Answer
(a) The magnitude of the acceleration is $17600~m/s^2$
(b) The magnitude of the acceleration is $4430~m/s^2$
Work Step by Step
(a) We can convert 4000 rpm to units of rad/s
$\omega = (4000~rpm)(\frac{1~min}{60~s})(\frac{2\pi~rad}{1~rev})$
$\omega = 419~rad/s$
We can find the centripetal acceleration at the end of a test tube.
$a_c = \omega^2~r$
$a_c = (419~rad/s)^2(0.10~m)$
$a_c = 17600~m/s^2$
The magnitude of the acceleration is $17600~m/s^2$
(b) We can find the speed just before hitting the floor.
$v^2= v_0^2+2ay$
$v = \sqrt{2ay} = \sqrt{(2)(9.8~m/s^2)(1.0~m)}$
$v = 4.43~m/s$
We can find the rate of deceleration after striking the floor. We can let $v_0 = 4.43~m/s$ for this part of the question.
$a = \frac{v-v_0}{t}$
$a = \frac{0-4.43~m/s}{1.0\times 10^{-3}~s}$
$a = -4430~m/s^2$
The magnitude of the acceleration is $4430~m/s^2$
