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 235.6 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 235.6 rad/s while it is being read, and then is allowed to come to rest over 0.167 seconds , what is the magnitude of the average angular acceleration of the disk? average angular acceleration: rad s 2 If the disk is 0.12 m in diameter, what is the magnitude of the tangential acceleration of a point 1 / 3 of the way out from the center of the disk? tangential acceleration

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RuslanHeatt 2 months 2021-07-31T09:25:45+00:00 1 Answers 0 views 0

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    2021-07-31T09:27:23+00:00

    Answer:

    Part(a): The angular acceleration is \bf{1410.78~rad/s^{2}}.

    Part(b): The linear acceleration is \bf{28.21~m/s^{2}}.

    Explanation:

    Given:

    The angular velocity, \omega  = 235.6~rad/s

    Time taken, t = 0.167~s

    Diameter of the disk, d = 0.12~m

    Radius of the concerned point, r_{p} = \dfrac{1}{3}(d/2).

    Part(a):

    The angular acceleration (\alpha) is given by

    \alpha &=& \dfrac{\omega}{t}\\&=& \dfrac{235.6~rad/s}{0.167~s}\\&=& 1410.78~rad/s^{2}

    Part(b):

    The radius (r_{p}) of the concerned point is given by

    r_{p} &=& \dfrac{1}{3}\dfrac{0.12~m}{2}\\&=& 0.02~m

    Linear acceleration (a) is give by

    a &=& \alpha \times r_{p}\\&=& (1410.78~rad/sec)(0.02~m)\\&=& 28.21~m/s^{2}

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