Question Number 47266 by MrW3 last updated on 07/Nov/18
Commented by MrW3 last updated on 07/Nov/18
$${Find}\:{the}\:{moment}\:{of}\:{initia}\:{of}\:{the} \\ $$$${solid}\:{body}\:{about}\:{its}\:{diagonal}\:{axis}. \\ $$
Answered by ajfour last updated on 08/Nov/18
![L_G ^ = ∫r^ ×(ω^ ×r^ )dm G being the centroid (mass center) and origin of x, y, z axes is G. L_G = ∫[(r^ .r^ )ω^ −(ω^ .r^ )r^ ]dm l^ = ai^ +bj^ +ck^ ; r^ =xi^ +yj^ +zk^ ω^ = ω((l^ /l)) L^ =(ω/l)∫[(ai^ +bj^ +ck^ )(x^2 +y^2 +z^2 ) −(xi^ +yj^ +zk^ )(ax+by+cz)]dm =(ω/l)∫ Σ{[a(y^2 +z^2 )−x(by+cz)]i^ }dm = ((𝛒 ω)/l)∫_(−c/2) ^( c/2) ∫_(−b/2) ^( b/2) ∫_(−a/2) ^( a/2) ( Σ{[a(y^2 +z^2 )−x(by+cz)]i^ )dxdydz =((ρ ω)/l)∫∫_(−b/2) ^( b/2) (Σ{a^2 (y^2 +z^2 )}i^ )dydz =((ρω)/l)∫_(−c/2) ^( c/2) (Σ{a^2 ((b^3 /(12))+bz^2 )}i^ )dz =((ρω)/l)Σ{a^2 (((b^3 c)/(12))+((bc^3 )/(12)))}i^ = ((ρ ω)/l)Σ{a[((abc)/(12))(b^2 +c^2 )]}i^ =((ρω)/l)×((abc)/(12)){a(b^2 +c^2 )i^ +b(c^2 +a^2 )j^ +c(a^2 +b^2 )k^ } L_G ^ = ((ρabcω)/(12(√(a^2 +b^2 +c^2 ))))Σ{a(b^2 +c^2 )i^ } and since ρabc = M L_G ^ = ((Mω)/(12(√(a^2 +b^2 +c^2 ))))Σ{a(b^2 +c^2 )i^ } Generally L_G ^ = ∫r^ ×(ω^ ×r^ )dm = ∫(xi^ +yj^ +zk^ )× determinant ((i^ ,j^ ,k^ ),(ω_x ,ω_y ,ω_z ),(x,y,z))dm =∫ determinant ((i^ ,j^ ,k^ ),(x,y,z),((ω_y z−ω_z y),(ω_z x−ω_x z),(ω_x y−ω_y x)))dm =Σ[∫{y(ω_x y−ω_y x)−z(ω_z x−ω_x z)}dm]i^ =Σ[∫{ω_x (y^2 +z^2 )−ω_y (xy)−ω_z (zx)}dm]i^ and since I_x =∫(y^2 +z^2 )dm I_(xy) = ∫xydm L_x = I_x ω_x −I_(xy) ω_y −I_(zx) ω_z L_y = −I_(xy) ω_x +I_y ω_y −I_(yz) ω_z L_z = −I_(zx) ω_x −I_(yz) ω_y +I_z ω_z Inertia tensor is ((I_x ,(−I_(xy) ),(−I_(zx) )),((−I_(xy) ),I_y ,(−I_(yz) )),((−I_(zx) ),(−I_(yz) ),I_z ) ) For this very question Inertia tensor is ((((M/(12))(b^2 +c^2 )),0,0),(0,((M/(12))(c^2 +a^2 )),0),(0,0,((M/(12))(a^2 +b^2 ))) ) If chosen coordinate system is Gxyz , the pricipal axes of inertia_(−−−−−−−−−−−−−−−−) , then the centroidal mass products of inertia of a body with respect to these prinxipal axes of inertia, are zero. I_(xy) = I_(yz) = I_(zx) = 0 . And I_x , I_y , I_z are called the principal centroidal moments of ineetia of the body. About any other point fixed point O L_O ^ = r^ ×mv_(cm) ^ +L_G ^ r^ being the position vector of G relative to O. _________________________.](https://www.tinkutara.com/question/Q47271.png)
$$\bar {{L}}_{{G}} \:=\:\int\bar {{r}}×\left(\bar {\omega}×\bar {{r}}\right){dm} \\ $$$$\:{G}\:{being}\:{the}\:{centroid}\:\left({mass}\:{center}\right) \\ $$$${and}\:{origin}\:{of}\:{x},\:{y},\:{z}\:\:{axes}\:{is}\:{G}. \\ $$$$\:\:\:\:{L}_{{G}} =\:\int\left[\left(\bar {{r}}.\bar {{r}}\right)\bar {\omega}\:−\left(\bar {\omega}.\bar {{r}}\right)\bar {{r}}\right]{dm} \\ $$$$\:\:\:\bar {{l}}\:=\:{a}\hat {{i}}+{b}\hat {{j}}+{c}\hat {{k}}\:;\:\bar {{r}}={x}\hat {{i}}+{y}\hat {{j}}+{z}\hat {{k}} \\ $$$$\:\:\bar {\omega}\:=\:\omega\left(\frac{\bar {{l}}}{{l}}\right) \\ $$$$\:\:\:\bar {{L}}=\frac{\omega}{{l}}\int\left[\left({a}\hat {{i}}+{b}\hat {{j}}+{c}\hat {{k}}\right)\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)\right. \\ $$$$\left.\:\:\:\:\:−\left({x}\hat {{i}}+{y}\hat {{j}}+{z}\hat {{k}}\right)\left({ax}+{by}+{cz}\right)\right]{dm} \\ $$$$\:\:=\frac{\omega}{{l}}\int\:\Sigma\left\{\left[{a}\left({y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)−{x}\left({by}+{cz}\right)\right]\hat {{i}}\right\}{dm} \\ $$$$\:=\:\frac{\boldsymbol{\rho}\:\omega}{{l}}\int_{−{c}/\mathrm{2}} ^{\:\:{c}/\mathrm{2}} \int_{−{b}/\mathrm{2}} ^{\:\:{b}/\mathrm{2}} \int_{−{a}/\mathrm{2}} ^{\:\:{a}/\mathrm{2}} \left(\:\Sigma\left\{\left[{a}\left({y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)−{x}\left({by}+{cz}\right)\right]\hat {{i}}\right){dxdydz}\right. \\ $$$$\:=\frac{\rho\:\omega}{{l}}\int\int_{−{b}/\mathrm{2}} ^{\:\:{b}/\mathrm{2}} \left(\Sigma\left\{{a}^{\mathrm{2}} \left({y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)\right\}\hat {{i}}\right){dydz} \\ $$$$\:=\frac{\rho\omega}{{l}}\int_{−{c}/\mathrm{2}} ^{\:\:{c}/\mathrm{2}} \left(\Sigma\left\{{a}^{\mathrm{2}} \left(\frac{{b}^{\mathrm{3}} }{\mathrm{12}}+{bz}^{\mathrm{2}} \right)\right\}\hat {{i}}\right){dz} \\ $$$$\:=\frac{\rho\omega}{{l}}\Sigma\left\{{a}^{\mathrm{2}} \left(\frac{{b}^{\mathrm{3}} {c}}{\mathrm{12}}+\frac{{bc}^{\mathrm{3}} }{\mathrm{12}}\right)\right\}\hat {{i}} \\ $$$$=\:\frac{\rho\:\omega}{{l}}\Sigma\left\{{a}\left[\frac{{abc}}{\mathrm{12}}\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)\right]\right\}\hat {{i}} \\ $$$$=\frac{\rho\omega}{{l}}×\frac{{abc}}{\mathrm{12}}\left\{{a}\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)\hat {{i}}+{b}\left({c}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)\hat {{j}}\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+{c}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\hat {{k}}\:\right\} \\ $$$$\bar {{L}}_{{G}} \:=\:\frac{\rho{abc}\omega}{\mathrm{12}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }}\Sigma\left\{{a}\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)\hat {{i}}\right\} \\ $$$${and}\:{since}\:\rho{abc}\:=\:{M} \\ $$$$\bar {{L}}_{{G}} \:=\:\:\frac{{M}\omega}{\mathrm{12}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }}\Sigma\left\{{a}\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)\hat {{i}}\right\} \\ $$$${Generally} \\ $$$$\bar {{L}}_{{G}} =\:\int\bar {{r}}×\left(\bar {\omega}×\bar {{r}}\right){dm} \\ $$$$ \\ $$$$\:\:=\:\int\left({x}\hat {{i}}+{y}\hat {{j}}+{z}\hat {{k}}\right)×\begin{vmatrix}{\hat {{i}}}&{\hat {{j}}}&{\hat {{k}}}\\{\omega_{{x}} }&{\omega_{{y}} }&{\omega_{{z}} }\\{{x}}&{{y}}&{{z}}\end{vmatrix}{dm} \\ $$$$\:\:=\int\begin{vmatrix}{\hat {{i}}}&{\hat {{j}}}&{\hat {{k}}}\\{{x}}&{{y}}&{{z}}\\{\omega_{{y}} {z}−\omega_{{z}} {y}}&{\omega_{{z}} {x}−\omega_{{x}} {z}}&{\omega_{{x}} {y}−\omega_{{y}} {x}}\end{vmatrix}{dm} \\ $$$$\:\:=\Sigma\left[\int\left\{{y}\left(\omega_{{x}} {y}−\omega_{{y}} {x}\right)−{z}\left(\omega_{{z}} {x}−\omega_{{x}} {z}\right)\right\}{dm}\right]\hat {{i}} \\ $$$$\:\:=\Sigma\left[\int\left\{\omega_{{x}} \left({y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)−\omega_{{y}} \left({xy}\right)−\omega_{{z}} \left({zx}\right)\right\}{dm}\right]\hat {{i}} \\ $$$$\:{and}\:{since} \\ $$$$\:\:\:\:\:{I}_{{x}} =\int\left({y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right){dm} \\ $$$$\:\:\:\:\:\:{I}_{{xy}} \:=\:\int{xydm} \\ $$$${L}_{{x}} \:=\:{I}_{{x}} \omega_{{x}} −{I}_{{xy}} \omega_{{y}} −{I}_{{zx}} \omega_{{z}} \\ $$$${L}_{{y}} \:=\:−{I}_{{xy}} \omega_{{x}} +{I}_{{y}} \omega_{{y}} −{I}_{{yz}} \omega_{{z}} \\ $$$${L}_{{z}} \:=\:−{I}_{{zx}} \omega_{{x}} −{I}_{{yz}} \omega_{{y}} +{I}_{{z}} \omega_{{z}} \\ $$$$\:\:\:\:{Inertia}\:{tensor}\:{is} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\begin{pmatrix}{{I}_{{x}} }&{−{I}_{{xy}} }&{−{I}_{{zx}} }\\{−{I}_{{xy}} }&{{I}_{{y}} }&{−{I}_{{yz}} }\\{−{I}_{{zx}} }&{−{I}_{{yz}} }&{{I}_{{z}} }\end{pmatrix} \\ $$$${For}\:{this}\:{very}\:{question} \\ $$$$\:\:\:\:{Inertia}\:{tensor}\:{is} \\ $$$$\:\begin{pmatrix}{\frac{{M}}{\mathrm{12}}\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)}&{\mathrm{0}}&{\mathrm{0}}\\{\mathrm{0}}&{\frac{{M}}{\mathrm{12}}\left({c}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{0}}&{\frac{{M}}{\mathrm{12}}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)}\end{pmatrix}\: \\ $$$${If}\:{chosen}\:{coordinate}\:{system}\:{is} \\ $$$${Gxyz}\:,\:{the}\:\underset{−−−−−−−−−−−−−−−−} {{pricipal}\:{axes}\:{of}\:{inertia}}, \\ $$$${then}\:{the}\:{centroidal}\:{mass}\:{products} \\ $$$${of}\:{inertia}\:{of}\:{a}\:{body}\:{with}\:{respect} \\ $$$${to}\:{these}\:{prinxipal}\:{axes}\:{of}\:{inertia}, \\ $$$${are}\:{zero}.\:{I}_{{xy}} =\:{I}_{{yz}} \:=\:{I}_{{zx}} =\:\mathrm{0}\:. \\ $$$${And}\:\:{I}_{{x}} ,\:{I}_{{y}} ,\:{I}_{{z}} \:{are}\:{called}\:{the} \\ $$$${principal}\:{centroidal}\:{moments} \\ $$$${of}\:{ineetia}\:{of}\:{the}\:{body}. \\ $$$${About}\:{any}\:{other}\:{point}\:{fixed}\:{point}\:{O} \\ $$$$\:\:\bar {{L}}_{{O}} \:=\:\bar {{r}}×{m}\bar {{v}}_{{cm}} +\bar {{L}}_{{G}} \\ $$$$\:\:\bar {{r}}\:{being}\:{the}\:{position}\:{vector}\:{of}\:{G} \\ $$$${relative}\:{to}\:{O}. \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_. \\ $$
Commented by MrW3 last updated on 07/Nov/18
$${thank}\:{you}\:{sir}! \\ $$$${Does}\:{it}\:{mean}\:{the}\:{MoI}\:{about}\:{the}\: \\ $$$${diagonal}\:{axis}\:{is}\:{I}={I}_{{x}} +{I}_{{y}} +{I}_{{z}} \\ $$$$=\frac{\mathrm{2}{M}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)}{\mathrm{3}}? \\ $$
Commented by ajfour last updated on 08/Nov/18
![No, MoI will be a tensor for asymmetric orientations (9 components) We need MoI to know angular momentum: For x- component L_x , we just need to attach ω_x , ω_y , and ω_z respectively to each element in first row of the inertia tensor found. similarly for L_y and L_z we shall rather use respectively the 2^(nd) and 3^(rd) row. L_x =Mω_x ( ((b^2 +c^2 )/(12))) L_x = ((Mω)/( (√(a^2 +b^2 +c^2 )))){a(((b^2 +c^2 )/(12)))} L_x =((Mωa(b^2 +c^2 ))/(12(√(a^2 +b^2 +c^2 )))) = ((Mω)/(12))(b^2 +c^2 )cos α α being angle of diagonal with x- axes. So L_(diagonal) =ΣL_x cos α = ((Mω)/(12))[((a^2 (b^2 +c^2 )+b^2 (c^2 +a^2 )+c^2 (a^2 +b^2 ))/(a^2 +b^2 +c^2 ))] If b=c=a we get I_(diagonal) = ((Ma^2 )/6) .](https://www.tinkutara.com/question/Q47281.png)
$${No},\:{MoI}\:{will}\:{be}\:{a}\:{tensor}\:{for} \\ $$$${asymmetric}\:{orientations} \\ $$$$\left(\mathrm{9}\:{components}\right) \\ $$$${We}\:{need}\:{MoI}\:{to}\:{know}\:{angular} \\ $$$${momentum}: \\ $$$${For}\:{x}-\:{component}\:{L}_{{x}} \:,\:{we}\:{just} \\ $$$${need}\:{to}\:{attach}\:\omega_{{x}} ,\:\omega_{{y}} ,\:{and}\:\omega_{{z}} \: \\ $$$${respectively}\:{to}\:{each}\:{element}\:{in} \\ $$$${first}\:{row}\:{of}\:{the}\:{inertia}\:{tensor} \\ $$$${found}. \\ $$$$\:{similarly}\:{for}\:{L}_{{y}} \:{and}\:{L}_{{z}} \:{we}\:{shall} \\ $$$${rather}\:{use}\:{respectively}\:{the}\:\mathrm{2}^{{nd}} \\ $$$${and}\:\mathrm{3}^{{rd}} \:{row}. \\ $$$$\:{L}_{{x}} \:={M}\omega_{{x}} \left(\:\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{12}}\right) \\ $$$$\:{L}_{{x}} =\:\frac{{M}\omega}{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }}\left\{{a}\left(\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{12}}\right)\right\} \\ $$$$\:{L}_{{x}} =\frac{{M}\omega{a}\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)}{\mathrm{12}\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }}\:\:\:=\:\frac{{M}\omega}{\mathrm{12}}\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)\mathrm{cos}\:\alpha \\ $$$$\:\alpha\:{being}\:{angle}\:{of}\:{diagonal}\:{with} \\ $$$$\:\:{x}-\:{axes}. \\ $$$${So}\:{L}_{{diagonal}} =\Sigma{L}_{{x}} \mathrm{cos}\:\alpha \\ $$$$\:=\:\frac{{M}\omega}{\mathrm{12}}\left[\frac{{a}^{\mathrm{2}} \left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)+{b}^{\mathrm{2}} \left({c}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)+{c}^{\mathrm{2}} \left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }\right]\: \\ $$$$\:\:{If}\:{b}={c}={a}\:\:\:{we}\:{get} \\ $$$$\:\:{I}_{{diagonal}} =\:\frac{{Ma}^{\mathrm{2}} }{\mathrm{6}}\:\:\:. \\ $$
Commented by mr W last updated on 07/Nov/18
$${thanks}\:{alot}\:{sir}! \\ $$
Commented by mr W last updated on 07/Nov/18
$${It}'{s}\:{a}\:{new}\:{ID}\:{sir}.\:{I}\:{can}\:{not}\:{get}\:{the}\:{oldest} \\ $$$${back}\:{any}\:{more}. \\ $$
Commented by ajfour last updated on 08/Nov/18
$${Previously}\:{i}\:{had}\:{obtained}\:{MoI} \\ $$$${about}\:{a}\:{corner}\:{point}\:{on}\:{the}\: \\ $$$${diagonal}\:{axis}.\:{Now}\:{its}\:{about} \\ $$$${G}\left({mass}\:{center}\right)\:{that}\:{still}\:{lies} \\ $$$${on}\:{the}\:{diagonal}\:{axis}\:\left({in}\:{this}\right. \\ $$$$\left.{question}\right). \\ $$