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A rod of length 1.00m has a mass per unit length given by λ = 2.00 + 1.00x^2, where λ is in kg/m. The rod is placed on the x axis going from x = 0.00 m to x = 1.00 m. What is the moment of inertia of the rod in kg•m^2 about the y axis?

Respuesta :

Answer:

[tex]I=\frac{13}{15} =0.867\ kg.m^2[/tex]

Explanation:

Given:

  • length of rod, [tex]l=1\ m[/tex]
  • mass density of rod, [tex]\lambda=2+x^2\ kg.m^{-1}[/tex]
  • initial point of rod length, [tex]x_0=0\ m[/tex]
  • final point of rod length, [tex]x=1\ m[/tex]

We know, moment of inertia from its basic definition:

[tex]I=\int\limits^x_{x_0} {x^2}\lambda \, dx[/tex]

[tex]I=\int\limits^1_0 {x^2}(2+x^2) \, dx[/tex]

[tex]I=\int\limits^1_0 (2x^2+x^4) \, dx[/tex]

[tex]I=[\frac{2}{3} x^3+\frac{x^5}{5} ]\limi_0^1[/tex]

[tex]I=\frac{13}{15} =0.867\ kg.m^2[/tex]

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