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

šƒ’Ž‡ ͵ǣ ’”‘˜‡ –Šƒ–ǡ –Š‡ –‹‡ ‘ˆ ”‘ŽŽ‹‰ ‹• ‘– ‹†‡’‡†‡– ‘ˆ –Š‡ ƒ••Ǥ

‘Ž—–‹‘ǣ ˆ‘” ƒ Š‘‘‰‡‘—• ›Ž‹†‡”ǡ ‹– ‹• –”—‡ –Šƒ–ǡ –Š‡ –‹‡ ‘ˆ ”‘ŽŽ‹‰ ‹• ‹†‡’‡†‡– ‘ˆ ƒ••Ǥ
‘™‡˜‡”ǡ ˆ‘” ‘’‘•‹–‡ ‘„Œ‡ –•ǡ –‹‡ ‘ˆ ”‘ŽŽ‹‰ ‹• †‡’‡†‡– ‘ †‡•‹–› ”ƒ–‹‘ ‘ˆ –Š‡ ‘’‘‡–• ‘ˆ
–Š‡ ‘’‘•‹–‡ ‘„Œ‡ –Ǥ
‡– —• ‘•‹†‡” ƒ •‹’Ž‡ ƒ•‡ ‘ˆ ƒ Š‘‘’ ‹–‡”ˆ‡”‡ ‡ ϐ‹––‡† ™‹–Š ƒ ›Ž‹†‡”ǡ ƒ• •Š‘™ ‹ ȏ ‹‰ ʹǤ͸ͷȐǤ

[Fig 2.65 : a composite cylinder rolling down an inclined plane]

The acceleration term that determines the time of travel is given as follows:

g sin θ
a =
1 + I
mr 2
I
The term , is independent of mass for a homogenous cylinder.
mr 2
Let us verify, if the same holds for a composite cylinder.

The mass moment of inertia of the composite cylinder is given as follows:

2
m r 2 ρ πr ∗ L ∗ r 2
c
c
2
2
I composite = I cylinder + I hoop = + m r = + ρ 2πr ∗ t ∗ L ∗ r
h
h
2 2
Now that we have found ‘I’ for the composite cylinder, let us compute ‘mr ’.
2
2
2
2
mr 2 composite = ρ πr ∗ L ∗ r + ρ 2πr ∗ t ∗ L ∗ r
c
h

Page 76 QP No. SSC/Q4401, Version 1.0, NSQF Level 7, Compliant with Aero and Auto Industries,