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Document Title
Fundamentals of Stress and Vibration 2. Engineering Mechanics Chapter
[A Practical guide for aspiring Designers / Analysts]

Solution 6: the sphere could be deemed upon as a pile up of plates. Each plate is a thin disc with
2
MMOI of [dm r 2]. The summation of MMOI’s of all such plates about the axis of rotation gives the
MMOI of the sphere. Therefore, we have:

R R
2
dm r 2 πr ∗ dy ∗ ρ ∗ r 2
I = =
sphere
2 2
−R −R
Since the integration is with respect to ‘dy’, let us represent ‘r’ in equation (2.31) in terms of ‘y’.
Therefore, from [Fig 2.33] we have: r = R − y
2
2
2 2
R π R − y ∗ dy ∗ ρ ∗ R − y
2
2
2
2

I
sphere = 2
−R

R
2
2 2
π R − y ∗ dy ∗ ρ
I =
sphere
2
−R
R R
2 2
4
4
π R + y − 2R y ∗ dy ∗ ρ ρπ
4
2 2
4
I sphere = = + y − 2R y dy
2 2
−R −R
2 3
ρπ y 5 R 2R y R ρπ 2R 5 4R 5 1 2
4 R
5
5
I sphere = yR + − = 2R + − = ρπR 1 + −
2 −R 5 3 2 5 3 5 3
−R −R
6 2 8ρπR 5
5
Therefore, I = ρπR − = - - - - (2.32)
sphere
5 3 15
4
3
We know that the mass of a sphere is πR ρ
3
4 2 2
3
2
2
Rewriting equation . we get: I = πR ρ ∗ R = mR
sphere
3 5 5
It can be noted that the axis of rotation for a sphere could be along any diameter. Due to symmetry,
the mass moment of inertia of a sphere about any axis of rotation passing through the CG is the
same.

QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries, P
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