A uniform thin rod of mass M = 3.11 kg M=3.11 kg pivots about an axis through its center and perpendicular to its length. Two small bodies,

Question

A uniform thin rod of mass M = 3.11 kg M=3.11 kg pivots about an axis through its center and perpendicular to its length. Two small bodies, each of mass m = 0.213 kg m=0.213 kg , are attached to the ends of the rod. What must the length L L of the rod be so that the moment of inertia of the three-body system with respect to the described axis is I = 0.831 kg ⋅ m 2 I=0.831 kg·m2 ?

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Thu Giang 2 days 2021-07-22T06:16:57+00:00 1 Answers 0 views 0

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    2021-07-22T06:18:39+00:00

    Answer:

    L=1.49 m

    Explanation:

    We are given that

    M=3.11 kg

    m_1=m_2=0.213 kg

    I=0.831 kgm^2

    We have to find the length L of the rod

    Moment of inertia of the system

    I=\frac{ML^2}{12}+\frac{mL^2}{4}+\frac{mL^2}{4}

    0.831=L^2(\frac{M}{12}+\frac{2m}{4})=L^2(\frac{3.11}{12}+\frac{0.213}{2})

    (\frac{3.11+1.278}{12}L^2=0.831

    L^2=\frac{0.813\times 12}{3.11+1.278}

    L=\sqrt{\frac{0.813\times 12}{3.11+1.278}}

    L=1.49 m

    Hence, the length of rod=L=1.49 m

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