Answer
a) $a_{cp}=0.95g$
b) $v_t=8.9\frac{m}{s}$
Work Step by Step
(a) We know that
$a_{cp}=r\omega^2$
We plug in the known values to obtain:
$a_{cp}=(28ft\times 0.305\frac{m}{ft})(10\frac{rev}{min}\times 2\pi \times \frac{1min}{60s})^2$
$a_{cp}=0.95g$
(b) We can find the linear speed as
$v_t=r\omega$
We plug in the known values to obtain:
$v_t=(28ft\times 0.305\frac{m}{ft})(10\frac{rev}{min}\times 2\pi \times \frac{1min}{60s})=8.9\frac{m}{s}$
