An optical disk drive in your computer can spin a disk up to 10,000 rpm (about 1045 rad/s ). If a particular disk is spun at 411.5 rad/s whi

Question

An optical disk drive in your computer can spin a disk up to 10,000 rpm (about 1045 rad/s ). If a particular disk is spun at 411.5 rad/s while it is being read, and then is allowed to come to rest over 0.569 seconds , what is the magnitude of the average angular acceleration of the disk

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Philomena 3 years 2021-08-11T19:40:31+00:00 1 Answers 12 views 0

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  1. kimchi2
    0
    2021-08-11T19:42:00+00:00
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    Answer:

    723.2 rad/s^2

    Explanation:

    The angular acceleration of an object in rotation is equal to the rate of change of its angular velocity.

    Mathematically:

    \alpha = \frac{\omega_f - \omega_i}{t}

    where

    \alpha is the angular acceleration

    \omega_i is the initial angular velocity

    \omega_f is the final angular velocity

    t is the time elapsed

    For the optical disk in this problem, we have:

    \omega_i = 411.5 rad/s is the initial angular velocity

    \omega_f=0 is the final angular velocity (because the disk comes to rest)

    t = 0.569 s is the time elapsed

    So, the angular acceleration is

    \alpha=\frac{0-411.5}{0.569}=-723.2rad/s^2

    And since we are only interested in the magnitude, then the magnitude is 723.2 rad/s^2

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