# 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 angularacceleration is constant. The disk's moment of inertia...

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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### 1 Answer

you need to know the following equations inorder to do this calculation.

`omega = omega_0+alpha*t`

`tau= I*alpha`

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 ;

tau= I*alpha

= `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.**

**Sources:**