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

Solution 2: using the general formula, with limits [-(L/2) to (L/2)] we have:

L L
2 2 3 2 L 2
2 m 2 m x mL
I = dmx = dx x = L = 12
L
3
L L − L 2
− 2 − 2

Situation 3: consider a thick cylinder rotating about an axis through the CG and parallel to the
length of the cylinder, as shown in [Fig 2.29]. Find its MMOI.

[Fig 2.29: Thick cylinder rotating about an axis of symmetry
perpendicular to its cross-section]

Solution 3: the mass of the elemental cylinder at a distance ‘r’ from the CG is given by:

dm = 2πr ∗ dr ∗ L ∗ ρ and the elemental MMOI is given by: dI = dm ∗ r
2
Therefore, the total mass moment of inertia of the cylinder is given by:

R R
2
r 4 2πLρR 4 πR ∗ L ∗ ρ ∗ R 2 mR 2
3
I = 2πLρ r dr = 2πLρ = = =
4 4 2 2
0 0

Page 40 QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries,