Many 2.5-in-diameter (6.35 cm) disks spin at a constant 7200 rpm operating speed

Question

Many 2.5-in-diameter (6.35 cm) disks spin at a constant 7200 rpm operating speed. The disks have a mass of about 7.5 g and are essentially uniform throughout with a very small hole at the center. If they reach their operating speed 3 s after being turned on, what average torque does the disk drive supply to the disk during the acceleration? A computer disk that has a diameter d 6.35 cm is initially at rest, ar-0. The disk drive applies a torque to it, and the disk ends up spinning at a final angular speed of = 7200 rpm, we can calculate the average torque the drive exerts on the disk through Newton's second law for rotation. The average acceleration can be calculated from the angular speeds and the time interval. We can model the computer disk as a thin disk of radius (d/2) spinning about an axis through its center

Explanation / Answer

21. I = m r^2 / 2 = 3.78 x 10^-6 kg m^2

w = 7200 x 2pi rad / 60 s = 754 rad/s

Applying wf = wi + alpha t

754 = 0 + 3 alpha

alpha = 251.3 rad/s^2

torque = I alpha = 9.5 x 10^-4 Nm

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