2. Locate the centroid of the shaded area6 in.3 in6 in.Area (A), in2(6)(3) = 9(6) (6) = 36(6)(9) = 27AT = 726 inShapeFor xΣαχx ==ATx= -1 in-72 in ³72 in²Examplex, iny, inAx, in ³Ay, in ³3= -2-18633+6=7-10810854ΣA, = -72636613||: 2612113(9) = 6For yΣαγy ==ATy = 4.625 in333 in ³72 in²162ΣΑ, = 333 Compute for the moments of inertia and radius of gyration of the shaded area withrespect to the x - y axes.y112040*Dimensions in mm4030-x3020 2050€20

Now first of all we need to separate the shapes

Answered: Compute for the moments of inertia and…

Compute for the moments of inertia and radius of gyration of the shaded area with respect to the x - y axes. y 1 120- 40 * Dimensions in mm 40 30 30 20 20 50 20 -x

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

100%

2. Locate the centroid of the shaded area
6 in.
3 in
6 in.
Area (A), in2
(6)(3) = 9
(6) (6) = 36
(6)(9) = 27
AT = 72
6 in
Shape
For x
Σαχ
x =
=
AT
x= -1 in
-72 in ³
72 in²
Example
x, in
y, in
Ax, in ³
Ay, in ³
3
= -2
-18
63
3+6=7
-108
108
54
ΣA, = -72
6
3
6
613
||
: 2
612
11
3
(9) = 6
For y
Σαγ
y =
=
AT
y = 4.625 in
333 in ³
72 in²
162
ΣΑ, = 333

Compute for the moments of inertia and radius of gyration of the shaded area with
respect to the x - y axes.
y
1
120
40
*
Dimensions in mm
40
30
-x
30
20 20
50
€20

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