2}+a_{t}^{2}} \approx a_{r}=31 \mathrm{m} / \mathrm{s}^{2} .
$$
Work Step by Step
The angular speed of the flywheel is
$$
\omega=(75 \mathrm{rev} / \mathrm{min})(2 \pi \mathrm{rad} / \mathrm{rev})(1 \min / 60 \mathrm{s})=7.85 \mathrm{rad} / \mathrm{s}
$$
With $r=0.50 \mathrm{m},$ the radial (or centripetal) acceleration is given by Eq. $10-23 :$
$$
a_{r}=\omega^{2} r=(7.85 \mathrm{rad} / \mathrm{s})^{2}(0.50 \mathrm{m}) \approx 31 \mathrm{m} / \mathrm{s}^{2}
$$
which is much bigger than $a_{t}$ . Consequently, the magnitude of the acceleration is
$$
|\vec{a}|=\sqrt{a_{r}^{2}+a_{t}^{2}} \approx a_{r}=31 \mathrm{m} / \mathrm{s}^{2} .
$$
