Starting from rest, a 12-cm-diameter compact disk takes 3.0 s to reach its operating angular velocity of 2000 rpm. Assume that the angular
acceleration is constant. The disk's moment of inertia is 2.5 x 10^-5 kg m^2. How much torque is applied to the disk? How many revolutions does it make before reaching full speed?
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you need to know the following equations inorder to do this calculation.
`omega = omega_0+alpha*t`
For our question;
omega_0 = 0 because it starts from rest.
omega = 2000/60*2*pi/3 = 69.813rad/s
t = 3s
Then `alpha = (omega-omega_0)/t`
`= 69.813/3` `rad(s^-2)`
torque applied ;
= `2.5 x 10^-5xx69.813/3`
= `5.818xx10^-4 `
The angular velocity of the disk = 2000rpm = 2000/60 rps
when achieving full velocity the time taken to that is 3s.
So at 3s the number of revolutions= 2000/60*3 = 100
So it take 100 revolutions to achieve max angular velocity.
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