Example 6

Problem :
The mass per unit length of a non-uniform rod of mass “M” and length “L” is given as “ax”, where “x” is measured from left end and “a” is a constant. What is its moment of inertia about an axis perpendicular to the rod and passing through its left end ( point “O”) ?

Solution : Let us now consider a small length “dx” at a distance “x” from the axis of rotation. The elemental mass is,

ⅆ
m
=
λ
ⅆ
x
=
a
x
ⅆ
x
ⅆ
m
=
λ
ⅆ
x
=
a
x
ⅆ
x

In order to evaluate the constant “a”, we integrate the elemental mass for the whole length of the rod and equate the same to its mass as :

M
=
∫
ⅆ
m
=
∫
a
x
ⅆ
x
M
=
∫
ⅆ
m
=
∫
a
x
ⅆ
x

Putting the appropriate limit,

⇒
M
=
a
[
x
2
3
]
0
L
=
a
L
2
2
⇒
M
=
a
[
x
2
3
]
0
L
=
a
L
2
2

⇒
a
=
2
M
L
2
⇒
a
=
2
M
L
2

Thus, elemental mass is :

⇒
ⅆ
m
=
(
2
M
L
2
)
x
ⅆ
x
⇒
ⅆ
m
=
(
2
M
L
2
)
x
ⅆ
x

The moment of inertia of the elemental mass about the axis is :

ⅆ
I
=
r
2
ⅆ
m
=
2
M
L
2
)
x
ⅆ
x
x
2
ⅆ
I
=
r
2
ⅆ
m
=
2
M
L
2
)
x
ⅆ
x
x
2

Integrating between the limits “0” and “L”, we have :

⇒
I
=
∫
I
=
∫
2
M
L
2
)
x
ⅆ
x
x
2
⇒
I
=
∫
I
=
∫
2
M
L
2
)
x
ⅆ
x
x
2

⇒
I
=
∫
I
=
∫
2
M
L
2
[
x
4
4
]
0
L
⇒
I
=
∫
I
=
∫
2
M
L
2
[
x
4
4
]
0
L

⇒
I
=
M
L
2
2
⇒
I
=
M
L
2
2

Example 7

Problem :
A circular disk of radius “R” and mass “M” has non-uniform surface mass density, given by
σ
=
a
r
2
σ = a
r
2
, where “r” is the distance from the center of the disk and “a” is a constant. What is its MI about the perpendicular central axis ?

Solution : Let us now consider a small concentric area of thickness “dr” at a distance “r” from the axis of rotation. The elemental mass is,

ⅆ
m
=
σ
ⅆ
A
=
a
r
2
2
π
r
ⅆ
r
ⅆ
m
=
σ
ⅆ
A
=
a
r
2
2
π
r
ⅆ
r

In order to evaluate the constant “a”, we integrate the elemental mass for the whole of the circular disk and equate the same to its mass as :

M
=
∫
ⅆ
m
=
2
π
a
∫
r
3
ⅆ
r
M
=
∫
ⅆ
m
=
2
π
a
∫
r
3
ⅆ
r

Putting the appropriate limit,

⇒
M
=
2
π
a
[
r
4
4
]
0
L
=
π
a
R
4
2
⇒
M
=
2
π
a
[
r
4
4
]
0
L
=
π
a
R
4
2

⇒
a
=
2
M
π
R
4
⇒
a
=
2
M
π
R
4

Thus, elemental mass is :

⇒
ⅆ
m
=
(
2
M
π
R
4
)
x
2
π
r
3
ⅆ
r
⇒
ⅆ
m
=
(
2
M
π
R
4
)
x
2
π
r
3
ⅆ
r

The moment of inertia of the ring of elemental mass about the axis is :

ⅆ
I
=
(
2
M
π
R
4
)
x
2
π
r
5
ⅆ
r
ⅆ
I
=
(
2
M
π
R
4
)
x
2
π
r
5
ⅆ
r

Integrating between the limits “0” and “R”, we have :

⇒
I
=
∫
I
=
∫
4
M
R
4
∫
r
5
ⅆ
r
⇒
I
=
∫
I
=
∫
4
M
R
4
∫
r
5
ⅆ
r

⇒
I
=
4
M
R
4
[
r
6
6
]
0
R
⇒
I
=
4
M
R
4
[
r
6
6
]
0
R

⇒
I
=
4
M
R
4
x
R
6
6
=
2
M
R
2
3
⇒
I
=
4
M
R
4
x
R
6
6
=
2
M
R
2
3