# Scaling Intertia Tensor

by alexanderBuzz
Tags: intertia, scaling, tensor
 P: 2 Hi everyone, I have the following problem in my hands, which I don't know how exactly to address. Let's assume that from any CAD(Solidworks, Catia), I obtain the inertia tensor of my model (impossible to calculate by hand btw). I_full=[Ixx Ixy Ixz Ixy Iyy Iyz Ixz Iyz Izz] I know if I change the mass of my model, the inertia tensor will scale linearly with it. But what If I scale my model to half-size, all dimensions? The mass probably goes by 1/8, since it's proportional to volume. Maybe the other factor would be 1/4 (α r^2). So would the correct Inertia scaling factor be?: I_half=1/4*1/8*I_full Cheers!!
 Sci Advisor Thanks PF Gold P: 1,911 It must vary jointly with the mass and the distribution; for example consider how the simple inertia of a thick rod changes under your conditions: http://hyperphysics.phy-astr.gsu.edu/hbase/mi2.html
P: 2
 Quote by UltrafastPED It must vary jointly with the mass and the distribution; for example consider how the simple inertia of a thick rod changes under your conditions: http://hyperphysics.phy-astr.gsu.edu/hbase/mi2.html
Thanks you!

It proves my derivation, if assumed constant density between the scaled object and full-size object.

by definition:

$$I=\int_V \rho r^2 dV$$

if r1-> ar ( scaled by a factor a) dV1->a^3dV

replacing on the above equation:
$$I1=\int_V \rho a^2 r^2 a^3 dV = a^5 \int_V \rho r^2 dV -> I1=a^5 I$$

again, assuming that the mass distribution remains constant.

Correct?

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