A small rock with mass 0.12 kg is fastened to a massless string with length 0.80 m to form a pendulum. The pendulum is swinging so as to mak

Question

A small rock with mass 0.12 kg is fastened to a massless string with length 0.80 m to form a pendulum. The pendulum is swinging so as to make a maximum angle of 45 ∘ with the vertical. Air resistance is negligible. Part A What is the speed of the rock when the string passes through the vertical position

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Bơ 5 years 2021-08-28T04:47:07+00:00 1 Answers 281 views 0

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  1. Giakhanh
    -3
    2021-08-28T04:48:46+00:00
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    Answer:

    v = 3.33 m/s

    Explanation:

    In the position of 45 degrees, all the energy of the rock is gravitational, then we have:

    E = m*g*L*cos(angle)

    and in the vertical position of the string, all the energy is kinetic, so we have:

    E = m*v^2/2

    If there is no dissipation, both energies are equal, so we have:

    m*g*L*cos(45) = m*v^2/2

    9.81 * 0.8 * 0.7071 * 2 = v^2

    v^2 = 11.0986

    v = 3.33 m/s

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