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Prove-that-the-angular-momentum-H-G-of-a-rigid-body-about-its-mass-center-is-given-by-H-x-I-x-x-I-xy-y-I-xz-z-H-y-I-yx-x-I-y-y-I-yz-z-H-z-I-zx-




Question Number 27986 by ajfour last updated on 18/Jan/18
Prove that the angular momentum  H_G ^�  of a rigid body about its mass  center is given by :  H_x =I_x ^� ω_x −I_(xy) ^� ω_y −I_(xz) ^� ω_z   H_y =−I_(yx) ^� ω_x +I_y ^� ω_y −I_(yz) ^� ω_z   H_z =−I_(zx) ^� ω_x −I_(zy) ^� ω_y +I_z ^� ω_z       where  I_x ^� =∫(y^2 +z^2 )dm        I_(xy) ^� =∫xy dm ...and so on..
$${Prove}\:{that}\:{the}\:{angular}\:{momentum} \\ $$$$\bar {\boldsymbol{{H}}}_{{G}} \:{of}\:{a}\:{rigid}\:{body}\:{about}\:{its}\:{mass} \\ $$$${center}\:{is}\:{given}\:{by}\:: \\ $$$${H}_{{x}} =\bar {{I}}_{{x}} \omega_{{x}} −\bar {{I}}_{{xy}} \omega_{{y}} −\bar {{I}}_{{xz}} \omega_{{z}} \\ $$$${H}_{{y}} =−\bar {{I}}_{{yx}} \omega_{{x}} +\bar {{I}}_{{y}} \omega_{{y}} −\bar {{I}}_{{yz}} \omega_{{z}} \\ $$$${H}_{{z}} =−\bar {{I}}_{{zx}} \omega_{{x}} −\bar {{I}}_{{zy}} \omega_{{y}} +\bar {{I}}_{{z}} \omega_{{z}} \\ $$$$\:\:\:\:{where}\:\:\bar {{I}}_{{x}} =\int\left({y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right){dm} \\ $$$$\:\:\:\:\:\:\bar {{I}}_{{xy}} =\int{xy}\:{dm}\:…{and}\:{so}\:{on}.. \\ $$
Commented by ajfour last updated on 22/Mar/19
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