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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] caballero
+++ 183_notes:moment_of_inertia_ex [2014/11/03 00:08] (current) – [An Example: Moment of Inertia for a Rod Spun About its Center] caballero
@@ Line 35: / Line 35: @@
 The integral is then calculated to find the momentum of inertia for the rod about its center,
-$$I = \dfrac{M}{L} \int_{-L/2}^{+L/2} x^2\,dx = \dfrac{M}{L} \dfrac{x^3}{3}\big|_{-L/2}^{+L/2} = \dfrac{M}{3L}\left[\left(\dfrac{L}{2}\right)^3 - \left(\dfrac{-L}{2}\right)^3\right]$$
+$$I = \dfrac{M}{L} \int_{-L/2}^{+L/2} x^2\,dx = \dfrac{M}{L} \dfrac{x^3}{3}\Big|_{-L/2}^{+L/2} = \dfrac{M}{3L}\left[\left(\dfrac{L}{2}\right)^3 - \left(\dfrac{-L}{2}\right)^3\right]$$
  $I = \dfrac{M}{3L}\left[\dfrac{L^3}{8} + \dfrac{L^3}{8}\right] = \dfrac{M}{3L}\dfrac{2L^3}{8} = \dfrac{1}{12}ML^2$
+which is precisely the moment of inertia of a rod about it's center.