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An optical disk drive in a computer can spin a disk at up to 10000 rpm. If a particular disk is spun at 4210 rpm while it is being read, and then is allowed to come to rest over 0.31 s, what is the magnitude of the average angular acceleration of the disk? If the disk is 12.0 cm in diameter, what is the magnitude of the tangential linear acceleration of a point on the disk's edge?

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Final answer:

The disk's average angular acceleration is -1424.77 rad/s², and the magnitude of the tangential linear acceleration of a point on the disk's edge is 85.49 m/s².

Step-by-step explanation:

To calculate the magnitude of the average angular acceleration of the disk, we use the equation α = Δω / Δt, where α is the angular acceleration, Δω is the change in angular velocity, and Δt is the change in time. First, we need to convert the angular velocity from rpm to rad/s. The conversion factor is 2π rad/s per revolution, which gives us an initial angular velocity of ωi = 4210 rpm * (2π rad/s) / (60 s/min) = 441.68 rad/s. Since the disk comes to rest, the final angular velocity, ωf, is 0 rad/s. The change in angular velocity is Δω = ωf - ωi = -441.68 rad/s, and the change in time is Δt = 0.31 s. Thus, the average angular acceleration is α = -441.68 rad/s / 0.31 s = -1424.77 rad/s².

The magnitude of the tangential linear acceleration at a point on the disk's edge can be found using the equation at = α*r, where at is the tangential acceleration, α is the angular acceleration, and r is the radius of the disk. Given that the disk is 12.0 cm in diameter, the radius r is 6.0 cm or 0.06 m. Thus, the tangential linear acceleration is at = 1424.77 rad/s² * 0.06 m = 85.49 m/s².

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