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When you have learned to integrate, derive the formula for the moment of inertia of a disk. |
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Answer» `(dm)/M=(2pirdr)/(piR^2)` , form which `dm= (2Mrdr)/(R^2)` The differential of the moment of inertia is equal to the moment of inertia of this ring: `dI=dm.r^2=(2Mr^3dr)/(R^2)` Hence `I = int_0^Rdm.r^2=(2M)/(R^2)int_0^Rr^3dr=(2M)/R^2=(2M)/(R^2)[r^4/4]_0^R=(2MR^4)/(R^2xx4)=(MR^2)/2` |
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