2π rads (about 6.28)=1 revolution. So 1 rad=1/2π revolution (about 0.16 rev).
1 rev/sec=60 rpm=120π rads/min (about 377).
10 seconds after startup the disk reaches an angular speed of 1900 rads/sec=114000 rads/min=114000/2π=18144 rpm. This exceeds 7200 maximum rpm. So the disk has already reached its maximum steady speed before 10 seconds have elapsed.
7200rpm=7200/60=120rps=240π rads/sec. Time to reach this speed is 240π/190=3.968 secs approx.
No further acceleration takes place after this so the disk spins at constant speed for 10-3.968=6.032 secs in which time it has performed 6.032*120 revolutions=723.8 revolutions.
Now we need to work out how many revolutions it takes to reach maximum speed. The disk starts from rest so its initial speed is zero. After time t seconds the angular speed is 190t rads/s. The average speed over this period of time is ½190t=95t and the angular distance is 95t² rads=95t²/2π revolutions.
We know that t=240π/190 so 95(240π/190)²/2π=238.1 revs.
Therefore the total number of revolutions is 238.1+723.8=961.9 revs approx.
ALGEBRAIC SOLUTION
Let ⍺=angular acceleration in rads/s², ⍵=angular velocity in rads/s, t₁=time in secs for the acceleration period and t₂=constant speed time, s=angular distance in rads, R=number of revolutions.
t₁+t₂=10, ⍵=⍺t₁, s=½⍺t₁²+⍵t₂=½⍵²/⍺+⍵(10-⍵/⍺), R=s/2π.
Therefore R=(½⍵²/⍺+⍵(10-⍵/⍺))/2π.
⍵=7200rpm=120rps=240πrads/s; ⍺=190/s².
R = 1496.02+4547.77)/2π = 6043.79/2π = 961.9 revs.