The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 3.0 rev/s in 13.0 s. At this point, th

Question

The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 3.0 rev/s in 13.0 s. At this point, the person doing the laundry opens the lid, and a safety switch turns off the washer. The tub slows to rest in 12.0 s. Through how many revolutions does the tub turn during this 25 s interval

in progress 0
Thạch Thảo 2 days 2021-07-20T10:59:29+00:00 1 Answers 4 views 0

Answers ( )

    0
    2021-07-20T11:01:16+00:00

    Answer:

    The tub turns 37.520 revolutions during the 25-second interval.

    Explanation:

    The total number of revolutions done by the tub of the washer (\Delta n), in revolutions, is the sum of the number of revolutions done in the acceleration (\Delta n_{1}), in revolutions, and deceleration stages (\Delta n_{2}), in revolutions:

    \Delta n = \Delta n_{1} + \Delta n_{2} (1)

    Then, we expand the previous expression by kinematic equations for uniform accelerated motion:

    \Delta n = \frac{1}{2}\cdot ( \ddot n_{1}\cdot t_{1}^{2} - \ddot n_{2} \cdot t_{2}^{2}) (1b)

    Where:

    \ddot n_{1}, \ddot n_{2} – Angular accelerations for acceleration and deceleration stages, in revolutions per square second.

    t_{1}, t_{2} – Acceleration and deceleration times, in seconds.

    And each acceleration is determined by the following formulas:

    Acceleration

    \ddot n_{1} = \frac{\dot n}{t_{1}} (2)

    Deceleration

    \ddot n_{2} = -\frac{\dot n}{t_{2} } (3)

    Where \dot n is the maximum angular velocity of the tub of the washer, in revolutions per second.

    If we know that \dot n = 3\,\frac{rev}{s}, t_{1} = 13\,s and t_{2} = 12\,s, then the quantity of revolutions done by the tub is:

    \ddot n_{1} = \frac{3\,\frac{rev}{s} }{13\,s}

    \ddot n_{1} = 0.231\,\frac{rev}{s^{2}}

    \ddot n_{2} = -\frac{3\,\frac{rev}{s} }{12\,s}

    \ddot n_{2} = -0.25\,\frac{rev}{s^{2}}

    \Delta n = \frac{1}{2}\cdot ( \ddot n_{1}\cdot t_{1}^{2} + \ddot n_{2} \cdot t_{2}^{2})

    \Delta n = \frac{1}{2}\cdot \left[\left(0.231\,\frac{rev}{s^{2}} \right)\cdot (13\,s)^{2}-\left(-0.25\,\frac{rev}{s^{2}} \right)\cdot (12\,s)^{2}\right]

    \Delta n = 37.520\,rev

    The tub turns 37.520 revolutions during the 25-second interval.

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )