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InterviewSolution
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State and prove parallel axis theorem. |
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Answer» SOLUTION : Parallel axis theorem : The moment of inertia of a rigid body about an axis passing through a point is the sum of moment of inertia about parallel axis passing through centre of mass `(I_G)` and mass of the body multiplied by square of distance`(MR^2)`between the axes i.e., `I = I_G + MR^2` Proof : Consider a rigid body of mass M with 'G' as its centre of mass. `I_G`the moment of inertia about an axis passing through centre of mass. I = The moment of inertia about an axis passing through the point 'O 'in that plane. Let perpendicular distance between the axes is OG = R (say) Consider point P in the given plane. Join OP and GP. Extend the line OG and drop a normal from 'P' on to it as shown in figure.  The moment of inertia about the axis passing through centre of mass G. `(I_G) = Sigma m GP^2 "" ....(1)` M.O.I of the body about an axis passing through O (I) `= Sigma mOP^2 "" ...(2)` From triangle OPD `OP^2 = OD^2 + DP^2` `RARR OD = OG +GD` `therefore OD^2 = (OG +GD)^2 = OG^2 + GD^2 + 2OG, GD ""...(3)` From equations (2) and (3) ` I = Sigmam OP^2 = Sigma m [(OG^2 + GD^2 + 2OG . GD)+ DP^2]` `therefore I = Sigma m {GD^2 + DP^2 + OG^2 + 2OG . GN}` But ` GD^2 + DP^2 = GP^2` `therefore I = Sigma m { GP^2 + OG^2 + 2OG . GP}` `therefore I = Sigma m GP^2 + Sigma m OG^2 + 2OG Sigma mGD "" ....(4)` But the terms `Sigma mGP^2 = I_G` `Sigma mOG^2 = MR^2 (becuase Sigmam = " M and OG = R)` The term `2OG Sigma mGD` = 0 , Because it REPRESENT sum of moment of masses about centre of mass . Hence its VALUE is zero . ` therefore I = I_G +MR^2` Hence parallel axis theorem is proved . |
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