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--- 183_notes:moment_of_inertia_ex [2014/11/03 00:00] – [An Example: Moment of Inertia for a Rod] 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 19: / Line 19: @@
 The integral that defines this calculation is,
- $\int_{\mathrm{whole\:rod}} r^2\,dm$
+$$I = \int_{\mathrm{whole\:rod}} r^2\,dm$$
 If you assume the rod is much thinner than it is long, you only have to worry about adding up the bits of mass along its length. So you end up with the one-dimensional integral,
- $\int_{-L/2}^{+L/2} x^2\,dm$
+$$I = \int_{-L/2}^{+L/2} x^2\,dm$$
+What's this little  $dm$ ? It's the little bit of mass that exists are some arbitrary location  $x$ . It is of length  $dx$  and therefore it's mass is,
+ $dm = \dfrac{M}{L}dx$
+Using this information, you can rewrite the integral,
+ $I = \int_{-L/2}^{+L/2} x^2\,\dfrac{M}{L}dx = \dfrac{M}{L} \int_{-L/2}^{+L/2} x^2\,dx$
+The integral is then calculated to find the momentum of inertia for the rod about its center,
+ $I = \dfrac{M}{L} \int_{-L/2}^{+L/2} x^2\,dx = \dfrac{M}{L} \dfrac{x^3}{3}\Big|_{-L/2}^{+L/2} = \dfrac{M}{3L}\left[\left(\dfrac{L}{2}\right)^3 - \left(\dfrac{-L}{2}\right)^3\right]$
+ $I = \dfrac{M}{3L}\left[\dfrac{L^3}{8} + \dfrac{L^3}{8}\right] = \dfrac{M}{3L}\dfrac{2L^3}{8} = \dfrac{1}{12}ML^2$
+which is precisely the moment of inertia of a rod about it's center.