The paraboloid is formed by revolving the shaded area around the x axis. Determine the radius of gyration kx. The density of the material is p = 5Mg/m³.. y y² = 50x 200 mm 100 mm X

Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
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where does the dm come from?

The paraboloid is formed by revolving the shaded area around the
x axis. Determine the radius of gyration kx.
The density of the material is p = 5Mg/m³..
y
y² = 50x
200 mm
100 mm
X
Transcribed Image Text:The paraboloid is formed by revolving the shaded area around the x axis. Determine the radius of gyration kx. The density of the material is p = 5Mg/m³.. y y² = 50x 200 mm 100 mm X
Solution
17.2
* Disk element (N/B that you need
edv
2/M =
Since
-
Lue
racing
Moment
of Isc
=
Ix=
radius
kx
زه
=
etty?dx (see page
need
moment
of inertia
½y²dm
=
I
411 of Hibbler)
of inertial for disk of
2 m r
thus
sabout the
x-axis, is
Sub dm, we
get
dIx = 55xp it ydx]
50
2500
6
don
جدنا
c
5 x π 50x dx
2500
:200
Joft x² dx
200
епов
3333333333 ETT
Since
Since y² = 50x
1200
Jo 50 pix dx
-
K₂ = 57, 735 mm
200
25 PTT x ² /
епоч
moment
this approach )
25 (200)² ett
е
of gyration Koc
3333333333
25(20012
ر
v=y
= 503C
ett
1
of inerting
y
y² = Sox
·abac,
300
M
Transcribed Image Text:Solution 17.2 * Disk element (N/B that you need edv 2/M = Since - Lue racing Moment of Isc = Ix= radius kx زه = etty?dx (see page need moment of inertia ½y²dm = I 411 of Hibbler) of inertial for disk of 2 m r thus sabout the x-axis, is Sub dm, we get dIx = 55xp it ydx] 50 2500 6 don جدنا c 5 x π 50x dx 2500 :200 Joft x² dx 200 епов 3333333333 ETT Since Since y² = 50x 1200 Jo 50 pix dx - K₂ = 57, 735 mm 200 25 PTT x ² / епоч moment this approach ) 25 (200)² ett е of gyration Koc 3333333333 25(20012 ر v=y = 503C ett 1 of inerting y y² = Sox ·abac, 300 M
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