Question

Pilots can be tested for the stresses of flying high-speed jets in a whirling "human centrifuge," which takes 1.2min to turn through 23 complete revolutions before reaching its final speed.
a. What was its angular acceleration (assumed constant)? I got 32 rev/min 2 and it was rightB. What was its final angular speed in rpm? I tried 27 and 32 rpm but they are both wrong. Any ideas?

Answer 1

Sure.
Can I use your answer to part-'a' ?

If the angular acceleration is actually 32 rev/min², than
after 1.2 min, it has reached the speed of

                 (32 rev/min²) x (1.2 min)  =  38.4 rev/min .

Check:

If the initial speed is zero and the final speed is 38.4 rpm,
then the average speed during the acceleration period is

                   (1/2) (0 + 38.4)  =  19.2 rpm  average

At an average speed of  19.2 rpm for 1.2 min,
it covers

                   (19.2 rev/min) x (1.2 min)  =  23.04 revs .

That's pretty close to the "23" in the question, so I think that
everything here is in order.

Answer 2

Answer:

a). Angular acceleration 32 $\frac{rev}{min^{2} }$

b). Final angular speed in rpm = 38.4 rpm

Explanation:

To calculated angular speed the distance is give 23 revolutions and time of 1.2 minutes so using equations:

a).

$S_{f}= S_{o} + W_{o}*t +\frac{1}{2} * a * t^{2} \\W_{o}= 0 \frac{rev}{min}\\ S_{o}= 0 rev\\S_{f}= \frac{1}{2} * a * t^{2}\\a= \frac{S_{f}*2}{t^{2} }\\ a= \frac{23rev*2}{1.2min^{2} }=\frac{46 rev}{1.44 min^{2} } \\a= 31.94 \frac{rev}{min^{2} }$

a ≅ 32 $\frac{rev}{min^{2} }$

b).

$w_{f} = w_{o}+ a*t\\w_{f} = 0+ 3.2 \frac{rev}{min^{2} } *1.2 min\\w_{f} = 38.4 \frac{rev}{min} = 38.4$rpm

Comprobation:

Using a different equation but replacing a= 32 $\frac{rev}{min^{2} }$, $S_{f}= 23 rev$, $V_{o}= 0 rev$, $v_{f}= 38.4 rev$

$w_{f} ^{2} =w_{o} ^{2}+2*a (S_{f} -S_{o})\\38.4 ^{2} =0^{2}+2*32 (23 -0)\\38.4=\sqrt{2*32*23} \\38.4=\sqrt{1472} \\38.4=38.4$