978-0133915426 Chapter 10 Part 6

10–94.

SOLUTION

The mass moment of inertia of this element about the yaxis is

.

Mass: The mass of the solid can be determined by integrating dm.Thus,

The mass of the solid is .Thus,

Mass Moment of Inertia: Integrating ,

Substituting into Iy,

Ans.Iy=

32p

7a375

pb=1.71(103)kg#m2

r=

375

p

kg>m3

Iy=LdIy=L4m

0

rp

512 y6dy =

rp

512

¢

y7

7

≤

`

4m

0

=

32p

7r

dIy

1500 =4pr

r=

375

p

kg>m3

m=1500 kg

m=Ldm =L4m

0

rp

16 y3dy =

rp

16

¢

y4

4

≤

`

4m

0

=4pr

dIy=

1

2dmr2=

1

2

A

rpr2dy

B

r2=

rp

2r4dy =

rp

2a1

4y3>2b4

dy =

rp

512 y6dy

4y3>2b2

16 y3dyr =z=

4y3>2

Determ

i

ne t

h

e mass moment of

i

nert

i

a of t

h

e so

lid

formed by revolving the shaded area around the axis.The

total mass of the solid is .1500 kg

y

Iy

y

x

z

4m

2m

z

2

y

3

1

––

16

O

10–95.

The slender rods have a mass of 4 kg

>

m. Determine the

moment of inertia of the assembly about an axis

perpendicular to the page and passing through point A.

A

200 mm

*10–96.

The pendulum consists of a 8-kg circular disk A, a 2-kg

circular disk B, and a 4-kg slender rod. Determine the radius

of gyration of the pendulum about an axis perpendicular to

the page and passing through point O.

SOLUTION

O

0.5 m

1 m

0.4 m 0.2 m

A

B

10–97.

Determine the moment of inertia of the frustum of the

cone which has a conical depression. The material has a

density of 200 kg>m3.

I

z

SOLUTION

–

3

10 [1

3p(0.4)2(0.6)(200)](0.4)2

–

3

10 [1

3p(0.2)2(0.8)(200)](0.2)2

Iz=

3

10[1

3p(0.4)2(1.6)(200)](0.4)2

z

0.8 m

0.6 m

0.2 m

0.4m

10–98.

SOLUTION

Ans.=4.45 kg #m2

=1

12 (3)(2)2+3(1.781 –1)2+1

12 (5)(0.52+12)+5(2.25 –1.781)2

IG=©IG¿+md2

©ym

The pendulum consists of the 3-kg slender rod and the 5-kg

thin plate.Determine the location of the center of mass G

of the pendulum; then find the mass moment of inertia of

the pendulum about an axis perpendicular to the page and

passing through G.

y

G

2m

1m

0.5 m

y

O

10–99.

Determine the mass moment of inertia of the thin plate

about an axis perpendicular to the page and passing

through point O.The material has a mass per unit area of

.20 kg>m2

SOLUTION

Composite Parts: The plate can be subdivided into the segments shown in Fig. a.

Here, the four similar holes of which the perpendicular distances measured from

their centers of mass to point Care the same and can be grouped as segment (2).

150 mm

O

50 mm

150 mm

*10–100.

SOLUTION

Ans.

k

O=

A

IO

m

=

A

4.917

0.4969

=3.15ft

m

=a4

32.2b+a12

32.2b=0.4969 slug

=4.917slug#ft2

=1

12 a4

32.2b(5)2+a4

32.2b(0.5)2+1

12a12

32.2b(12+12)+a12

32.2b(3.5)2

I

O=©IG+md2

T

he pendulum consists of a plate having a weight of 12 lb

a

nd a slender rod having a weight of 4 lb.Determine the

r

adius of gyration of the pendulum about an axis

p

erpendicular to the page and passing through point O.

O

2ft

3ft

1ft

10–101.

SOLUTION

distance of from point Ocan be grouped as segment (2).

Mass Moment of Inertia: First, we will compute the mass moment of inertia of the

wheel about an axis perpendicular to the page and passing through point O.

The mass moment of inertia of the wheel about an axis perpendicular to the page and

passing through point Acan be found using the parallel-axis theorem ,

where and .Thus,

Ans.IA=84.94 +8.5404(42)=221.58 slug #ft2=222 slug #ft2

d=4ftm=100

32.2 +8

¢

20

32.2

≤

+15

32.2 =8.5404 slug

IA=IO+md2

=84.94 slug #ft2

IO=

¢

100

32.2

≤

(42)+8

B

1

12

¢

20

32.2

≤

(32)+

¢

20

32.2

≤

(2.52)

R

+

¢

15

32.2

≤

(12)

a1+3

2bft =2.5 ft

If the large ring,small ring and each of the spokes weigh

100 lb,15 lb,and 20 lb,respectively,determine the mass

moment of inertia of the wheel about an axis perpendicular

to the page and passing through point A.

O

1ft

4ft

10–102.

Determine the mass moment of inertia of the assembly

about the zaxis.The density of the material is 7.85 Mg m3.

SOLUTION

Mass: The mass of each segment is calculated as

Mass Moment of Inertia: Since the zaxis is parallel to the axis of the cone and the

hemisphere and passes through their center of mass,the mass moment of inertia can be

computed from and .Thus,

Ans. =29.4 kg #m2

=3

10(158.9625p)(0.32)+2

5(141.3p)(0.32)–3

10(5.8875p)(0.12)

Iz=©(Iz)i

3

10 m3r2

3

(Iz)1=3

10 m1r2

1, (Iz)2=2

5 m2r2

2,

m3=rV3=ra1

3pr2hb=7.85(103)c1

3p(0.12)(0.225) d=5.8875p kg

m2=rV2=ra2

3pr3b=7.85(103)c2

3p(0.33)d=141.3p kg

m1=rV1=ra1

3pr2hb=7.85(103)c1

3p(0.32)(0.675) d=158.9625p kg

0.45 +z=0.1

0.3

>z

y

x

300 mm

100 mm

1116

10–103.

Each of the three slender rods has a mass m. Determine the

moment of inertia of the assembly about an axis that is

perpendicular to the page and passes through the center

point O.

O

a

2

*10–104.

SOLUTION

distances measured from the centroid of each segment to the yaxis are indicated in

Fig. a.The mass moment of inertia of each segment about the yaxis can be

determined using the parallel-axis theorem.

Ans.=0.144 kg#m2

=2c1

12 (1.6)(0.42)+1.6(0.22)d–2c1

4(0.1p)(0.12)+0.1p(0.22)d

Iy=©

A

Iy

B

G+md2

The thin plate has a mass per unit area of .

Determine its mass moment of inertia about the yaxis.

10 kg

>

m

2

200 mm

z

100 mm

10–105.

The thin plate has a mass per unit area of .

Determine its mass moment of inertia about the zaxis.

10 kg>m

2

SOLUTION

moment of inertia of each segment about the zaxis can be determined using the

parallel-axis theorem.

200 mm

z

100 mm

=1

12 (1.6)(0.42)+c1

12 (1.6)(0.42+0.42)+1.6(0.22)d–1

4(0.1p)(0.12)–c1

2(0.1p)(0.12)+0.1p(0.22)d

Iz=©

A

Iz

B

G+md2

1119

10–106.

SOLUTION

Ans.

=

118 slug

#

ft2

+1

2ca 90

32.2 bp(2)2(0.25) d(2)2–1

2ca 90

32.2 bp(1)2(0.25) d(1)2

I

G=1

2ca 90

32.2 bp(2.5)2(1) d(2.5)2–1

2ca 90

32.2 bp(2)2(1) d(2)2

Determine the moment of inertia of the assembly about an

axis

that is perpendicular to the page and passes through

the

center of mass G.The material has a specific weight of

.

g

=90 lb>ft3

O

1ft

2ft

G

0.25 ft

Ans:

I

G=

118

slug

#

ft

2

10–107.

Determine the moment of inertia of the assembly about an

axis that is perpendicular to the page and passes through

point O.The material has a specific weight of .g=90 lb>ft3

SOLUTION

O

1ft

2ft

0.5 ft

G

0.25 ft

*10–108.

SOLUTION

Ans.

Ans.IG=5.61 kg #m2

–1

2[p(0.1)2(12)(0.1)2–[p(0.1)2(12)](1.8 –0.8878)2

+1

2[p(0.3)2(12)(0.3)2+[p(0.3)2(12)](1.8 –0.8878)2

+1

12 (1.5)(3)(1.5)2+1.5(3)(0.75 –0.8878)2

IG=1

12 (0.8)(3)(0.8)2+0.8(3)(0.8878)2

=0.8878 m =0.888 m

y=1.5(3)(0.75) +p(0.3)2(12)(1.8) –p(0.1)2(12)(1.8)

1.5(3) +p(0.3)2(12) –p(0.1)2(12) +0.8(3)

T

h

e pen

d

u

l

um cons

i

sts of two s

l

en

d

er ro

d

s AB an

d

OC

which have a mass of The thin plate has a mass of

Determine the location of the center of mass G

of the pendulum, then calculate the moment of inertia of

the pendulum about an axis perpendicular to the page and

passing through G.

y

12 kg>m2.

3kg>m.

AB

0.1 m

0.3 m

C

G

O

1.5

m

0.4 m 0.4 m

_

y

10–109.

Determine the moment of inertia of the frustrum of the

cone which has a conical depression. The material has a

density of 200 kg>m3.

I

z

SOLUTION

The mass moment of inertia of each cone about zaxis can be

determined using

Ans.=34.2 kg#m2

–3

10 cp

310.22210.6212002d10.222

–3

10 cp

310.22210.333212002d10.222

I

z=©1I

z2i=3

10 cp

310.82211.333212002d10.822

I

z=3

10 mr2.

z=0.333 m.

z

200 mm

600 mm

400 mm