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

a). Angular acceleration 32
`(rev)/(min^(2) )`

b). Final angular speed in rpm = 38.4 rpm

Step-by-step 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 +(1)/(2) * a * t^(2) \\W_(o)= 0 (rev)/(min)\\ S_(o)= 0 rev\\S_(f)=  (1)/(2) * a * t^(2)\\a= (S_(f)*2)/(t^(2) )\\ a= (23rev*2)/(1.2min^(2) )=(46 rev)/(1.44 min^(2) ) \\a= 31.94 (rev)/(min^(2) ) \\`

a ≅ 32
`(rev)/(min^(2) )`

b).

`w_(f) = w_(o)+ a*t\\w_(f) = 0+ 3.2 (rev)/(min^(2) ) *1.2 min\\w_(f) = 38.4 (rev)/(min) = 38.4`rpm

Comprobation:

Using a different equation but replacing a= 32
`(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=√(2*32*23) \\38.4=√(1472) \\38.4=38.4`