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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 ?

Respuesta :

lublana

Answer:

L=1.49 m

Explanation:

We are given that

[tex]M=3.11 kg[/tex]

[tex]m_1=m_2=0.213 kg[/tex]

[tex]I=0.831 kgm^2[/tex]

We have to find the length L of the rod

Moment of inertia of the system

[tex]I=\frac{ML^2}{12}+\frac{mL^2}{4}+\frac{mL^2}{4}[/tex]

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

[tex](\frac{3.11+1.278}{12}L^2=0.831[/tex]

[tex]L^2=\frac{0.813\times 12}{3.11+1.278}[/tex]

[tex]L=\sqrt{\frac{0.813\times 12}{3.11+1.278}}[/tex]

[tex]L=1.49 m[/tex]

Hence, the length of rod=L=1.49 m

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