Resolução Marion [Mec.Clássica] - Capítulo 11

Resolução Marion [Mec.Clássica] - Capítulo 11

(Parte 1 de 9)

Dynamics of Rigid Bodies

1-1. The calculation will be simplified if we use spherical coordinates: sin cos

sin sin cos xr yr zr x Using the definition of the moment of inertia,

() 2 ij ij k i jk we have

Ir z dv r r dr d dρ ρ θθ=−=− ∫


1c os cos

R Ir dr d

The mass of the sphere is


Since the sphere is symmetrical around the origin, the diagonal elements of {I} are equal:

A typical off-diagonal element is

Ix y dv r dr dρ dρ θφ φ θ=−=− ∫

This vanishes because the integral with respect to φ is zero. In the same way, we can show that all terms except the diagonal terms vanish. Therefore, the secular equation is

From (9) and (7), we have

a) Moments of inertia with respect to the x axes: i x = x′

R h CM

It is easily seen that for i ≠ j. Then the diagonal elements become the principal moments I, which we now calculate. 0ijI = iiI

The computation can be simplified by noting that because of the symmetry, . Then, 12III=≠3

DYNAMICS OF RIGID BODIES 355 which, in cylindrical coordinates, can be written as hR z h


Performing the integration and substituting for ρ, we find

from which

b) Moments of inertia with respect to the xi′ axes:

xd v z h

Then, using Eq. (1.49),

I Mh M R h

I Mh M R h I MR

1-3. The equation of an ellipsoid is

which can be written in normalized form if we make the following substitutions:

This is the equation of a sphere in the (ξ,η,ζ) system.

If we denote by dv the volume element in the system and by dτ the volume element in the

(ξ,η,ζ ) system, we notice that the volume of the ellipsoid is ix

Vd v dx dx dx abc d d d abc d abc ξ η ζτπ == ===

because dτ∫ is just the volume of a sphere of unit radius.

The rotational inertia with respect to the passing through the center of mass of the ellipsoid (we assume the ellipsoid to be homogeneous), is given by 3-axisx

M Ix x dv

M abc a b d

In order to evaluate this integral, consider the following equivalent integral in which z = r cos θ : ( )2 2 sin

cos sin

R az dv a z r dr r d d ad d r d


Since the same analysis can be applied for any axis, the other moments of inertia are


IM b c


The linear density of the rod is

For the origin at one end of the rod, the moment of inertia is

If all of the mass were concentrated at the point which is at a distance a from the origin, the moment of inertia would be

This is the radius of gyration.

(Parte 1 de 9)