1. For the shaded area shown, determine by integrationa. The location of the centroidb. the centroidal moments of inertia0.6 m0.3 m1 m Table 16.1Moment of Inertia of Common ShapesShapeMoment of InertiaRadius of GyrationRectanglebh312bh3I,12hk, =V3hbTrianglebhV18hkz36bh312Circle4J =2SemicircleI, =I,k, - k, -5%3D%3D8.2Ī, = 0.11r= 0.264rQuarter circleI, = I,k, =%3D16Ī, =Ï,=0.055rªk = ky = 0.264r

Given Data: The corner points are (0,0), (1,0), (1,0.6) and (0,0.3). The line of equation AB can…

Answered: 1. For the shaded area shown, determine…

1. For the shaded area shown, determine by integration a. The location of the centroid b. the centroidal moments of inertia

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Concept explainers

Moment of Inertia

Inertia can be termed as a resistance offered by a body to change its state. A rigid body initially at rest will apply resistance when acted by an external force that tends to change its state of rest or velocity, or a body initially at motion will provide resistance in changing its state of velocity or motion for coming to a state of rest. In other words, Newton's first law of motion describes the law of inertia of a rigid body.

Question

1. For the shaded area shown, determine by integration
a. The location of the centroid
b. the centroidal moments of inertia
0.6 m
0.3 m
1 m

Table 16.1
Moment of Inertia of Common Shapes
Shape
Moment of Inertia
Radius of Gyration
Rectangle
bh3
12
bh3
I,
12
h
k, =
V3
h
b
Triangle
bh
V18
h
kz
36
bh3
12
Circle
4
J =
2
Semicircle
I, =I,
k, - k, -5
%3D
%3D
8.
2
Ī, = 0.11r
= 0.264r
Quarter circle
I, = I,
k, =
%3D
16
Ī, =Ï,=0.055rª
k = ky = 0.264r

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