The tub of a washer goes into its spin cycle, starting from rest and gaining angular speed steadily for 7.00 s, at which time it is turning

Question

The tub of a washer goes into its spin cycle, starting from rest and gaining angular speed steadily for 7.00 s, at which time it is turning at 5.00 rev/s. At this point, the person doing the laundry opens the lid, and a safety switch turns off the washer. The tub smoothly slows to rest in 14.0 s. Through how many revolutions does the tub turn while it is in motion?

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RI SƠ 4 years 2021-07-29T05:48:20+00:00 1 Answers 36 views 0

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  1. Maris
    0
    2021-07-29T05:49:22+00:00
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    Answer:

    The no. of revolutions does the tub turn while it is in motion is = 52.51 revolutions

    Explanation:

    Given data

    \omega_1 = 0

    \omega_2 = 5 \frac{rev}{sec}

    Time taken = 7 sec

    (1). The angular acceleration is given by

    \alpha = \frac{d \omega}{dt}

    \alpha = \frac{5}{7}

    \alpha = 0.714 \frac{rev}{s^{2} }

    We know that from the equation of motion

    \omega_2^{2} = \omega_1^{2} + 2 \alpha \theta_1

    5^{2} = 0 + 2 (0.714) \theta_1

    \theta_1 = 17.5 \ rev  ——– (1)

    (2). The angular acceleration is given by

    \alpha = \frac{d \omega}{dt}

    \alpha = - \frac{5}{14}

    \alpha = – 0.357 \frac{rev}{s^{2} }

    We know that from the equation of motion

    \omega_2^{2} = \omega_1^{2} + 2 \alpha \theta_2

    0^{2} = 5^{2} + 2 (-0.357) \theta_2

    \theta_2 = 35.01 rev  ——-  (2)

    Total no of revolution made by the machine is

    \theta = \theta_1 + \theta_2

    \theta = 17.5 + 35.01

    \theta = 52.51 rev

    Therefore the no. of revolutions does the tub turn while it is in motion is = 52.51  rev

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