Mechanical Engineering Chapter 21 Homework where r is the distance from the origin O to dm

1109

21–1.

SOLUTION

Show that the sum of the moments of inertia of a body,

, is independent of the orientation of the

x,y,zaxes and thus depends only on the location of its

Ixx +Iyy +Izz

origin.

1110

21–2.

Determine the moment of inertia of the cone with respect to

avertical axis passing through the cone’s center of mass.

What is the moment of inertia about a parallel axis that

passes through the diameter of the base of the cone? The

cone has a mass m.

y¿

y

SOLUTION

The mass of the differential element is .

Using the parallel axis theorem:

dm =rdV =r(py2)dx =rpa2

h2x2dx h

x

y

a

yy¿

–

Ans:

1111

21–3.

SOLUTION

. si tnemele laitnereffid eht fo ssam ehT

dIy=1

2dmy2+dmx2

dm =rdV =r

A

py2

B

dx =rpxdx

Determine moment of inertia of the solid formed by

revolving the shaded area around the xaxis. The density of

the material is r=12 slug/ft3.

Iy

4ft

2ft

y

x

y2=x

1112

*21–4.

SOLUTION

The mass of the differential element is .

Ans.

Ix=LdIx=L2

A

5y2+10y3

B

dy =53.3 slug #ft2

dm =rdV =r

A

pz2

B

dy =2rpydy

Determine the moments of inertia and of the paraboloid

of revolution. The mass of the paraboloid is 20 slug.

Iy

Ix

2ft

z

x

y

z22y

Ans:

1113

21–5.

Determine by direct integration the product of inertia for

the homogeneous prism. The density of the material is .

Express the result in terms of the total mass mof the prism.

r

I

yz

SOLUTION

The mass of the differential element is .

Using the parallel axis theorem:

dIyz =(dIy¿z¿)G+dmyGzG

dm =rdV =rhxdy =rh(a–y)dy

a

x

y

z

a

h

Ans:

1114

21–6.

SOLUTION

The mass of the differential element is .

Using the parallel axis theorem:

dIxy =(dIx¿y¿)G+dmxGyG

dm =rdV =rhxdy =rh(a–y)dy

Determine by direct integration the product of inertia for

the homogeneous prism. The density of the material is .

Express the result in terms of the total mass mof the prism.

r

Ixy

a

x

y

z

a

h

Ans:

1115

21–7.

Determine the product of inertia of the object formed

by revolving the shaded area about the line

Express the result in terms of the density of the material, r.

x=5 ft.

Ixy

SOLUTION

L3

0

dm =r2pL3

0

(5 –x)ydx=r2pL3

0

(5 –x)23xdx =38.4rp

3ft2ft

y

x

y

2

3x

Ans:

1116

*21–8.

SOLUTION

Mass of body;

Also,

Iy¿=L3

0

r2dm

m=L3

0

rp(5 –x)2dy –m¿

Iy¿=L3

0

1

2dm r2–1

2(m¿)(2)2

Determine the moment of inertia of the object formed by

revolving the shaded area about the line Express

the result in terms of the density of the material, r.

x=5ft.

Iy

3ft2ft

y

x

y

2

3x

21–9.

Determine the moment of inertia of the cone about the

axis.The weight of the cone is 15 lb, the height is

and the radius is r=0.5 ft.h=1.5 ft,

z¿

SOLUTION

u=tan–1(0.5

1.5)=18.43°

z¿

z

r

h

Ans:

1118

21–10.

Determine the radii of gyration and for the solid

formed by revolving the shaded area about the yaxis.The

density of the material is r.

ky

kx

SOLUTION

For :The mass of the differential element is .

For :

dI¿

x=1

4dmx2+dmy2

0.25 ft 6y…4ftkx

dm =rdV =r(px2)dy =rpdy

y2

ky

4ft

0.25 ft

y

x

xy 1

Ans:

1119

21–11.

Determine the moment of inertia of the cylinder with

respect to the a–a axis of the cylinder.The cylinder has a

mass m.

SOLUTION

The mass of the differential element is .

dIaa =1

4dma2+dm

A

y2

B

dm =rdV =r

A

pa2

B

dy

a

h

Ans:

*21–12.

SOLUTION

Horizontial plate:

Vertical plates:

Using Eq. 21–5,

lxx¿=0.707,

lxy¿=0.707,

lxz¿=0

Determine the moment of inertia of the composite plate

assembly.The plates have a specific weight of 6lb>ft2.

Ix

0.5 ft

z

y

x

0.25 ft

Ans:

1121

21–13.

SOLUTION

Due to symmetry,

0.5 ft

z

y

x

0.25 ft

Determine the product of inertia of the composite

plate assembly. The plates have a weight of 6 lb>ft

2.

Iyz

Ans:

1122

21–14.

Determine the products of inertia ,,and ,of the

thin plate.The material has a density per unit area of

.50 kg>m2

Ixz

Iyz

Ixy

SOLUTION

The masses of segments 1 and 2 shown in Fig. aare

and . Due to symmetry for

segment 1 and for segment 2 .

Ixy =©Ix¿y¿+mxGyG

Ix–y–=Iy–z–=Ix–z–=0

Ix¿y¿=Iy¿z¿=Ix¿z¿=0m2=50(0.4)(0.2) =4kg

m1=50(0.4)(0.4) =8kg

200 mm

400 mm

z

y

x

Ans:

#

2

21–15.

SOLUTION

Due to symmetry

Ixy =Iyz +Izx =0

Determine the moment of inertia of both the 1.5-kg rod

and 4-kg disk about the axis.z¿

300 mm z

z’

100 mm

Ans:

1124

*21–16.

The bent rod has a mass of

3 kg>m

. Determine the moment

of inertia of the rod about the O–a axis.

SOLUTION

The bent rod is subdivided into three segments and the location of center of mass for

each segment is indicated in Fig. a. The mass of each segments is

m1=3(1) =3 kg

,

m2=3(0.5) =1.5 kg

m3=3(0.3) =0.9 kg

Ixy =[0 +0] +[0 +1.5(0.25)(–1)] +[0 +0.9(0.5)(–1)] =–0.825 kg

#

m

2

The unit vector that denes the direction of the

Oa

axis is

0.5i–1j+0.3k

0.5

1

0.3

xy

0.5 m

0.3 m

1 m

a

O

z

1125

*21–16. Continued

Then

IOa

=Ixx ux

2

+Iyy u

2

y+Izz u

2

z–2Ixy uxuy–2Iyz uyuz–2Izx uzux

1126

21–17.

SOLUTION

Due to symmetry

Iy¿=2c1

12 a1.5

32.2 b(1)2+a1.5

32.2 b(0.667 –0.5)2d+a1.5

32.2 b(1 –0.667)2

The bent rod has a weight of Locate the center of

gravity G() and determine the principal moments of

inertia and of the rod with respect to the

axes.z¿

y¿,x¿,Iz¿

Iy¿,Ix¿,

y

x,

1.5 lb

>

ft.

x

y

z

x¿

y¿

z¿

1ft

G

A

_

x

_

y

Ans:

y=0.5 ft

21–18.

Determine the moment of inertia of the rod-and-disk

assembly about the xaxis.The disks each have a weight of

12 lb.The two rods each have a weight of 4 lb, and their

ends extend to the rims of the disks.

SOLUTION

For a rod:

Iy¿=0

2ft

1ft

x

1ft

Ans:

21–19.

SOLUTION

uax =0

uaz =0.707

Determine the moment of inertia of the composite body

about the aa axis.The cylinder weighs 20 lb, and each

hemisphere weighs 10 lb.

2ft

a

Ans: