Why does the third term in the expression (the contraction between the time derivative of the inerti

Why does the third term in the expression (the contraction between the time derivative of the inertia tensor and the angular velocity) vanish?

The third term in the expression for I˙ijωj\dot{I}{ij}\omega_j(where I˙ij\dot{I}{ij} is the time derivative of the inertia tensor contracted with the angular velocity ωj\omega_j) vanishes because the integral it contains is the first moment of mass, and the problem specifies that the rigid body is rotating about its center of mass.

Here is the breakdown of the vanishing term and the physical reason:

From Step 3 of the proof on the web page, the third term in the expanded expression for I˙ijωj\dot{I}_{ij}\omega_j is:

Third Term=VρxiϵjpqωpxqωjdV \text{Third Term} = -\int_V \rho x_i \epsilon_{jpq} \omega_p x_q \omega_j dV

This term is manipulated to separate the angular velocity and position components:

Third Term=xiϵjpqωpωjVρxqdV \text{Third Term} = -x_i \epsilon_{jpq} \omega_p \omega_j \int_V \rho x_q dV

The integral part of this term is:

VρxqdV \int_V \rho x_q dV

  • ρ\rho is the mass density.

  • dVdV is the infinitesimal volume element.

  • xqx_q is the qq-th component of the position vector x\vec{x}.

This integral, VρxdV\int_V \rho \vec{x} dV, represents the first moment of mass (or static moment) of the body.

The integral VρxqdV\int_V \rho x_q dV is related to the center of mass (R\vec{R}) by the definition:

MR=VρxdV M\vec{R} = \int_V \rho \vec{x} dV

where MM is the total mass of the rigid body.

The proof explicitly states that the rotation is about the center of mass (CM). If the origin of the coordinate system is chosen to be the center of mass, then the position vector of the center of mass is R=0\vec{R} = 0.

Therefore, since the rotation is about the center of mass:

MR=M(0)=0 M\vec{R} = M(0) = 0

Which means:

VρxdV=0VρxqdV=0 \int_V \rho \vec{x} dV = 0 \quad \Rightarrow \quad \int_V \rho x_q dV = 0

Because this integral is zero, the entire third term vanishes, significantly simplifying the expression for I˙ijωj\dot{I}_{ij}\omega_j.

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