where does the dm come from? The paraboloid is formed by revolving the shaded area around thex axis. Determine the radius of gyration kx.The density of the material is p = 5Mg/m³..yy² = 50x200 mm100 mmX Solution17.2* Disk element (N/B that you neededv2/M =Since-LueracingMomentof Isc=Ix=radiuskxزه=etty?dx (see pageneedmomentof inertia½y²dm=I411 of Hibbler)of inertial for disk of2 m rthussabout thex-axis, isSub dm, wegetdIx = 55xp it ydx]5025006donجدناc5 x π 50x dx2500:200Joft x² dx200епов3333333333 ETTSinceSince y² = 50x1200Jo 50 pix dx-K₂ = 57, 735 mm20025 PTT x ² /епочmomentthis approach )25 (200)² ettеof gyration Koc333333333325(20012رv=y= 503Cett1of inertingyy² = Sox·abac,300M

Given: The density of material, ρ = 5Mg/m³

Answered: The paraboloid is formed by revolving…

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

International Edition---engineering Mechanics: Statics, 4th Edition

4th Edition

ISBN:9781305501607

Author:Andrew Pytel And Jaan Kiusalaas

Publisher:Andrew Pytel And Jaan Kiusalaas

Chapter8: Centroids And Distributed Loads

Section: Chapter Questions

Problem 8.75P

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Concept explainers

Center of Gravity and Centroid

The term center of gravity indicates the position or location of a point about which the whole weight of a body or system is supposed to be concentrated, which means the whole weight of the body or system works through this point. It is closely related to mass distribution. For all positions of the body, the center of gravity would be a single point.

Question

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

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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