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6 August, 00:03

The 3.00 kg cube in fig. 15-47 has edge lengths d 6.00 cm and is mounted on an axle through its center. a spring (k 1200 n/m con - nects the cube's upper corner to a rigid wall. initially the spring is at its rest length. if the cube is rotated 3 and released, what is the period of the resulting shm?

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  1. 6 August, 00:42
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    The Period of the resulting shm will be T=39.7

    Explanation:

    Given data

    m=3kg

    d=.06m

    k=1200 N/m

    Θ=3 °

    T=?

    we have the formulas,

    I = (1/6) Md2

    F = ma

    F = - kx = - (mω2x)

    k = mω2 τ = - d (FgsinΘ)

    T=2 x 3.14 / √ (m/k)

    Solution for the given problem would be,

    F=-Kx (where x = dsin Θ)

    F=-k dsin Θ

    F = - (1200) (.06) sin (3 °)

    F=-10.16N

    By newton's second law.

    F = ma

    a = F/m

    a = (-10.16N) / 3

    a=3.38

    using the k=mω value

    k=mω

    ω=k/m

    ω=1200/3

    ω=400

    Using F = - kx value

    x = F/-k

    x = (-10.16) / 1200

    x=0.00847m

    Restoring the torque value

    τ = - dmgsinΘ where (τ = Iα so.) ... Iα = - dmgsinΘ α = - (.06) (4) α =

    α = (.06) (4) (9.81) sin (4°)

    α=-1.781

    Rotational to linear form

    a = αr

    r =.1131 m

    a=-1.781 x. 1131 m

    a=-0.2015233664

    Time Period

    T=2 x 3.14 / √ (m/k)

    T=6.28/√ (3/1200)

    T=6.28/0.158

    T=39.7
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