Place the cube with corners at (0,0,0) to (a,a,a). The inertia tensor components are: [ I_{xx} = \int (y^2+z^2) , dm ] Mass density ( \rho = M/a^3 ). Integrate: [ I_{xx} = \rho \int_0^a \int_0^a \int_0^a (y^2+z^2) , dx,dy,dz ] The x-integral gives factor ( a ), so: [ I_{xx} = \rho a \int_0^a\int_0^a (y^2+z^2) dy dz ] Compute: ( \int_0^a y^2 dy = a^3/3 ), similarly for z². Thus: [ \int_0^a \int_0^a y^2 dy dz = a \cdot a^3/3 = a^4/3 ] Same for z² term. Sum: ( a^4/3 + a^4/3 = 2a^4/3 ). Multiply by ( \rho a ): [ I_{xx} = \rho a \cdot \frac{2a^4}{3} = \frac{2\rho a^5}{3} ] But ( \rho a^3 = M ), so ( \rho a^5 = M a^2 ). Hence: [ I_{xx} = \frac{2}{3} M a^2 ] By symmetry, ( I_{yy} = I_{zz} = \frac{2}{3} M a^2 ).
where q is the generalized coordinate and q̇ is the generalized velocity. goldstein classical mechanics solutions chapter 4
: Proving that rotations can be represented by matrices where the transpose is the inverse ( ), preserving the length of vectors. Place the cube with corners at (0,0,0) to (a,a,a)
Solving these equations, we get:
Surfx Technologies Limited Copyright DLS Source © 2026. All rights reserved.
Terms & Conditions Privacy Policy