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Moment of inertia of sphere.. ✍✍✍✍✍✍✍✍✍ |
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Answer» Hollow sphere is 2/3MR^2 And Solid sphere is 2/5MR^2 The moment of inertia of a sphere expression is obtained in two ways.\tFirst, we take the solid sphere and slice it up into infinitesimally thin\xa0solid cylinders.\tThen we have to sum the moments of exceedingly small thin disks in a given axis from left to right.We will look and understand the derivation below.First, we take the moment of inertia of a disc that is thin. It is given as;I = ½ MR2In this case, we write it as;dI (infinitesimally moment of inertia element) = ½ r2dmFind the dm and dv using;dm = p dVp = moment of a thin disk of mass dmdv = expressing mass dm in terms of density and volumedV = π r2\xa0dxNow we replace dV into dm. We get;dm = p π r2\xa0dxAnd finally, we replace dm with dI.dI = ½ p π r4\xa0dxThe next step involves adding x to the equation. If we look at the diagram we see that r, R and x forms a triangle. Now we will use Pythagoras theorem which gives us;r2\xa0= R2\xa0– x2Now if we substitute the values we get;dI = ½ p π (R2\xa0– x2)2\xa0dxThis leads to:I = ½ p π\xa0-R∫R\xa0(R2\xa0– x2)2\xa0dxAfter integration and expanding we get;I = ½ pπ 8/15 R5Additionally, we have to find the density as well. For that we use;p = m / Vp = m / m/v πR3If we substitute all the values;I = 8/15 [m / m/v πR3\xa0] R5I = ⅖ MR2 |
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