Consider-a-uniform-square-plate-of-side-a-and-mass-m-The-moment-of-inertia-of-this-plate-about-an-axis-perpendicular-to-its-plane-and-passing-through-one-of-its-corners-is-

Question Number 24730 by Tinkutara last updated on 25/Nov/17

$$\mathrm{Consider}\:\mathrm{a}\:\mathrm{uniform}\:\mathrm{square}\:\mathrm{plate}\:\mathrm{of}\:\mathrm{side} \\ $$$${a}\:\mathrm{and}\:\mathrm{mass}\:{m}.\:\mathrm{The}\:\mathrm{moment}\:\mathrm{of}\:\mathrm{inertia} \\ $$$$\mathrm{of}\:\mathrm{this}\:\mathrm{plate}\:\mathrm{about}\:\mathrm{an}\:\mathrm{axis}\:\mathrm{perpendicular} \\ $$$$\mathrm{to}\:\mathrm{its}\:\mathrm{plane}\:\mathrm{and}\:\mathrm{passing}\:\mathrm{through}\:\mathrm{one}\:\mathrm{of} \\ $$$$\mathrm{its}\:\mathrm{corners}\:\mathrm{is} \\ $$

Commented by ajfour last updated on 25/Nov/17

Commented by ajfour last updated on 25/Nov/17

I_x =I_y =((ma^2 )/(12)) so I_z =I_x +I_y =((ma^2 )/6) I_(z′) =I_z +m((a/( (√2))))^2 =((ma^2 )/6)+((ma^2 )/2) =((2ma^2 )/3) .

$${I}_{{x}} ={I}_{{y}} =\frac{{ma}^{\mathrm{2}} }{\mathrm{12}} \\ $$$${so}\:\:\:{I}_{{z}} ={I}_{{x}} +{I}_{{y}} =\frac{{ma}^{\mathrm{2}} }{\mathrm{6}} \\ $$$${I}_{{z}'} ={I}_{{z}} +{m}\left(\frac{{a}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} =\frac{{ma}^{\mathrm{2}} }{\mathrm{6}}+\frac{{ma}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\:\:\:=\frac{\mathrm{2}{ma}^{\mathrm{2}} }{\mathrm{3}}\:. \\ $$

Commented by mrW1 last updated on 25/Nov/17

I_(z′) =∫_0 ^( a) ∫_0 ^( a) ρ(x^2 +y^2 )dxdy =ρ∫_0 ^( a) ((a^3 /3)+ay^2 )dy =ρ((a^3 /3)×a+a×(a^3 /3)) =((2ρa^4 )/3) =((2ma^2 )/3)

$${I}_{{z}'} =\int_{\mathrm{0}} ^{\:{a}} \int_{\mathrm{0}} ^{\:{a}} \rho\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right){dxdy} \\ $$$$=\rho\int_{\mathrm{0}} ^{\:{a}} \left(\frac{{a}^{\mathrm{3}} }{\mathrm{3}}+{ay}^{\mathrm{2}} \right){dy} \\ $$$$=\rho\left(\frac{{a}^{\mathrm{3}} }{\mathrm{3}}×{a}+{a}×\frac{{a}^{\mathrm{3}} }{\mathrm{3}}\right) \\ $$$$=\frac{\mathrm{2}\rho{a}^{\mathrm{4}} }{\mathrm{3}} \\ $$$$=\frac{\mathrm{2}{ma}^{\mathrm{2}} }{\mathrm{3}} \\ $$

Commented by Tinkutara last updated on 25/Nov/17

But I_x =I_y =4×((ma^2 )/(12))=((ma^2 )/3) ?

$${But}\:{I}_{{x}} ={I}_{{y}} =\mathrm{4}×\frac{{ma}^{\mathrm{2}} }{\mathrm{12}}=\frac{{ma}^{\mathrm{2}} }{\mathrm{3}} \\ $$$$? \\ $$

Commented by Tinkutara last updated on 25/Nov/17

$${But}\:{whats}\:{wrong}\:{with}\:{which}\:{I}\:{posted}? \\ $$

Commented by ajfour last updated on 25/Nov/17

$${why}\:{the}\:\mathrm{4}\:? \\ $$

Commented by Tinkutara last updated on 25/Nov/17

Because of symmetry there are 4 rods and each have ((ml^2 )/(12))

$${Because}\:{of}\:{symmetry}\:{there}\:{are}\:\mathrm{4} \\ $$$${rods}\:{and}\:{each}\:{have}\:\frac{{ml}^{\mathrm{2}} }{\mathrm{12}} \\ $$

Commented by ajfour last updated on 25/Nov/17

$${its}\:{a}\:{plate}\:{or}\:{slab},\:{there}\:{are}\:{not} \\ $$$${any}\:{rods}.\:{further}\:{see}\:{image}\:{below}. \\ $$

Commented by ajfour last updated on 25/Nov/17

Commented by Tinkutara last updated on 25/Nov/17

$${Yes}\:{but}\:{how}\:{this}\:\frac{{ml}^{\mathrm{2}} }{\mathrm{12}}? \\ $$

Commented by ajfour last updated on 25/Nov/17

Commented by ajfour last updated on 25/Nov/17

dm=(m/(Lbt))(btdr)=(m/L)dr I=∫r^2 dm = (m/L)∫_(−L/2) ^( L/2) r^2 dr =(m/L)((L^3 /(24))+(L^3 /(24))) = ((mL^2 )/(12)) .

$${dm}=\frac{{m}}{{Lbt}}\left({btdr}\right)=\frac{{m}}{{L}}{dr} \\ $$$${I}=\int{r}^{\mathrm{2}} {dm}\:=\:\frac{{m}}{{L}}\int_{−{L}/\mathrm{2}} ^{\:\:\:{L}/\mathrm{2}} \:{r}^{\mathrm{2}} {dr} \\ $$$$\:\:\:\:=\frac{{m}}{{L}}\left(\frac{{L}^{\mathrm{3}} }{\mathrm{24}}+\frac{{L}^{\mathrm{3}} }{\mathrm{24}}\right)\:=\:\frac{{mL}^{\mathrm{2}} }{\mathrm{12}}\:. \\ $$

Commented by Tinkutara last updated on 26/Nov/17

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{Sir}! \\ $$

Facebook Tweet Pin

Consider-a-uniform-square-plate-of-side-a-and-mass-m-The-moment-of-inertia-of-this-plate-about-an-axis-perpendicular-to-its-plane-and-passing-through-one-of-its-corners-is-

Leave a Reply Cancel reply