Parallel Axis Theorem
The moment of inertia of a body about any axis is equal to the sum of moment of inertia about a parallel axis through its center-of-mass, and the product of mass of the body to square of distance between two axes. That is,
I = ICM + Mh2
where h is distance between two (parallel) axes.
Proof: Given a body and an axis, consider the body as made up of slices cut perpendicular to the given axis. Suppose one slice lies in the plane of paper. Hence the axis is perpendicular to paper and suppose it passes through point O. Let C be the CM of this slice and imagine a parallel axis at C.
Consider a mass element dm at point P, at position r and r’ from O and C. Hence, we have
r = h + r’
where h denotes vector distance between the two axes. The moment of inertia dl0 of mass element dm about axis at O is given by
dI0 = (dm) r2 = (dm) (r’ + h) . (r’ + h)
= dIc + h2 dm + 2 r’ . h dm
where dIc + r2 dm is MI of mass element about axis at C.
The moment of inertia of entire slice, therefore, is
∫dI0 = ∫dIc + h2 ∫dm + 2 h . ∫r’ dm
By definition of CM the last term vanishes. Hence, for the slice, we get
I0 = Ic + mh2
where, m = ∫dm is mass of the slice. The MI of the entire body is obtained by summing the contributions from all slices. We shall illustrate use of parallel-axes theorem by considering a few examples.
Thin rod: The MI of the rod about an axis passing through one end of the rod and perpendicular to its length is given by
I = ICM + M (L/M)2
where, ICM = 1/12 ML2, and L/2 is the distance between parallel axes.
Hence, we get
I = 1/3 ML2
Solid sphere: The MI of a uniform solid sphere about any tangent (to its surface) at any point is determined by considered the diameter parallel to the tangent. Hence, we get
I = ICM + MR2 = 7/5 MR2
where, ICM = 2/5 MR2 is MI of the sphere about any diameter and R is distance between two parallel axes.
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