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For-an-axis-passing-through-the-centre-of-mass-of-a-rectangular-plate-along-its-length-Show-that-its-moment-of-inertia-is-ML-2-12-and-the-radius-of-gyration-is-L-2-3-




Question Number 26179 by NECx last updated on 21/Dec/17
For an axis passing through the  centre of mass of a rectangular  plate(along its length).Show that  its moment of inertia is ((ML^2 )/(12)) and  the radius of gyration is (L/(2(√3).))
$${For}\:{an}\:{axis}\:{passing}\:{through}\:{the} \\ $$$${centre}\:{of}\:{mass}\:{of}\:{a}\:{rectangular} \\ $$$${plate}\left({along}\:{its}\:{length}\right).{Show}\:{that} \\ $$$${its}\:{moment}\:{of}\:{inertia}\:{is}\:\frac{{ML}^{\mathrm{2}} }{\mathrm{12}}\:{and} \\ $$$${the}\:{radius}\:{of}\:{gyration}\:{is}\:\frac{{L}}{\mathrm{2}\sqrt{\mathrm{3}}.} \\ $$
Answered by mrW1 last updated on 22/Dec/17
Let′s say the plate has the size of  B×L, the mass of unit area is  ρ=(M/(B×L))  I=2∫_0 ^(L/2) ρBx^2 dx=2ρB((((L/2))^3 )/3)=((ρBL×L^2 )/(12))=((ML^2 )/(12))  r_g =(√(I/M))=(L/( (√(12))))=(L/(2(√3)))
$${Let}'{s}\:{say}\:{the}\:{plate}\:{has}\:{the}\:{size}\:{of} \\ $$$${B}×{L},\:{the}\:{mass}\:{of}\:{unit}\:{area}\:{is} \\ $$$$\rho=\frac{{M}}{{B}×{L}} \\ $$$${I}=\mathrm{2}\int_{\mathrm{0}} ^{\frac{{L}}{\mathrm{2}}} \rho{Bx}^{\mathrm{2}} {dx}=\mathrm{2}\rho{B}\frac{\left(\frac{{L}}{\mathrm{2}}\right)^{\mathrm{3}} }{\mathrm{3}}=\frac{\rho{BL}×{L}^{\mathrm{2}} }{\mathrm{12}}=\frac{{ML}^{\mathrm{2}} }{\mathrm{12}} \\ $$$${r}_{{g}} =\sqrt{\frac{{I}}{{M}}}=\frac{{L}}{\:\sqrt{\mathrm{12}}}=\frac{{L}}{\mathrm{2}\sqrt{\mathrm{3}}} \\ $$

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