183_notes:moment_of_inertia_ex

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183_notes:moment_of_inertia_ex [2014/11/03 00:07] – [An Example: Moment of Inertia for a Rod Spun About its Center] caballero183_notes:moment_of_inertia_ex [2014/11/03 00:08] (current) – [An Example: Moment of Inertia for a Rod Spun About its Center] caballero
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 The integral is then calculated to find the momentum of inertia for the rod about its center, The integral is then calculated to find the momentum of inertia for the rod about its center,
  
-$$I = \dfrac{M}{L} \int_{-L/2}^{+L/2} x^2\,dx = \dfrac{M}{L} \dfrac{x^3}{3}\big|_{-L/2}^{+L/2} = \dfrac{M}{3L}\left[\left(\dfrac{L}{2}\right)^3 - \left(\dfrac{-L}{2}\right)^3\right]$$+$$I = \dfrac{M}{L} \int_{-L/2}^{+L/2} x^2\,dx = \dfrac{M}{L} \dfrac{x^3}{3}\Big|_{-L/2}^{+L/2} = \dfrac{M}{3L}\left[\left(\dfrac{L}{2}\right)^3 - \left(\dfrac{-L}{2}\right)^3\right]$$
 $$I = \dfrac{M}{3L}\left[\dfrac{L^3}{8} + \dfrac{L^3}{8}\right] = \dfrac{M}{3L}\dfrac{2L^3}{8} = \dfrac{1}{12}ML^2$$ $$I = \dfrac{M}{3L}\left[\dfrac{L^3}{8} + \dfrac{L^3}{8}\right] = \dfrac{M}{3L}\dfrac{2L^3}{8} = \dfrac{1}{12}ML^2$$
 +
 +which is precisely the moment of inertia of a rod about it's center.
  
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