Moment of inertia
1. Let R be the triangle with vertices (0, 0), (1, 0), (1,
polar moment of inertia.
√
3) and density δ = 1. Find the
Answer: The region R is a 30, 60 , 90 triangle.
y
√
3
r = sec θ
r
x
1
The polar moment of inertia is the moment of inertia around the origin (that is, the z-axis).
The figure shows the triangle and a small square piece within R. If the piece has area dA
then its polar moment of inertia is dI = r2 δ dA. Summing the contributions of all such
pieces and using δ = 1, dA = r dr dθ, we get the total moment of inertia is
��
��
��
I=
r2 δ dA =
r2 r dr dθ =
r3 dr dθ.
R
R
R
Next we find the limits of integration in polar coordinates. The line
x = 1 ⇔ r cos θ = 1 ⇔ r = sec θ.
So, using radial stripes, the limits are: (inner) r from 0 to sec θ; (outer) θ from 0 to π/3.
Thus,
�
π/3 � sec θ
I=
0
Inner integral:
sec4 θ
r3 dr dθ.
0
.
4
Outer integral: Use sec4 θ = sec2 θ sec2 θ = (1 + tan2 θ) d(tan θ) ⇒ the outer integral is
�
√ � √
�
��π/3
1
tan3 θ ��
1 √
( 3)3
3
tan θ +
=
3+
=
.
�
4
3
4
3
2
0
√
The polar moment of inertia is
3
.
2
MIT OpenCourseWare
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18.02SC Multivariable Calculus
Fall 2010
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