User:Prof McCarthy/parallel axis theorem

In physics, the parallel axis theorem or Huygens–Steiner theorem can be used to determine the second moment of area or the mass moment of inertia of a rigid body about any axis, given the body's moment of inertia about a parallel axis through the object's centre of mass and the perpendicular distance (r) between the axes.

Polar moment of inertia for planar dynamics

The mass properties of a rigid body that is constrained to move parallel to a plane are defined by its center of mass R=(x, y) in this plane, and its polar moment of inertia I_R about this point. The parallel axis theorem provides a convenient relationship between the moment of inertia I_S and the moment of inertia about the center of mass I_R.

Recall that the center of mass R has the property

$\int _{V}\rho (\mathbf {r} )(\mathbf {r} -\mathbf {R} )dV=0,$,

where r is integrated over the volume V of the body. The polar moment of inertia of a body undergoing planar movement can be computed relative to any reference point S,

$I_{S}=\int _{V}\rho (\mathbf {r} )(\mathbf {r} -\mathbf {S} )\cdot (\mathbf {r} -\mathbf {S} )dV,$

where S is constant and r is integrated over the volume V.

In order to obtain the moment of inertia I_S in terms of the moment of inertia I_R, introduce the vector d from S to the center of mass R,

$I_{S}=\int _{V}\rho (\mathbf {r} )(\mathbf {r} -\mathbf {R} +\mathbf {d} )\cdot (\mathbf {r} -\mathbf {R} +\mathbf {d} )dV=\int _{V}\rho (\mathbf {r} )(\mathbf {r} -\mathbf {R} )\cdot (\mathbf {r} -\mathbf {R} )dV+2\mathbf {d} \cdot (\int _{V}\rho (\mathbf {r} )(\mathbf {r} -\mathbf {R} )dV)+(\int _{V}\rho (\mathbf {r} )dV)\mathbf {d} \cdot \mathbf {d} ,$

The first term is the moment of inertia I_R, the second term is zero by definition of the center of mass, and the last term is the total mass of the body times the square magnitude of the vector d. Thus,

$I_{S}=I_{R}+Md^{2},$

which is known as the parallel axis theorem.

Let a rigid assembly of points P_i be constrained to move in trajectories parallel to a reference plane.

If the movement of a rigid body is constrained so that all if its point trajectories line in planes parallel to a reference plane, the body is said to undergo planar movement. The dynamics of a rigid body in planar movement is defined by Newton's second laws restricted to the plane,

$\mathbf {F} =M\mathbf {A} ,\quad T=I_{R}\alpha ,$

where the resultant force has two components, F=(F_x, F_y), and the resultant toque T is a scalar. Similarly, A=(a_x, a_y) is the acceleration of the center of mass R=(x, y) and the scalar I_R is the mass moment of inertia of the body about the center of mass.

Recall that the center of mass R of the body is defined by the requirement that the relative position coordinates r-R of points in the volume V of the body weighted by density ρ(r) sum to zero, that is,

$\int _{V}\rho (\mathbf {r} )(\mathbf {r} -\mathbf {R} )dV=0.$

For planar movement of a rigid body, the moment of inertia I_S about an arbitrary reference point S is defined by

$I_{S}=\int _{V}\rho (\mathbf {r} )(\mathbf {r} -\mathbf {S} )\cdot (\mathbf {r} -\mathbf {S} )dV,$

where S is constant and r is integrated over the volume V.

The parallel axis theorem provides a convenient relationship between the moment of inertia I_S and the moment of inertia about the center of mass I_R. Let d be the vector from S to the center of mass, so S=R-d, then the moment of inertia about S becomes

$I_{S}=\int _{V}\rho (\mathbf {r} )(\mathbf {r} -\mathbf {R} +\mathbf {d} )\cdot (\mathbf {r} -\mathbf {R} +\mathbf {d} )dV=\int _{V}\rho (\mathbf {r} )(\mathbf {r} -\mathbf {R} )\cdot (\mathbf {r} -\mathbf {R} )dV+2\mathbf {d} \cdot (\int _{V}\rho (\mathbf {r} )(\mathbf {r} -\mathbf {R} )dV)+(\int _{V}\rho (\mathbf {r} )dV)\mathbf {d} \cdot \mathbf {d} ,$

$I_{z}=I_{cm}+mr^{2},\,$

where:

$I_{cm}\!$ is the moment of inertia of the object about an axis passing through its centre of mass;

$m\!$ is the object's mass;

$r\!$ is the perpendicular distance between the axis of rotation and the axis that would pass through the centre of mass.

This rule can be applied with the stretch rule and perpendicular axis theorem to find moments of inertia for a variety of shapes.

Parallel axes rule for area moment of inertia

The parallel axes rule also applies to the second moment of area (area moment of inertia) for a plane region D:

$I_{z}=I_{x}+Ar^{2},\,$

where:

$I_{z}\!$ is the area moment of inertia of D relative to the parallel axis;

$I_{x}\!$ is the area moment of inertia of D relative to its centroid;

$A\!$ is the area of the plane region D;

$r\!$ is the distance from the new axis z to the centroid of the plane region D.

Note: The centroid of D coincides with the centre of gravity (CG) of a physical plate with the same shape that has uniform density.

Proof

We may assume, without loss of generality, that in a Cartesian coordinate system the perpendicular distance between the axes lies along the x-axis and that the centre of mass lies at the origin. The moment of inertia relative to the z-axis, passing through the centre of mass, is:

$I_{cm}=\int {(x^{2}+y^{2})}dm$

The moment of inertia relative to the new axis, perpendicular distance r along the x-axis from the centre of mass, is:

$I_{z}=\int {((x-r)^{2}+y^{2})}dm$

If we expand the brackets, we get:

$I_{z}=\int {(x^{2}+y^{2})}dm+r^{2}\int dm-2r\int {x}dm$

The first term is I_cm, the second term becomes mr², and the final term is zero since the origin is at the centre of mass. So, this expression becomes:

$I_{z}=I_{cm}+mr^{2}\,$

In classical mechanics

In classical mechanics, the Parallel axis theorem (also known as Huygens-Steiner theorem) can be generalized to calculate a new inertia tensor J_ij from an inertia tensor about a centre of mass I_ij when the pivot point is a displacement a from the centre of mass:

$J_{ij}=I_{ij}+m(\|{\boldsymbol {a}}\|^{2}\delta _{ij}-a_{i}a_{j})\!$

where

${\boldsymbol {a}}=a_{1}{\boldsymbol {\hat {x}}}+a_{2}{\boldsymbol {\hat {y}}}+a_{3}{\boldsymbol {\hat {z}}}\!$

is the displacement vector from the centre of mass to the new axis, and

$\delta _{ij}\!$

is the Kronecker delta.

We can see that, for diagonal elements (when i = j), displacements perpendicular to the axis of rotation results in the above simplified version of the parallel axis theorem.

See also

• Perpendicular axis theorem
• Stretch rule
• Jakob Steiner
• Christiaan Huygens

References

• Parallel axis theorem