A potter’s wheel is rotating around a vertical axis through its center at a frequency of 2.0 rev/s. The wheel can be considered a uniform di

Question

A potter’s wheel is rotating around a vertical axis through its center at a frequency of 2.0 rev/s. The wheel can be considered a uniform disk of mass 4.9 kg and diameter 0.40m. The potter then throws a 2.7 kg chucnk of clay, approximately shaped as a flat disk of radius 8.0 cm, onto the center of the rotating wheel.What is the frequency of the wheel after the clay sticks to it?

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Đan Thu 5 years 2021-07-28T18:27:34+00:00 1 Answers 21 views 0

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  1. Jezebel
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    2021-07-28T18:28:37+00:00
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    Answer:

    1.84 rev/s

    Explanation:

    Given

    Initial frequency, f1 = 2 rev/s, w1 = 2πf1 = 12.568 rad/s

    Mass of the wheel, m(w) = 4.9 kg

    Diameter of wheel, d(w) = 0.4 m

    Mass of clay, m(c) = 2.7 kg

    Radius of clay, r(c) = 8 cm = 0.08 m

    Moment of inertia of wheel,

    I(w) = 1/2m(w)R(w)²

    I(w) = 1/2 * 4.9 * (0.4/2)²

    I(w) = 1/2 * 4.9 * 0.04

    I(w) = 0.098 kgm²

    Moment of inertia of clay

    I(c) = 1/2m(c)R(c)²

    I(c) = 1/2 * 2.7 * 0.08²

    I(c) = 1/2 * 2.7 * 0.0064

    I(c) = .00864 kgm²

    Since chunk of clay attached to the wheel, then perfectly inelastic collision occurred, and as such, momentum is conserved.

    I(i)w(i) = I(f)w(f)

    I(i)w(i) = (I(c) + I(w)) * w(f), on substituting, we have

    0.098 * 12.568 = (0.00864 + 0.098) * w(f)

    1.232 = 0.10664 * w(f)

    w(f) = 1.232 / 0.10664

    w(f) = 11.55 rad/s

    remember, w = 2πf, thus

    f = w/2π

    f = 11.55/6.284

    f = 1.84 rev/s

    thus, final frequency of the wheel is 1.84 rev/s

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