Write the equations of motion of a pendulum consisting of a thin, 2 kg stick of length suspended from a pivot. How long should the rod be in

Question

Write the equations of motion of a pendulum consisting of a thin, 2 kg stick of length suspended from a pivot. How long should the rod be in order for the period to be exactly 1 sec

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Thạch Thảo 4 years 2021-08-20T00:11:20+00:00 1 Answers 23 views 0

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  1. kimcuc
    0
    2021-08-20T00:12:52+00:00
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    Answer:

    3g/(8π²) ≈ 0.372 m

    Explanation:

    Draw a free body diagram.  There is a weight force at the center of the pendulum.

    Sum of the torques about the pivot:

    ∑τ = Iα

    mg (½ L sin θ) = (⅓ mL²) α

    3g sin θ = 2L α

    α = 3g/(2L) sin θ

    For small θ, sin θ ≈ θ.

    α = 3g/(2L) θ

    θ” = 3g/(2L) θ

    The solution of this differential equation is:

    θ = θ₀ cos(√(3g/(2L)) t)

    So the period is:

    T = 2π / √(3g/(2L))

    If the period is 1 second:

    1 = 2π / √(3g/(2L))

    √(3g/(2L)) = 2π

    3g/(2L) = 4π²

    L = 3g/(8π²)

    L ≈ 0.372 m

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