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•21–1. Show that the sum of the moments of inertia of a

body, Ixx + Iyy + Izz, is independent of the orientation of

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

origin.

Ixx + Iyy + Izz =

Lm

= 2

(y2 + z2)dm +

Lm

(x2 + z2)dm +

(x2 + y2)dm

Lm

Lm

(x2 + y2 + z2)dm

However, x2 + y2 + z2 = r2, where r is the distance from the origin O to dm. Since

ƒ r ƒ is constant, it does not depend on the orientation of the x, y, z axis. Consequently,

Ixx + Iyy + Izz is also indepenent of the orientation of the x, y, z axis.

Q.E.D.

925

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21–2. Determine the moment of inertia of the cone with

respect to a vertical y axis that passes through the cone’s

center of mass.What is the moment of inertia about a parallel

axis y¿ that passes through the diameter of the base of the

cone? The cone has a mass m.

–y

y

y¿

a

x

The mass of the differential element is dm = rdV = r(py2) dx =

dIv =

=

Iv =

L

h2

h

x2dx.

1

dmy2 + dmx2

4

r pa2

a 2

1 rpa2 2

B 2 x dx R a xb + ¢ 2 x2 ≤ x2 dx

4

h

h

h

r pa2

=

rpa2

(4h2 + a2) x4 dx

4h4

dIv =

rpa2

4

4h

h

(4h2 + a2)

x4dx =

r pa2h

(4h2 + a2)

20

x2 dx =

r pa2h

3

L0

However,

m =

Lm

dm =

h

r pa2

2

h

L0

Hence,

Iy =

3m

(4h2 + a2)

20

Using the parallel axis theorem:

Iv = Iv + md2

3h 2

3m

(4h2 + a2) = Iv + ma b

20

4

Iv =

3m 2

(h + 4a2)

80

Ans.

Iv = Iy + md2

=

h 2

3m 2

(h + 4a2) + ma b

80

4

=

m

(2h2 + 3a2)

20

Ans.

926

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z

21–3. Determine the moments of inertia Ix and Iy of the

paraboloid of revolution. The mass of the paraboloid is m.

a

m = r

L0

a

pz2 dy = rp

L0

a

r2

r2

bydy = rra b a

a

2

a

Iv =

2

r

z2 ϭ — y

a

r

a

1

1

1

r4

r4

z4dy = rp a 2 b

v2dy = rp a ba

dm z2 = rp

2

2

6

a L0

Lm 2

L0

y

x

a

Thus,

Ix =

1

mr2

3

a

Ix =

=

Ans.

a

1

1

a dm z2 + dm y2 b = rp

z4 dy + r

pz2 y2 dy

4

L0

Lm 4

L0

a

a

rpr 2a 3

rpr 4a

1

r4

r2

1

1

r pa 2 b

y2dy + rpa b

y3dy =

+

= mr 2 + ma 2

a L0

4

12

4

6

2

a L0

Ix =

m 2

(r + 3a2)

6

Ans.

z

*21–4. Determine by direct integration the product of

inertia Iyz for the homogeneous prism. The density of the

material is r. Express the result in terms of the total mass m

of the prism.

a

a

The mass of the differential element is dm = rdV = rhxdy = rh(a - y)dy.

h

a

ra2h

m =

dm = rh

(a - y)dy =

2

Lm

L0

x

Using the parallel axis theorem:

dIyz = (dIy¿z¿)G + dmyGzG

h

= 0 + (rhxdy) (y) a b

2

Iyz =

=

rh2

xydy

2

=

rh2

(ay - y2) dy

2

a

rh2

ra3h2

1 ra2h

m

= a

b(ah) =

ah

(ay - y2) dy =

2 L0

12

6

2

6

927

Ans.

y

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z

•21–5. Determine by direct integration the product of

inertia Ixy for the homogeneous prism. The density of the

material is r. Express the result in terms of the total mass m

of the prism.

a

a

The mass of the differential element is dm = rdV = rhxdy = rh(a - y)dy.

h

a

y

ra2h

m =

dm = rh

(a - y)dy =

2

Lm

L0

x

Using the parallel axis theorem:

dIxy = (dIx¿y¿)G + dmxGyG

x

= 0 + (rhxdy)a b(y)

2

=

rh2 2

x ydy

2

=

rh2 3

(y - 2ay2 + a 2 y) dy

2

Ixy =

a

rh

(y3 - 2ay2 + a 2 y) dy

2 L0

=

ra 4h

1 ra 2h 2

m 2

=

a

ba =

a

24

12

2

12

Ans.

z

21–6. Determine the product of inertia Ixy for the

homogeneous tetrahedron. The density of the material is r.

Express the result in terms of the total mass m of the solid.

Suggestion: Use a triangular element of thickness dz and

then express dIxy in terms of the size and mass of the

element using the result of Prob. 21–5.

a

y

r

1

dm = r dV = rc (a - z)(a - z) ddz = (a - z)2 dz

2

2

m =

a

a

r

ra3

(a2 - 2az + z2)dz =

2 L0

6

a

x

From Prob. 21–5 the product of inertia of a triangular prism with respect to the xz

ra 4h

rdz

(a - z)4. Hence,

and yz planes is Ixy =

. For the element above, dIxy =

24

24

Ixy =

a

r

(a4 - 4a3z + 6z 2a 2 - 4az 3 + z 4)dz

24 L0

Ixy =

ra 5

120

Ixy =

ma 2

20

or,

Ans.

928

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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

21–7. Determine the moments of inertia for the

homogeneous cylinder of mass m about the x¿ , y¿ , z¿ axes.

r

z¿

x¿

y

r

z

x

Due to symmetry

Ixy = Iyz = Izx = 0

Iy = Ix =

1

r 2

7mr 2

m(3r 2 + r 2) + ma b =

12

2

12

Iz =

1 2

mr

2

For x¿ ,

ux = cos 135° = -

1

22

uy = cos 90° = 0,

,

1

uz = cos 135° = -

22

Ix = Ix u2x + Iy u2y + Iz u2z - 2Ixy ux uy - 2Iyz uy uz - 2Izx uz ux

=

7mr 2

1 2

1

1 2

b + 0 + mr 2 a b - 0-0-0

¢12

2

22

22

=

13

mr 2

24

Ans.

For y¿ ,

Iy¿ = Iy =

7mr 2

12

Ans.

For z¿ ,

ux = cos 135° = -

1

22

,

uy = cos 90° = 0,

uz = cos 45° =

1

22

Iz¿ = Ix u2x + Iy u2y + Iz u2z - 2Ixy ux uy - 2Iyz uy uz - 2Izx uz ux

=

7mr 2

1

1 2

1 2

b + 0 + mr 2 a b - 0-0-0

¢12

2

22

22

=

13

mr 2

24

Ans.

929

y¿

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z

*21–8. Determine the product of inertia Ixy of the

homogeneous triangular block. The material has a

density of r . Express the result in terms of the total mass

m of the block.

b

h

The mass of the differential rectangular volume element shown in Fig. a is

x

dm = rdV = rbzdy. Using the parallel - plane theorem,

dIxy = dI x¿y¿ + dmxGyG

b

= 0 + [rbzdy]a by

2

=

However, z =

rb2

zydy

2

h

(a - y). Then

a

dIxy =

rb2 h

rb2h

c (a - y)y ddy =

A ay - y2 B dy

2 a

2a

Thus,

Ixy =

L

dIxy =

a

rb 2h

A ay - y2 B dy

2a L0

=

y3 a

rb2h ay2

b2

¢

2a

2

3 0

=

1

ra 2b 2 h

12

1

1

However, m = rV = a ahb b = rabh. Then

2

2

Ixy =

1

m

1

ra 2b2h§

¥ = mab

12

1

6

rabh

2

Ans.

930

y

a

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z

•21–9. The slender rod has a mass per unit length of

6 kg>m. Determine its moments and products of inertia

with respect to the x, y, z axes.

2m

x

The mass of segments (1), (2), and (3) shown in Fig. a is

m1 = m2 = m3 = 6(2) = 12 kg. The mass moments of inertia of the bent rod about

the x, y, and z axes are

1

1

(12) A 22 B + 12 A 12 + 02 B d + c (12) A 22 B + 12 C 22 +

12

12

= 80 kg # m2

A -1 B 2 D d

Ans.

Iy = ©Iy¿ + m A xG 2 + zG 2 B

= c

1

1

(12) A 22 B + 12 A 12 + 02 B d + c0 + 12 A 22 + 02 B d + c (12) A 22 B + 12 C 22 +

12

12

= 128 kg # m2

A -1 B 2 D d

Ans.

Iz = ©Iz¿ + m A xG 2 + yG 2 B

= c

1

1

(12) A 22 B + 12 A 12 + 02 B d + c (12) A 22 B + 12 A 22 + 12 B d + c0 + 12 A 22 + 22 B d

12

2

= 176 kg # m2

Ans.

Due to symmetry, the products of inertia of segments (1), (2), and (3) with respect to

their centroidal planes are equal to zero. Thus,

Ixy = ©Ix¿y¿ + mxGyG

= C 0 + 12(1)(0) D + C 0 + 12(2)(1) D + C 0 + 12(2)(2) D

= 72 kg # m2

Ans.

Iyz = ©Iy¿z¿ + myGzG

= C 0 + 12(0)(0) D + C 0 + 12(1)(0) D + C 0 + 12(2)(-1) D

= -24 kg # m2

Ans.

Ixz = ©Ix¿z¿ + mxGzG

= C 0 + 12(1)(0) D + C 0 + 12(2)(0) D + C 0 + 12(2)(-1) D

= -24 kg # m2

Ans.

931

y

2m

2m

Ix = ©Ix¿ + m A yG 2 + zG 2 B

= A0 + 0B + c

O

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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

z

21–10. Determine the products of inertia Ixy, Iyz, and Ixz

of the homogeneous solid. The material has a density of

7.85 Mg>m3.

200 mm

200 mm

100 mm

x

The masses of segments (1) and (2) shown in Fig. a are m1 = r V1

= 7850(0.4)(0.4)(0.1) = 125.6 kg and m2 = r V2 = 7850(0.2)(0.2)(0.1) = 31.4 kg.

Due

to

symmetry

for

segment

(1)

and

Ix¿y¿ = Iy¿z¿ = Ix¿z¿ = 0

Ix–y– = Iy–z– = Ix–z– = 0 for segment (2). Since segment (2) is a hole, it should be

considered as a negative segment. Thus

Ixy = ©Ix¿y¿ + mxGyG

= C 0 + 125.6(0.2)(0.2) D - C 0 + 31.4(0.3)(0.1) D

= 4.08 kg # m2

Ans.

Iyz = ©Iy¿z¿ + myGzG

= C 0 + 125.6(0.2)(0.05) D - C 0 + 31.4(0.1)(0.05) D

= 1.10 kg # m2

Ans.

Ixz = ©Ix¿z¿ + mxGzG

= C 0 + 125.6(0.2)(0.05) D - C 0 + 31.4(0.3)(0.05) D

= 0.785 kg # m2

Ans.

932

200 mm

200 mm

y

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z

21–11. The assembly consists of two thin plates A and B

which have a mass of 3 kg each and a thin plate C which has

a mass of 4.5 kg. Determine the moments of inertia Ix , Iy

and Iz .

B

0.4 m

60Њ

A

C

0.4 m

0.3 m

x

Ix¿ = Iz¿ =

Iy¿ =

1

(3)(0.4)2 = 0.04 kg # m2

12

1

(3)[(0.4)2 + (0.4)2] = 0.08 kg # m2

12

Ix¿y¿ = Iz¿y¿ = Iz¿x¿ = 0

For zG,

ux¿ = 0

uy¿ = cos 60° = 0.50

uz¿ = cos 30° = 0.8660

IzG = 0 + 0.08(0.5)2 + 0.04(0.866)2 - 0 - 0 - 0

= 0.05 kg # m2

IxG = Ix¿ = 0.04 kg # m2

For yG,

ux¿ = 0

uy¿ = cos 30° = 0.866

uz¿ = cos 120° = - 0.50

IyG = 0 + 0.08(0.866)2 + 0.04(-0.5)2 - 0 - 0 - 0

= 0.07 kg # m2

Ix =

1

(4.5)(0.6)2 + 2[0.04 + 3{(0.3 + 0.1)2 + (0.1732)2}]

12

Ix = 1.36 kg # m2

Iy =

Ans.

1

(4.5)(0.4)2 + 2[0.07 + 3(0.1732)2]

12

Iy = 0.380 kg # m2

Iz =

Ans.

1

(4.5)[(0.6)2 + (0.4)2] + 2[0.05 + 3(0.3 + 0.1)2]

12

Iz = 1.26 kg # m2

Ans.

933

60Њ

0.3 m

0.2 m

0.2 m

y

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z

*21–12. Determine the products of inertia Ixy, Iyz, and Ixz,

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

50 kg>m2.

The masses of segments (1) and (2) shown in Fig. a are m1 = 50(0.4)(0.4) = 8 kg

and m2 = 50(0.4)(0.2) = 4 kg. Due to symmetry Ix¿y¿ = Iy¿z¿ = Ix¿z¿ = 0 for

segment (1) and Ix–y– = Iy–z– = Ix–z– = 0 for segment (2).

400 mm

200 mm

Ixy = ©Ix¿y¿ + mxGyG

= C 0 + 8(0.2)(0.2) D + C 0 + 4(0)(0.2) D

x

= 0.32 kg # m2

y

400 mm

Ans.

Iyz = ©Iy¿z¿ + myGzG

= C 0 + 8(0.2)(0) D + C 0 + 4(0.2)(0.1) D

= 0.08 kg # m2

Ans.

Ixz = ©Ix¿z¿ + mxGzG

= C 0 + 8(0.2)(0) D + C 0 + 4(0)(0.1) D

= 0

Ans.

z¿

•21–13. The bent rod has a weight of 1.5 lb>ft. Locate the

center of gravity G(x, y) and determine the principal

moments of inertia Ix¿ , Iy¿ , and Iz¿ of the rod with respect

to the x¿, y¿, z¿ axes.

z

1 ft

Due to symmetry

1 ft

A

_

y

x¿

x

y = 0.5 ft

Ans.

x =

Ans.

(-1)(1.5)(1) + 2 C ( -0.5)(1.5)(1) D

©x W

=

= -0.667 ft

©w

3 C 1.5(1) D

Ix¿ = 2c a

1.5

1

1.5

b(0.5)2 d +

a

b(1)2

32.2

12 32.2

= 0.0272 slug # ft2

Iy¿ = 2 c

Ans.

1 1.5

1.5

1.5

a

b (1)2 + a

b (0.667 - 0.5)2 d + a

b(1 - 0.667)2

12 32.2

32.2

32.2

= 0.0155 slug # ft2

Iz¿ = 2 c

+

Ans.

1 1.5

1.5

a

b (1)2 + a

b(0.52 + 0.16672) d

12 32.2

32.2

1

1.5

1.5

a

b(1)2 + a

b(0.3333)2

12 32.2

32.2

= 0.0427 slug # ft2

Ans.

934

G

y¿

_

x

y

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z

21–14. The assembly consists of a 10-lb slender rod and a

30-lb thin circular disk. Determine its moment of inertia

about the y¿ axis.

2 ft

The mass of moment inertia of the assembly about the x, y, and z axes are

1 ft

1 30

30

1 10

10

b A 12 B +

b A 22 B +

Ix = Iz = c a

A 22 B d + c a

A 12 B d

4 32.2

32.2

12 32.2

32.2

A

= 4.3737 slug # ft

Iy =

y

B

2

x

1 30

a

b A 12 B + 0 = 0.4658 slug # ft2

2 32.2

y¿

Due to symmetry, Ixy = Iyz = Ixz = 0. From the geometry shown in Fig. a,

1

u = tan-1 a b = 26.57°. Thus, the direction of the y¿ axis is defined by the unit

2

vector

u = cos 26.57°j - sin 26.57°k = 0.8944j - 0.4472k

Thus,

ux = 0

uy = 0.8944

uz = -0.4472

Then

Iy¿ = Ix ux 2 + Iy uy 2 + Iz uz 2 - 2Ixy ux uy - 2Iyz uy uz - 2Ixzux uz

= 4.3737(0) + 0.4658(0.8944)2 + 4.3737(-0.4472)2 - 0 - 0 - 0

= 1.25 slug # ft2

Ans.

z

21–15. The top consists of a cone having a mass of 0.7 kg

and a hemisphere of mass 0.2 kg. Determine the moment of

inertia Iz when the top is in the position shown.

Ix¿ = Iy¿ =

30 mm

2

3

3

(0.7) C (4)(0.3)2 + (0.1)2 D + (0.7)c (0.1) d

80

4

+ a

100 mm

30 mm

2

83

3

b (0.2)(0.03)2 + (0.2)c (0.03) + (0.1) d = 6.816 A 10-3 B kg # m2

320

8

y

3

2

Iz¿ = a b(0.7)(0.03)2 + a b (0.2)(0.03)2

10

5

45Њ

Iz = 0.261 A 10-3 B kg # m2

ux = cos 90° = 0,

Iz =

Ix¿ u2x¿

+

Iy¿ u2y¿

uy¿ = cos 45° = 0.7071,

+

Iz¿ u2z¿

uz¿ = cos 45° = 0.7071

x

- 2Ix¿y¿ux¿ uy¿ - 2Iy¿z¿ uy¿ uz¿ - 2Ix¿z¿ ux¿ uz¿

= 0 + 6.816 A 10-3 B (0.7071)2 + (0.261) A 10-3 B (0.7071)2 - 0 - 0 - 0

Iz = 3.54 A 10-3 B kg # m2

Ans.

935

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Page 936

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z

*21–16. Determine the products of inertia Ixy, Iyz, and

Ixz of the thin plate. The material has a mass per unit area

of 50 kg>m2.

200 mm

200 mm

200 mm

The masses of segments (1), (2), and (3) shown in Fig. a are m1 = m2

= 50(0.4)(0.4) = 8 kg and m3 = 50cp(0.1)2 d = 0.5p kg.

200 mm

Due to symmetry Ix¿y¿ = Iy¿z¿ = Ix¿z¿ = 0 for segment (1), Ix–y– = Iy–z– = Ix–z– = 0

for segment (2), and Ix‡y‡ = Iy‡z‡ = Ix‡z‡ = 0 for segment (3). Since segment (3) is

a hole, it should be considered as a negative segment. Thus

Ixy = ©Ix¿y¿ + mxGyG

x

= C 0 + 8(0.2)(0.2) D + C 0 + 8(0)(0.2) D - C 0 + 0.5p(0)(0.2) D

= 0.32 kg # m2

100 mm

400 mm

y

400 mm

Ans.

Iyz = ©Iy¿z¿ + myGzG

= C 0 + 8(0.2)(0) D + C 0 + 8(0.2)(0.2) D - C 0 + 0.5p(0.2)(0.2) D

= 0.257 kg # m2

Ans.

Ixz = ©Ix¿z¿ + mxGzG

= C 0 + 8(0.2)(0) D + C 0 + 8(0)(0.2) D - C 0 + 0.5p(0)(0.2) D

= 0 kg # m2

Ans.

z

•21–17. Determine the product of inertia Ixy for the bent

rod. The rod has a mass per unit length of 2 kg>m.

Product of Inertia: Applying Eq. 21–4. we have

Ixy = © A Ix¿y¿ B G + mxG yG

400 mm

= [0 + 0.4 (2) (0) (0.5)] + [0 + 0.6 (2) (0.3) (0.5)] + [0 + 0.5 (2) (0.6) (0.25)]

= 0.330 kg # m2

y

Ans.

600 mm

500 mm

x

936

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z

21–18. Determine the moments of inertia Ixx, Iyy, Izz for

the bent rod. The rod has a mass per unit length of 2 kg>m.

400 mm

y

600 mm

Moments of Inertia: Applying Eq. 21–3, we have

500 mm

Ixx = ©(Ix¿x¿)G + m(y2G + z2G)

= c

x

1

(0.4) (2) A 0.42 B + 0.4 (2) A 0.52 + 0.22 B d

12

+ C 0 + 0.6 (2) A 0.52 + 02 B D

+ c

1

(0.5) (2) A 0.52 B + 0.5 (2) A 0.252 + 02 B d

12

= 0.626 kg # m2

Ans.

Ixy = ©(Iy¿y¿)G + m(x2G + z2G)

= c

1

(0.4) (2) A 0.42 B + 0.4 (2) A 02 + 0.22 B d

12

+ c

1

(0.6) (2) A 0.62 B + 0.6 (2) A 0.32 + 02 B d

12

+ C 0 + 0.5(2) A 0.62 + 02 B D

= 0.547 kg # m2

Ans.

Izz = ©(Iz¿z¿)G + m(x2G + y2G)

= C 0 + 0.4 (2) A 02 + 0.52 B D

+ c

1

(0.6) (2) A 0.62 B + 0.6 (2) A 0.32 + 0.52 B d

12

+ c

1

(0.5) (2) A 0.52 B + 0.5 (2) A 0.62 + 0.252 B d

12

= 1.09 kg # m2

Ans.

937

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z

21–19. Determine the moment of inertia of the rod-andthin-ring assembly about the z axis. The rods and ring have

a mass per unit length of 2 kg>m.

A

O

500 mm

400 mm

For the rod,

D

ux¿ = 0.6,

Ix = Iy =

uy¿ = 0,

uz¿ = 0.8

B

1

[(0.5)(2)](0.5)2 = 0.08333 kg # m2

3

120Њ

120Њ

x

Ix¿ = 0

Ix¿y¿ = Iy¿z¿ = Ix¿z¿ = 0

From Eq. 21–5,

Iz = 0.08333(0.6)2 + 0 + 0 - 0 - 0 - 0

Iz = 0.03 kg # m2

For the ring,

The radius is r = 0.3 m

Thus,

Iz = mR2 = [2 (2p)(0.3)](0.3)2 = 0.3393 kg # m2

Thus the moment of inertia of the assembly is

Iz = 3(0.03) + 0.339 = 0.429 kg # m2

Ans.

938

y

120Њ

C

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z

*21–20. If a body contains no planes of symmetry, the

principal moments of inertia can be determined

mathematically. To show how this is done, consider the rigid

body which is spinning with an angular velocity V , directed

along one of its principal axes of inertia. If the principal

moment of inertia about this axis is I, the angular momentum

can be expressed as H = IV = Ivx i + Ivy j + Ivz k. The

components of H may also be expressed by Eqs. 21–10,

where the inertia tensor is assumed to be known. Equate the

i, j, and k components of both expressions for H and consider

vx, vy, and vz to be unknown. The solution of these three

equations is obtained provided the determinant of the

coefficients is zero. Show that this determinant, when

expanded, yields the cubic equation

V

O

x

I 3 - (Ixx + Iyy + Izz)I 2

+ (IxxIyy + IyyIzz + IzzIxx - I2xy - I2yz - I2zx)I

- (IxxIyyIzz - 2IxyIyzIzx - IxxI 2yz

- IyyI 2zx - IzzI 2xy) = 0

The three positive roots of I, obtained from the solution of

this equation, represent the principal moments of inertia

Ix, Iy, and Iz.

H = Iv = Ivx i + Ivy j + Ivz k

Equating the i, j, k components to the scalar equations (Eq. 21–10) yields

(Ixx - I) vx - Ixy vy - Ixz vz = 0

-Ixx vx + (Ixy - I) vy - Iyz vz = 0

-Izx vz - Izy vy + (Izz - I) vz = 0

Solution for vx, vy, and vz requires

3

(Ixx - I)

-Iyx

-Izx

-Ixy

(Iyy - I)

-Izy

-Ixz

-Iyz 3 = 0

(Izz - I)

Expanding

I3 - (Ixx + Iyy + Izz)I2 + A Ixx Iyy + Iyy Izz + Izz Ixx - I2xy - I2yz - I2zx B I

- A Ixx Iyy Izz - 2Ixy Iyz Izx - Ixx I2yz - IyyI2zx - Izz I2xy B = 0 Q.E.D.

939

y

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•21–21. Show that if the angular momentum of a body is

determined with respect to an arbitrary point A, then H A

can be expressed by Eq. 21–9. This requires substituting

R A = R G + R G>A into Eq. 21–6 and expanding, noting

that 1 R G dm = 0 by definition of the mass center and

vG = vA + V : R G>A.

Z

z

RG/A

G

RG

RA

P

y

A

Y

x

HA = a

Lm

rA dmb * vA +

= a

Lm

(rG + rG>A) dmb * vA +

= a

Lm

rG dmb * vA + (rG>A * vA)

+ a

Since

Lm

Lm

Lm

X

rA * (v * rA)dm

(rG + rG>A) * C v * rG + rG>A) D dm

Lm

dm +

rG * (v * rG) dm

Lm

Lm

rGdmb * (v * rG>A) + rG>A * a v *

rG dm = 0 and from Eq. 21–8 HG =

Lm

Lm

rG dm b + rG>A * (v * rG>A)

rG * (v * rG)dm

HA = (rG>A * vA)m + HG + rG>A * (v * rG>A)m

= rG>A * (vA + (v * rG>A))m + HG

= (rG>A * mvG) + HG

Q.E.D.

940

Lm

dm

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z

21–22. The 4-lb rod AB is attached to the disk and collar

using ball-and-socket joints. If the disk has a constant

angular velocity of 2 rad>s, determine the kinetic energy of

the rod when it is in the position shown. Assume the angular

velocity of the rod is directed perpendicular to the axis of

the rod.

B

1 ft

A

vA = vB + v * rA>B

i

vAi = -(1)(2)j + 3 vx

3

j

vy

-1

k

vz 3

-1

x

Expand and equate components:

vA = -vy + vz

(1)

2 = vx + 3 vz

(2)

0 = -vx - 3 vy

(3)

Also:

v # rA>B = 0

(4)

3vx - vy - vz = 0

Solving Eqs. (1)–(4):

vx = 0.1818 rad>s

vy = -0.06061 rad>s

vz = 0.6061 rad>s

vA = 0.667 ft>s

v is perpendicular to the rod.

vz = 0.1818 rad>s,

vy = -0.06061 rad>s,

vz = 0.6061 rad>s

vB = {-2j} ft>s

rA>B = {3i - 1j - 1k} ft

vG = vB + v *

vG

i

13

= -2j +

0.1818

2

3

rA>B

2

j

-0.06061

-1

2 rad/s

k

0.6061 3

-1

vG = {0.333i - 1j} ft>s

vG = 2(0.333)2 + (-1)2 = 1.054 ft>s

v = 2(0.1818)2 + (-0.06061)2 + (0.6061)2 = 0.6356 rad>s

1

4

1

4

1

b (1.054)2 + a b c

a

b(3.3166)2 d(0.6356)2

T = a ba

2 32.2

2 12 32.2

T = 0.0920 ft # lb

Ans.

941

3 ft

y

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z

21–23. Determine the angular momentum of rod AB in

Prob. 21–22 about its mass center at the instant shown.

Assume the angular velocity of the rod is directed

perpendicular to the axis of the rod.

B

2 rad/s

1 ft

A

x

vA = vB + v * rA>B

i

vAi = -(1)(2)j + 3 vx

3

j

vy

-1

k

vz 3

-1

Expand and equate components:

vA = -vy + vz

(1)

2 = vx + 3 vz

(2)

0 = -vx - 3 vy

(3)

Also:

v # rA>B = 0

(4)

3vx - vy - vz = 0

Solving Eqs. (1)–(4):

vx = 0.1818 rad>s

vy = -0.06061 rad>s

vz = 0.6061 rad>s

vA = 0.667 ft>s

v is perpendicular to the rod.

rA>B = 2(3)2 + (-1)2 + (-1)2 = 3.3166 ft

IG = a

1

4

ba

b (3.3166)2 = 0.1139 slug # ft2

12 32.2

HG = IG v = 0.1139 (0.1818i - 0.06061j + 0.6061k)

HG = {0.0207i - 0.00690j + 0.0690k) slug # ft2>s

Ans.

942

3 ft

y

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z

*21–24. The uniform thin plate has a mass of 15 kg. Just

before its corner A strikes the hook, it is falling with a

velocity of vG = 5-5k6 m>s with no rotational motion.

Determine its angular velocity immediately after corner A

strikes the hook without rebounding.

vG

A

G

x 200 mm

Referring to Fig. a, the mass moments of inertia of the plate about the x, y, and z

axes are

1

(15) A 0.42 B + 15 A 0.22 + 02 B = 0.8 kg # m2

Ix = Ix¿ + m A yG 2 + zG 2 B =

12

Iy = Iy¿ + m A xG 2 + zG 2 B =

1

(15) A 0.62 B + 15c(-0.3)2 + 02 d = 1.8 kg # m2

12

Iz = Iz¿ + m A xG 2 + yG 2 B =

Due to symmetry, Ix¿y¿

1

(15) A 0.42 + 0.62 B + 15c( -0.3)2 + 0.22 d = 2.6 kg # m2

12

= Iy¿z¿ = Ix¿z¿ = 0. Thus,

Ixy = Ix¿y¿ + mxGyG = 0 + 15(-0.3)(0.2) = -0.9 kg # m2

Iyz = Iy¿z¿ + myGzG = 0 + 15(0.2)(0) = 0

Ixz = Ix¿z¿ + mxGzG = 0 + 15( -0.3)(0) = 0

Since the plate falls without rotational motion just before the impact, its angular

momentum about point A is

(HA)1 = rG>A * mvG = (-0.3i + 0.2j) * 15(-5k)

= [-15i - 22.5j] kg # m2>s

Since the plate rotates about point A just after impact, the components of its

angular momentum at this instant can be determined from

C (HA)2 D x = Ixvx - Ixyvy - Ixz vz

= 0.8vx - (-0.9)vy - 0(vz)

= 0.8vx + 0.9vy

C (HA)2 D y = -Ixyvx + Iyvy - Iyz vz

= -(-0.9)vx + 1.8vy - 0(vz)

= 0.9vx + 1.8vy

C (HA)2 D z = -Ixz vx + Iyzvy - Iz vz

= 0(vx) - 0(vy) + 2.6vz

= 2.6vz

Thus,

(HA)2 = (0.8vx + 0.9vy)i + (0.9vx + 1.8vy)j + 2.6vz k

Referring to the free-body diagram of the plate shown in Fig. b, the weight W is a

nonimpulsive force and the impulsive force FA acts through point A. Therefore,

angular momentum of the plate is conserved about point A. Thus,

(HA)1 = (HA)2

-15i - 22.5j = (0.8vx + 0.9vy)i + (0.9vx + 1.8vy)j + 2.6vz k

Equating the i, j, and k components,

-15 = 0.8vx + 0.9vy

(1)

-22.5 = 0.9vx + 1.8vy

(2)

0 = 2.6vz

(3)

Solving Eqs. (1) through (3),

vx = -10.71 rad>s

vy = -7.143 rad>s

vz = 0

Thus,

v = [-10.7i - 7.14j] rad>s

Ans.

943

300 mm

300 mm

200 mm

y

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z

•21–25. The 5-kg disk is connected to the 3-kg slender

rod. If the assembly is attached to a ball-and-socket joint at

A and the 5-N # m couple moment is applied, determine the

angular velocity of the rod about the z axis after the

assembly has made two revolutions about the z axis starting

from rest. The disk rolls without slipping.

Mϭ5Nиm

A

x

Ix = Iz =

Iv =

1

1

(5)(0.2)2 + 5(1.5)2 + (3)(1.5)2 = 13.55

4

3

1

(5)(0.2)2 = 0.100

2

v = -vy j¿ + vz k = -vy¿ j¿ + vz sin 7.595°j¿ + vz cos 7.595°k¿

= (0.13216vz - vy¿)j¿ + 0.99123 vz k¿

Since points A and C have zero velocity,

vC = vA + v * rC>A

0 = 0 + C (0.13216 vz - vy¿)j¿ + 0.99123vz k¿ D * (1.5j¿ - 0.2k¿)

0 = -1.48684vz - 0.026433 vz + 0.2 vy¿

vy¿ = 7.5664 vz

Thus,

v = -7.4342 vz j¿ + 0.99123 vz k¿

T1 + ©U1 - 2 = T2

0 + 5(2p) (2) = 0 +

1

1

(0.100)(-7.4342 vz)2 + (13.55)(0.99123 vz)2

2

2

vz = 2 58 rad>s

Ans.

944

1.5 m

B

0.2 m

y

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z

21–26. The 5-kg disk is connected to the 3-kg slender rod.

If the assembly is attached to a ball-and-socket joint at A

and the 5-N # m couple moment gives it an angular velocity

about the z axis of vz = 2 rad>s, determine the magnitude

of the angular momentum of the assembly about A.

Mϭ5Nиm

A

x

Ix = Iz =

Ix =

1

1

(5)(0.2)2 + 5(1.5)2 + (3)(1.5)2 = 13.55

4

3

1

(5)(0.2)2 = 0.100

2

v = -vy j¿ + vz k = -vy¿ j¿ + vz sin 7.595°j¿ + vz cos 7.595°k¿

= (0.13216vz - vy )j¿ + 0.99123 vz k¿

Since points A and C have zero velocity,

vC = vA + v * rC>A

0 = 0 + C (0.13216 vz - vy ) j¿ + 0.99123 vz k¿ D * (1.5j¿ - 0.2k¿)

0 = -1 48684vz - 0.26433vz + 0.2vy

vy¿ = -7.5664 vz

Thus,

v = -7.4342 vz j¿ + 0.99123 vz k¿

Since vz = 2 rad>s

v = -14.868j¿ + 1 9825k¿

So that,

HA = Ix¿ vx i¿ + Iy¿ vy¿ j + Iz¿ vz¿ k¿ = 0 + 0.100(-14.868)j¿ + 13.55(1.9825) k¿

= -1.4868j¿ + 26.862k¿

HA = 2(-1.4868)2 + (26.862)2 = 26.9 kg # m2>s

Ans.

945

1.5 m

B

0.2 m

y

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21–27. The space capsule has a mass of 5 Mg and the

radii of gyration are kx = kz = 1.30 m and ky = 0.45 m .

If it travels with a velocity vG = 5400j + 200k6 m>s ,

compute its angular velocity just after it is struck by a

meteoroid having a mass of 0.80 kg and a velocity

vm = 5-300i + 200j - 150k6 m>s . Assume that the

meteoroid embeds itself into the capsule at point A and

that the capsule initially has no angular velocity.

vG

G

x

Conservation of Angular Momentum: The angular momentum is conserved about

the center of mass of the space capsule G. Neglect the mass of the meteroid after the

impact.

(HG)1 = (HG)2

rGA * mm vm = IG v

(0.8i + 3.2j + 0.9k) * 0.8(-300i + 200j - 150k)

= 5000 A 1.302 B vx i + 5000 A 0.452 B vy j + 5000 A 1.302 B vz k

-528i - 120j + 896k = 8450vx i + 1012.5vy j + 8450 vz k

Equating i, j and k components, we have

-528 = 8450vx

-120 = 1012.5vy

896 = 8450vz

z

vm

vx = -0.06249 rad>s

vy = -0.11852 rad>s

vz = 0.1060 rad>s

Thus,

v = {-0.0625i - 0.119j + 0.106k} rad>s

Ans.

946

A (0.8 m, 3.2 m, 0.9 m)

y

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z

*21–28. Each of the two disks has a weight of 10 lb. The

axle AB weighs 3 lb. If the assembly rotates about the z

axis at vz = 6 rad>s, determine its angular momentum

about the z axis and its kinetic energy. The disks roll

without slipping.

2 ft

2 ft

B

1 ft

A

x

1

6

=

vx

2

vx = 12 rad>s

vA = {-12i} rad>s

Hz = c

vB = {12i} rad>s

1 10

1 10

a

b (1)2 d(12i) + c a

b (1)2 d( -12i)

2 32.2

2 32.2

+ 0 + b 2c

3

1 10

10

1

a

b (1)2 +

(2)2 d(6) +

a

b(4)2(6) r k

4 32.2

32.2

12 32.2

Hz = {16.6k} slug # ft2>s

T =

=

Ans.

1

1

1

I v2 + Iy v2y + Iz v2z

2 x x

2

2

1

1 10

c2 a a

b (1)2 b d(12)2 + 0

2

2 32.2

+

1 10

10

1

3

1

b (1)2 +

(2)2 d +

a

b(4)2 r (6)2

b 2c a

2

4 32.2

32.2

12 32.2

= 72.1 lb # ft

Ans.

947

1 ft

vz ϭ 6 rad/s

y

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z

•21–29. The 10-kg circular disk spins about its axle with a

constant angular velocity of v1 = 15 rad>s. Simultaneously,

arm OB and shaft OA rotate about their axes with constant

angular velocities of v2 = 0 and v3 = 6 rad>s, respectively.

Determine the angular momentum of the disk about point O,

and its kinetic energy.

O

The mass moments of inertia of the disk about the centroidal x¿ , y¿ , and z¿ axes,

Fig. a, are

x

Ix¿ = Iy¿ =

1

1

mr2 = (10) A 0.152 B = 0.05625 kg # m2

4

4

Ix¿y¿ = Iy¿z¿ = Ix¿z¿ = 0

Here, the angular velocity of the disk can be determined from the vector addition of

v1 and v3. Thus,

v = v1 + v2 = [6i + 15k] rad>s

The angular momentum of the disk about its mass center G can be obtained by

applying

Hx = Ix¿vx = 0.05625(6) = 0.3375 kg # m2>s

Hy = Iy¿vy = 0.05625(0) = 0

Hz = Iz¿vz = 0.1125(15) = 1.6875 kg # m2>s

Thus,

HG = [0.3375i + 1.6875k] kg # m2>s

Since the mass center G rotates about the x axis with a constant angular velocity of

v3 = [6i] rad>s, its velocity is

vG = v3 * rC>O = (6i) * (0.6j) = [3.6k] m>s

Since the disk does not rotate about a fixed point O, its angular momentum must be

determined from

HO = rC>O * mvC + HG

= (0.6j) * 10(3.6k) + (0.3375i + 1.6875k)

= [21.9375i + 1.6875k] kg # m2>s

Ans.

The kinetic energy of the disk is therefore

T =

=

1 #

v HO

2

1

(6i + 15k) # (21.9375i + 1.6875k)

2

= 78.5 J

Ans.

948

V2

B

150 mm

y

Due to symmetry, the products of inertia of the disk with respect to its centroidal

planes are equal to zero.

= [21.9i + 1.69k] kg # m2>s

V3

V1

1

1

mr2 = (10) A 0.152 B = 0.1125 kg # m2

2

2

Iz¿ =

A

600 mm

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z

21–30. The 10-kg circular disk spins about its axle with a

constant angular velocity of v1 = 15 rad>s. Simultaneously,

arm OB and shaft OA rotate about their axes with constant

angular velocities of v2 = 10 rad>s and v3 = 6 rad>s,

respectively. Determine the angular momentum of the disk

about point O, and its kinetic energy.

O

The mass moments of inertia of the disk about the centroidal x¿ , y¿ , and z¿ axes.

Fig. a, are

x

Ix¿ = Iy¿ =

Iz¿ =

1

1

mr2 = (10) A 0.152 B = 0.1125 kg # m2

2

2

Ix¿y¿ = Iy¿z¿ = Ix¿z¿ = 0

Here, the angular velocity of the disk can be determined from the vector addition of

v1, v2, and v3. Thus,

v = v1 + v2 + v3 = [6i + 10j + 15k] rad>s

The angular momentum of the disk about its mass center G can be obtained by

applying

Hx = Ix¿vx = 0.05625(6) = 0.3375 kg # m2

Hy = Iy¿vy = 0.05625(10) = 0.5625 kg # m2

Hz = Iz¿vz = 0.1125(15) = 1.6875 kg # m2

Thus,

HG = [0.3375i + 0.5625j + 1.6875k] kg # m2

Since the mass center G rotates about the fixed point O with an angular velocity of

Æ = v2 + v3 = [6i + 10j], its velocity is

vG = Æ * rG>O = (6i + 10j) * (0.6j) = [3.6k] m>s

Since the disk does not rotate about a fixed point O, its angular momentum must be

determined from

HO = rC>O * mvG + HG

= (0.6j) * 10(3.6k) + (0.3375i + 0.5625j + 1.6875k)

= [21.9375i + 0.5625j + 1.6875k] kg # m2>s

Ans.

The kinetic energy of the disk is therefore

=

1 #

v HO

2

1

(6i + 10j + 15k) # (21.9375i + 0.5625j + 1.6875k)

2

= 81.3J

Ans.

949

V2

B

150 mm

y

Due to symmetry, the products of inertia of the disk with respect to its centroidal

planes are equal to zero.

T =

V3

V1

1

1

mr2 = (10) A 0.152 B = 0.05625 kg # m2

4

4

= [21.9i + 0.5625j + 1.69k] kg # m2>s

A

600 mm

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Page 925

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently

exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

•21–1. Show that the sum of the moments of inertia of a

body, Ixx + Iyy + Izz, is independent of the orientation of

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

origin.

Ixx + Iyy + Izz =

Lm

= 2

(y2 + z2)dm +

Lm

(x2 + z2)dm +

(x2 + y2)dm

Lm

Lm

(x2 + y2 + z2)dm

However, x2 + y2 + z2 = r2, where r is the distance from the origin O to dm. Since

ƒ r ƒ is constant, it does not depend on the orientation of the x, y, z axis. Consequently,

Ixx + Iyy + Izz is also indepenent of the orientation of the x, y, z axis.

Q.E.D.

925

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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

21–2. Determine the moment of inertia of the cone with

respect to a vertical y axis that passes through the cone’s

center of mass.What is the moment of inertia about a parallel

axis y¿ that passes through the diameter of the base of the

cone? The cone has a mass m.

–y

y

y¿

a

x

The mass of the differential element is dm = rdV = r(py2) dx =

dIv =

=

Iv =

L

h2

h

x2dx.

1

dmy2 + dmx2

4

r pa2

a 2

1 rpa2 2

B 2 x dx R a xb + ¢ 2 x2 ≤ x2 dx

4

h

h

h

r pa2

=

rpa2

(4h2 + a2) x4 dx

4h4

dIv =

rpa2

4

4h

h

(4h2 + a2)

x4dx =

r pa2h

(4h2 + a2)

20

x2 dx =

r pa2h

3

L0

However,

m =

Lm

dm =

h

r pa2

2

h

L0

Hence,

Iy =

3m

(4h2 + a2)

20

Using the parallel axis theorem:

Iv = Iv + md2

3h 2

3m

(4h2 + a2) = Iv + ma b

20

4

Iv =

3m 2

(h + 4a2)

80

Ans.

Iv = Iy + md2

=

h 2

3m 2

(h + 4a2) + ma b

80

4

=

m

(2h2 + 3a2)

20

Ans.

926

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Page 927

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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

z

21–3. Determine the moments of inertia Ix and Iy of the

paraboloid of revolution. The mass of the paraboloid is m.

a

m = r

L0

a

pz2 dy = rp

L0

a

r2

r2

bydy = rra b a

a

2

a

Iv =

2

r

z2 ϭ — y

a

r

a

1

1

1

r4

r4

z4dy = rp a 2 b

v2dy = rp a ba

dm z2 = rp

2

2

6

a L0

Lm 2

L0

y

x

a

Thus,

Ix =

1

mr2

3

a

Ix =

=

Ans.

a

1

1

a dm z2 + dm y2 b = rp

z4 dy + r

pz2 y2 dy

4

L0

Lm 4

L0

a

a

rpr 2a 3

rpr 4a

1

r4

r2

1

1

r pa 2 b

y2dy + rpa b

y3dy =

+

= mr 2 + ma 2

a L0

4

12

4

6

2

a L0

Ix =

m 2

(r + 3a2)

6

Ans.

z

*21–4. Determine by direct integration the product of

inertia Iyz for the homogeneous prism. The density of the

material is r. Express the result in terms of the total mass m

of the prism.

a

a

The mass of the differential element is dm = rdV = rhxdy = rh(a - y)dy.

h

a

ra2h

m =

dm = rh

(a - y)dy =

2

Lm

L0

x

Using the parallel axis theorem:

dIyz = (dIy¿z¿)G + dmyGzG

h

= 0 + (rhxdy) (y) a b

2

Iyz =

=

rh2

xydy

2

=

rh2

(ay - y2) dy

2

a

rh2

ra3h2

1 ra2h

m

= a

b(ah) =

ah

(ay - y2) dy =

2 L0

12

6

2

6

927

Ans.

y

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Page 928

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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

z

•21–5. Determine by direct integration the product of

inertia Ixy for the homogeneous prism. The density of the

material is r. Express the result in terms of the total mass m

of the prism.

a

a

The mass of the differential element is dm = rdV = rhxdy = rh(a - y)dy.

h

a

y

ra2h

m =

dm = rh

(a - y)dy =

2

Lm

L0

x

Using the parallel axis theorem:

dIxy = (dIx¿y¿)G + dmxGyG

x

= 0 + (rhxdy)a b(y)

2

=

rh2 2

x ydy

2

=

rh2 3

(y - 2ay2 + a 2 y) dy

2

Ixy =

a

rh

(y3 - 2ay2 + a 2 y) dy

2 L0

=

ra 4h

1 ra 2h 2

m 2

=

a

ba =

a

24

12

2

12

Ans.

z

21–6. Determine the product of inertia Ixy for the

homogeneous tetrahedron. The density of the material is r.

Express the result in terms of the total mass m of the solid.

Suggestion: Use a triangular element of thickness dz and

then express dIxy in terms of the size and mass of the

element using the result of Prob. 21–5.

a

y

r

1

dm = r dV = rc (a - z)(a - z) ddz = (a - z)2 dz

2

2

m =

a

a

r

ra3

(a2 - 2az + z2)dz =

2 L0

6

a

x

From Prob. 21–5 the product of inertia of a triangular prism with respect to the xz

ra 4h

rdz

(a - z)4. Hence,

and yz planes is Ixy =

. For the element above, dIxy =

24

24

Ixy =

a

r

(a4 - 4a3z + 6z 2a 2 - 4az 3 + z 4)dz

24 L0

Ixy =

ra 5

120

Ixy =

ma 2

20

or,

Ans.

928

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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

21–7. Determine the moments of inertia for the

homogeneous cylinder of mass m about the x¿ , y¿ , z¿ axes.

r

z¿

x¿

y

r

z

x

Due to symmetry

Ixy = Iyz = Izx = 0

Iy = Ix =

1

r 2

7mr 2

m(3r 2 + r 2) + ma b =

12

2

12

Iz =

1 2

mr

2

For x¿ ,

ux = cos 135° = -

1

22

uy = cos 90° = 0,

,

1

uz = cos 135° = -

22

Ix = Ix u2x + Iy u2y + Iz u2z - 2Ixy ux uy - 2Iyz uy uz - 2Izx uz ux

=

7mr 2

1 2

1

1 2

b + 0 + mr 2 a b - 0-0-0

¢12

2

22

22

=

13

mr 2

24

Ans.

For y¿ ,

Iy¿ = Iy =

7mr 2

12

Ans.

For z¿ ,

ux = cos 135° = -

1

22

,

uy = cos 90° = 0,

uz = cos 45° =

1

22

Iz¿ = Ix u2x + Iy u2y + Iz u2z - 2Ixy ux uy - 2Iyz uy uz - 2Izx uz ux

=

7mr 2

1

1 2

1 2

b + 0 + mr 2 a b - 0-0-0

¢12

2

22

22

=

13

mr 2

24

Ans.

929

y¿

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Page 930

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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

z

*21–8. Determine the product of inertia Ixy of the

homogeneous triangular block. The material has a

density of r . Express the result in terms of the total mass

m of the block.

b

h

The mass of the differential rectangular volume element shown in Fig. a is

x

dm = rdV = rbzdy. Using the parallel - plane theorem,

dIxy = dI x¿y¿ + dmxGyG

b

= 0 + [rbzdy]a by

2

=

However, z =

rb2

zydy

2

h

(a - y). Then

a

dIxy =

rb2 h

rb2h

c (a - y)y ddy =

A ay - y2 B dy

2 a

2a

Thus,

Ixy =

L

dIxy =

a

rb 2h

A ay - y2 B dy

2a L0

=

y3 a

rb2h ay2

b2

¢

2a

2

3 0

=

1

ra 2b 2 h

12

1

1

However, m = rV = a ahb b = rabh. Then

2

2

Ixy =

1

m

1

ra 2b2h§

¥ = mab

12

1

6

rabh

2

Ans.

930

y

a

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Page 931

© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently

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z

•21–9. The slender rod has a mass per unit length of

6 kg>m. Determine its moments and products of inertia

with respect to the x, y, z axes.

2m

x

The mass of segments (1), (2), and (3) shown in Fig. a is

m1 = m2 = m3 = 6(2) = 12 kg. The mass moments of inertia of the bent rod about

the x, y, and z axes are

1

1

(12) A 22 B + 12 A 12 + 02 B d + c (12) A 22 B + 12 C 22 +

12

12

= 80 kg # m2

A -1 B 2 D d

Ans.

Iy = ©Iy¿ + m A xG 2 + zG 2 B

= c

1

1

(12) A 22 B + 12 A 12 + 02 B d + c0 + 12 A 22 + 02 B d + c (12) A 22 B + 12 C 22 +

12

12

= 128 kg # m2

A -1 B 2 D d

Ans.

Iz = ©Iz¿ + m A xG 2 + yG 2 B

= c

1

1

(12) A 22 B + 12 A 12 + 02 B d + c (12) A 22 B + 12 A 22 + 12 B d + c0 + 12 A 22 + 22 B d

12

2

= 176 kg # m2

Ans.

Due to symmetry, the products of inertia of segments (1), (2), and (3) with respect to

their centroidal planes are equal to zero. Thus,

Ixy = ©Ix¿y¿ + mxGyG

= C 0 + 12(1)(0) D + C 0 + 12(2)(1) D + C 0 + 12(2)(2) D

= 72 kg # m2

Ans.

Iyz = ©Iy¿z¿ + myGzG

= C 0 + 12(0)(0) D + C 0 + 12(1)(0) D + C 0 + 12(2)(-1) D

= -24 kg # m2

Ans.

Ixz = ©Ix¿z¿ + mxGzG

= C 0 + 12(1)(0) D + C 0 + 12(2)(0) D + C 0 + 12(2)(-1) D

= -24 kg # m2

Ans.

931

y

2m

2m

Ix = ©Ix¿ + m A yG 2 + zG 2 B

= A0 + 0B + c

O

91962_12_s21_p0925-0987

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Page 932

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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

z

21–10. Determine the products of inertia Ixy, Iyz, and Ixz

of the homogeneous solid. The material has a density of

7.85 Mg>m3.

200 mm

200 mm

100 mm

x

The masses of segments (1) and (2) shown in Fig. a are m1 = r V1

= 7850(0.4)(0.4)(0.1) = 125.6 kg and m2 = r V2 = 7850(0.2)(0.2)(0.1) = 31.4 kg.

Due

to

symmetry

for

segment

(1)

and

Ix¿y¿ = Iy¿z¿ = Ix¿z¿ = 0

Ix–y– = Iy–z– = Ix–z– = 0 for segment (2). Since segment (2) is a hole, it should be

considered as a negative segment. Thus

Ixy = ©Ix¿y¿ + mxGyG

= C 0 + 125.6(0.2)(0.2) D - C 0 + 31.4(0.3)(0.1) D

= 4.08 kg # m2

Ans.

Iyz = ©Iy¿z¿ + myGzG

= C 0 + 125.6(0.2)(0.05) D - C 0 + 31.4(0.1)(0.05) D

= 1.10 kg # m2

Ans.

Ixz = ©Ix¿z¿ + mxGzG

= C 0 + 125.6(0.2)(0.05) D - C 0 + 31.4(0.3)(0.05) D

= 0.785 kg # m2

Ans.

932

200 mm

200 mm

y

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z

21–11. The assembly consists of two thin plates A and B

which have a mass of 3 kg each and a thin plate C which has

a mass of 4.5 kg. Determine the moments of inertia Ix , Iy

and Iz .

B

0.4 m

60Њ

A

C

0.4 m

0.3 m

x

Ix¿ = Iz¿ =

Iy¿ =

1

(3)(0.4)2 = 0.04 kg # m2

12

1

(3)[(0.4)2 + (0.4)2] = 0.08 kg # m2

12

Ix¿y¿ = Iz¿y¿ = Iz¿x¿ = 0

For zG,

ux¿ = 0

uy¿ = cos 60° = 0.50

uz¿ = cos 30° = 0.8660

IzG = 0 + 0.08(0.5)2 + 0.04(0.866)2 - 0 - 0 - 0

= 0.05 kg # m2

IxG = Ix¿ = 0.04 kg # m2

For yG,

ux¿ = 0

uy¿ = cos 30° = 0.866

uz¿ = cos 120° = - 0.50

IyG = 0 + 0.08(0.866)2 + 0.04(-0.5)2 - 0 - 0 - 0

= 0.07 kg # m2

Ix =

1

(4.5)(0.6)2 + 2[0.04 + 3{(0.3 + 0.1)2 + (0.1732)2}]

12

Ix = 1.36 kg # m2

Iy =

Ans.

1

(4.5)(0.4)2 + 2[0.07 + 3(0.1732)2]

12

Iy = 0.380 kg # m2

Iz =

Ans.

1

(4.5)[(0.6)2 + (0.4)2] + 2[0.05 + 3(0.3 + 0.1)2]

12

Iz = 1.26 kg # m2

Ans.

933

60Њ

0.3 m

0.2 m

0.2 m

y

91962_12_s21_p0925-0987

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Page 934

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exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

z

*21–12. Determine the products of inertia Ixy, Iyz, and Ixz,

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

50 kg>m2.

The masses of segments (1) and (2) shown in Fig. a are m1 = 50(0.4)(0.4) = 8 kg

and m2 = 50(0.4)(0.2) = 4 kg. Due to symmetry Ix¿y¿ = Iy¿z¿ = Ix¿z¿ = 0 for

segment (1) and Ix–y– = Iy–z– = Ix–z– = 0 for segment (2).

400 mm

200 mm

Ixy = ©Ix¿y¿ + mxGyG

= C 0 + 8(0.2)(0.2) D + C 0 + 4(0)(0.2) D

x

= 0.32 kg # m2

y

400 mm

Ans.

Iyz = ©Iy¿z¿ + myGzG

= C 0 + 8(0.2)(0) D + C 0 + 4(0.2)(0.1) D

= 0.08 kg # m2

Ans.

Ixz = ©Ix¿z¿ + mxGzG

= C 0 + 8(0.2)(0) D + C 0 + 4(0)(0.1) D

= 0

Ans.

z¿

•21–13. The bent rod has a weight of 1.5 lb>ft. Locate the

center of gravity G(x, y) and determine the principal

moments of inertia Ix¿ , Iy¿ , and Iz¿ of the rod with respect

to the x¿, y¿, z¿ axes.

z

1 ft

Due to symmetry

1 ft

A

_

y

x¿

x

y = 0.5 ft

Ans.

x =

Ans.

(-1)(1.5)(1) + 2 C ( -0.5)(1.5)(1) D

©x W

=

= -0.667 ft

©w

3 C 1.5(1) D

Ix¿ = 2c a

1.5

1

1.5

b(0.5)2 d +

a

b(1)2

32.2

12 32.2

= 0.0272 slug # ft2

Iy¿ = 2 c

Ans.

1 1.5

1.5

1.5

a

b (1)2 + a

b (0.667 - 0.5)2 d + a

b(1 - 0.667)2

12 32.2

32.2

32.2

= 0.0155 slug # ft2

Iz¿ = 2 c

+

Ans.

1 1.5

1.5

a

b (1)2 + a

b(0.52 + 0.16672) d

12 32.2

32.2

1

1.5

1.5

a

b(1)2 + a

b(0.3333)2

12 32.2

32.2

= 0.0427 slug # ft2

Ans.

934

G

y¿

_

x

y

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Page 935

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z

21–14. The assembly consists of a 10-lb slender rod and a

30-lb thin circular disk. Determine its moment of inertia

about the y¿ axis.

2 ft

The mass of moment inertia of the assembly about the x, y, and z axes are

1 ft

1 30

30

1 10

10

b A 12 B +

b A 22 B +

Ix = Iz = c a

A 22 B d + c a

A 12 B d

4 32.2

32.2

12 32.2

32.2

A

= 4.3737 slug # ft

Iy =

y

B

2

x

1 30

a

b A 12 B + 0 = 0.4658 slug # ft2

2 32.2

y¿

Due to symmetry, Ixy = Iyz = Ixz = 0. From the geometry shown in Fig. a,

1

u = tan-1 a b = 26.57°. Thus, the direction of the y¿ axis is defined by the unit

2

vector

u = cos 26.57°j - sin 26.57°k = 0.8944j - 0.4472k

Thus,

ux = 0

uy = 0.8944

uz = -0.4472

Then

Iy¿ = Ix ux 2 + Iy uy 2 + Iz uz 2 - 2Ixy ux uy - 2Iyz uy uz - 2Ixzux uz

= 4.3737(0) + 0.4658(0.8944)2 + 4.3737(-0.4472)2 - 0 - 0 - 0

= 1.25 slug # ft2

Ans.

z

21–15. The top consists of a cone having a mass of 0.7 kg

and a hemisphere of mass 0.2 kg. Determine the moment of

inertia Iz when the top is in the position shown.

Ix¿ = Iy¿ =

30 mm

2

3

3

(0.7) C (4)(0.3)2 + (0.1)2 D + (0.7)c (0.1) d

80

4

+ a

100 mm

30 mm

2

83

3

b (0.2)(0.03)2 + (0.2)c (0.03) + (0.1) d = 6.816 A 10-3 B kg # m2

320

8

y

3

2

Iz¿ = a b(0.7)(0.03)2 + a b (0.2)(0.03)2

10

5

45Њ

Iz = 0.261 A 10-3 B kg # m2

ux = cos 90° = 0,

Iz =

Ix¿ u2x¿

+

Iy¿ u2y¿

uy¿ = cos 45° = 0.7071,

+

Iz¿ u2z¿

uz¿ = cos 45° = 0.7071

x

- 2Ix¿y¿ux¿ uy¿ - 2Iy¿z¿ uy¿ uz¿ - 2Ix¿z¿ ux¿ uz¿

= 0 + 6.816 A 10-3 B (0.7071)2 + (0.261) A 10-3 B (0.7071)2 - 0 - 0 - 0

Iz = 3.54 A 10-3 B kg # m2

Ans.

935

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Page 936

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z

*21–16. Determine the products of inertia Ixy, Iyz, and

Ixz of the thin plate. The material has a mass per unit area

of 50 kg>m2.

200 mm

200 mm

200 mm

The masses of segments (1), (2), and (3) shown in Fig. a are m1 = m2

= 50(0.4)(0.4) = 8 kg and m3 = 50cp(0.1)2 d = 0.5p kg.

200 mm

Due to symmetry Ix¿y¿ = Iy¿z¿ = Ix¿z¿ = 0 for segment (1), Ix–y– = Iy–z– = Ix–z– = 0

for segment (2), and Ix‡y‡ = Iy‡z‡ = Ix‡z‡ = 0 for segment (3). Since segment (3) is

a hole, it should be considered as a negative segment. Thus

Ixy = ©Ix¿y¿ + mxGyG

x

= C 0 + 8(0.2)(0.2) D + C 0 + 8(0)(0.2) D - C 0 + 0.5p(0)(0.2) D

= 0.32 kg # m2

100 mm

400 mm

y

400 mm

Ans.

Iyz = ©Iy¿z¿ + myGzG

= C 0 + 8(0.2)(0) D + C 0 + 8(0.2)(0.2) D - C 0 + 0.5p(0.2)(0.2) D

= 0.257 kg # m2

Ans.

Ixz = ©Ix¿z¿ + mxGzG

= C 0 + 8(0.2)(0) D + C 0 + 8(0)(0.2) D - C 0 + 0.5p(0)(0.2) D

= 0 kg # m2

Ans.

z

•21–17. Determine the product of inertia Ixy for the bent

rod. The rod has a mass per unit length of 2 kg>m.

Product of Inertia: Applying Eq. 21–4. we have

Ixy = © A Ix¿y¿ B G + mxG yG

400 mm

= [0 + 0.4 (2) (0) (0.5)] + [0 + 0.6 (2) (0.3) (0.5)] + [0 + 0.5 (2) (0.6) (0.25)]

= 0.330 kg # m2

y

Ans.

600 mm

500 mm

x

936

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z

21–18. Determine the moments of inertia Ixx, Iyy, Izz for

the bent rod. The rod has a mass per unit length of 2 kg>m.

400 mm

y

600 mm

Moments of Inertia: Applying Eq. 21–3, we have

500 mm

Ixx = ©(Ix¿x¿)G + m(y2G + z2G)

= c

x

1

(0.4) (2) A 0.42 B + 0.4 (2) A 0.52 + 0.22 B d

12

+ C 0 + 0.6 (2) A 0.52 + 02 B D

+ c

1

(0.5) (2) A 0.52 B + 0.5 (2) A 0.252 + 02 B d

12

= 0.626 kg # m2

Ans.

Ixy = ©(Iy¿y¿)G + m(x2G + z2G)

= c

1

(0.4) (2) A 0.42 B + 0.4 (2) A 02 + 0.22 B d

12

+ c

1

(0.6) (2) A 0.62 B + 0.6 (2) A 0.32 + 02 B d

12

+ C 0 + 0.5(2) A 0.62 + 02 B D

= 0.547 kg # m2

Ans.

Izz = ©(Iz¿z¿)G + m(x2G + y2G)

= C 0 + 0.4 (2) A 02 + 0.52 B D

+ c

1

(0.6) (2) A 0.62 B + 0.6 (2) A 0.32 + 0.52 B d

12

+ c

1

(0.5) (2) A 0.52 B + 0.5 (2) A 0.62 + 0.252 B d

12

= 1.09 kg # m2

Ans.

937

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z

21–19. Determine the moment of inertia of the rod-andthin-ring assembly about the z axis. The rods and ring have

a mass per unit length of 2 kg>m.

A

O

500 mm

400 mm

For the rod,

D

ux¿ = 0.6,

Ix = Iy =

uy¿ = 0,

uz¿ = 0.8

B

1

[(0.5)(2)](0.5)2 = 0.08333 kg # m2

3

120Њ

120Њ

x

Ix¿ = 0

Ix¿y¿ = Iy¿z¿ = Ix¿z¿ = 0

From Eq. 21–5,

Iz = 0.08333(0.6)2 + 0 + 0 - 0 - 0 - 0

Iz = 0.03 kg # m2

For the ring,

The radius is r = 0.3 m

Thus,

Iz = mR2 = [2 (2p)(0.3)](0.3)2 = 0.3393 kg # m2

Thus the moment of inertia of the assembly is

Iz = 3(0.03) + 0.339 = 0.429 kg # m2

Ans.

938

y

120Њ

C

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z

*21–20. If a body contains no planes of symmetry, the

principal moments of inertia can be determined

mathematically. To show how this is done, consider the rigid

body which is spinning with an angular velocity V , directed

along one of its principal axes of inertia. If the principal

moment of inertia about this axis is I, the angular momentum

can be expressed as H = IV = Ivx i + Ivy j + Ivz k. The

components of H may also be expressed by Eqs. 21–10,

where the inertia tensor is assumed to be known. Equate the

i, j, and k components of both expressions for H and consider

vx, vy, and vz to be unknown. The solution of these three

equations is obtained provided the determinant of the

coefficients is zero. Show that this determinant, when

expanded, yields the cubic equation

V

O

x

I 3 - (Ixx + Iyy + Izz)I 2

+ (IxxIyy + IyyIzz + IzzIxx - I2xy - I2yz - I2zx)I

- (IxxIyyIzz - 2IxyIyzIzx - IxxI 2yz

- IyyI 2zx - IzzI 2xy) = 0

The three positive roots of I, obtained from the solution of

this equation, represent the principal moments of inertia

Ix, Iy, and Iz.

H = Iv = Ivx i + Ivy j + Ivz k

Equating the i, j, k components to the scalar equations (Eq. 21–10) yields

(Ixx - I) vx - Ixy vy - Ixz vz = 0

-Ixx vx + (Ixy - I) vy - Iyz vz = 0

-Izx vz - Izy vy + (Izz - I) vz = 0

Solution for vx, vy, and vz requires

3

(Ixx - I)

-Iyx

-Izx

-Ixy

(Iyy - I)

-Izy

-Ixz

-Iyz 3 = 0

(Izz - I)

Expanding

I3 - (Ixx + Iyy + Izz)I2 + A Ixx Iyy + Iyy Izz + Izz Ixx - I2xy - I2yz - I2zx B I

- A Ixx Iyy Izz - 2Ixy Iyz Izx - Ixx I2yz - IyyI2zx - Izz I2xy B = 0 Q.E.D.

939

y

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•21–21. Show that if the angular momentum of a body is

determined with respect to an arbitrary point A, then H A

can be expressed by Eq. 21–9. This requires substituting

R A = R G + R G>A into Eq. 21–6 and expanding, noting

that 1 R G dm = 0 by definition of the mass center and

vG = vA + V : R G>A.

Z

z

RG/A

G

RG

RA

P

y

A

Y

x

HA = a

Lm

rA dmb * vA +

= a

Lm

(rG + rG>A) dmb * vA +

= a

Lm

rG dmb * vA + (rG>A * vA)

+ a

Since

Lm

Lm

Lm

X

rA * (v * rA)dm

(rG + rG>A) * C v * rG + rG>A) D dm

Lm

dm +

rG * (v * rG) dm

Lm

Lm

rGdmb * (v * rG>A) + rG>A * a v *

rG dm = 0 and from Eq. 21–8 HG =

Lm

Lm

rG dm b + rG>A * (v * rG>A)

rG * (v * rG)dm

HA = (rG>A * vA)m + HG + rG>A * (v * rG>A)m

= rG>A * (vA + (v * rG>A))m + HG

= (rG>A * mvG) + HG

Q.E.D.

940

Lm

dm

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z

21–22. The 4-lb rod AB is attached to the disk and collar

using ball-and-socket joints. If the disk has a constant

angular velocity of 2 rad>s, determine the kinetic energy of

the rod when it is in the position shown. Assume the angular

velocity of the rod is directed perpendicular to the axis of

the rod.

B

1 ft

A

vA = vB + v * rA>B

i

vAi = -(1)(2)j + 3 vx

3

j

vy

-1

k

vz 3

-1

x

Expand and equate components:

vA = -vy + vz

(1)

2 = vx + 3 vz

(2)

0 = -vx - 3 vy

(3)

Also:

v # rA>B = 0

(4)

3vx - vy - vz = 0

Solving Eqs. (1)–(4):

vx = 0.1818 rad>s

vy = -0.06061 rad>s

vz = 0.6061 rad>s

vA = 0.667 ft>s

v is perpendicular to the rod.

vz = 0.1818 rad>s,

vy = -0.06061 rad>s,

vz = 0.6061 rad>s

vB = {-2j} ft>s

rA>B = {3i - 1j - 1k} ft

vG = vB + v *

vG

i

13

= -2j +

0.1818

2

3

rA>B

2

j

-0.06061

-1

2 rad/s

k

0.6061 3

-1

vG = {0.333i - 1j} ft>s

vG = 2(0.333)2 + (-1)2 = 1.054 ft>s

v = 2(0.1818)2 + (-0.06061)2 + (0.6061)2 = 0.6356 rad>s

1

4

1

4

1

b (1.054)2 + a b c

a

b(3.3166)2 d(0.6356)2

T = a ba

2 32.2

2 12 32.2

T = 0.0920 ft # lb

Ans.

941

3 ft

y

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z

21–23. Determine the angular momentum of rod AB in

Prob. 21–22 about its mass center at the instant shown.

Assume the angular velocity of the rod is directed

perpendicular to the axis of the rod.

B

2 rad/s

1 ft

A

x

vA = vB + v * rA>B

i

vAi = -(1)(2)j + 3 vx

3

j

vy

-1

k

vz 3

-1

Expand and equate components:

vA = -vy + vz

(1)

2 = vx + 3 vz

(2)

0 = -vx - 3 vy

(3)

Also:

v # rA>B = 0

(4)

3vx - vy - vz = 0

Solving Eqs. (1)–(4):

vx = 0.1818 rad>s

vy = -0.06061 rad>s

vz = 0.6061 rad>s

vA = 0.667 ft>s

v is perpendicular to the rod.

rA>B = 2(3)2 + (-1)2 + (-1)2 = 3.3166 ft

IG = a

1

4

ba

b (3.3166)2 = 0.1139 slug # ft2

12 32.2

HG = IG v = 0.1139 (0.1818i - 0.06061j + 0.6061k)

HG = {0.0207i - 0.00690j + 0.0690k) slug # ft2>s

Ans.

942

3 ft

y

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z

*21–24. The uniform thin plate has a mass of 15 kg. Just

before its corner A strikes the hook, it is falling with a

velocity of vG = 5-5k6 m>s with no rotational motion.

Determine its angular velocity immediately after corner A

strikes the hook without rebounding.

vG

A

G

x 200 mm

Referring to Fig. a, the mass moments of inertia of the plate about the x, y, and z

axes are

1

(15) A 0.42 B + 15 A 0.22 + 02 B = 0.8 kg # m2

Ix = Ix¿ + m A yG 2 + zG 2 B =

12

Iy = Iy¿ + m A xG 2 + zG 2 B =

1

(15) A 0.62 B + 15c(-0.3)2 + 02 d = 1.8 kg # m2

12

Iz = Iz¿ + m A xG 2 + yG 2 B =

Due to symmetry, Ix¿y¿

1

(15) A 0.42 + 0.62 B + 15c( -0.3)2 + 0.22 d = 2.6 kg # m2

12

= Iy¿z¿ = Ix¿z¿ = 0. Thus,

Ixy = Ix¿y¿ + mxGyG = 0 + 15(-0.3)(0.2) = -0.9 kg # m2

Iyz = Iy¿z¿ + myGzG = 0 + 15(0.2)(0) = 0

Ixz = Ix¿z¿ + mxGzG = 0 + 15( -0.3)(0) = 0

Since the plate falls without rotational motion just before the impact, its angular

momentum about point A is

(HA)1 = rG>A * mvG = (-0.3i + 0.2j) * 15(-5k)

= [-15i - 22.5j] kg # m2>s

Since the plate rotates about point A just after impact, the components of its

angular momentum at this instant can be determined from

C (HA)2 D x = Ixvx - Ixyvy - Ixz vz

= 0.8vx - (-0.9)vy - 0(vz)

= 0.8vx + 0.9vy

C (HA)2 D y = -Ixyvx + Iyvy - Iyz vz

= -(-0.9)vx + 1.8vy - 0(vz)

= 0.9vx + 1.8vy

C (HA)2 D z = -Ixz vx + Iyzvy - Iz vz

= 0(vx) - 0(vy) + 2.6vz

= 2.6vz

Thus,

(HA)2 = (0.8vx + 0.9vy)i + (0.9vx + 1.8vy)j + 2.6vz k

Referring to the free-body diagram of the plate shown in Fig. b, the weight W is a

nonimpulsive force and the impulsive force FA acts through point A. Therefore,

angular momentum of the plate is conserved about point A. Thus,

(HA)1 = (HA)2

-15i - 22.5j = (0.8vx + 0.9vy)i + (0.9vx + 1.8vy)j + 2.6vz k

Equating the i, j, and k components,

-15 = 0.8vx + 0.9vy

(1)

-22.5 = 0.9vx + 1.8vy

(2)

0 = 2.6vz

(3)

Solving Eqs. (1) through (3),

vx = -10.71 rad>s

vy = -7.143 rad>s

vz = 0

Thus,

v = [-10.7i - 7.14j] rad>s

Ans.

943

300 mm

300 mm

200 mm

y

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z

•21–25. The 5-kg disk is connected to the 3-kg slender

rod. If the assembly is attached to a ball-and-socket joint at

A and the 5-N # m couple moment is applied, determine the

angular velocity of the rod about the z axis after the

assembly has made two revolutions about the z axis starting

from rest. The disk rolls without slipping.

Mϭ5Nиm

A

x

Ix = Iz =

Iv =

1

1

(5)(0.2)2 + 5(1.5)2 + (3)(1.5)2 = 13.55

4

3

1

(5)(0.2)2 = 0.100

2

v = -vy j¿ + vz k = -vy¿ j¿ + vz sin 7.595°j¿ + vz cos 7.595°k¿

= (0.13216vz - vy¿)j¿ + 0.99123 vz k¿

Since points A and C have zero velocity,

vC = vA + v * rC>A

0 = 0 + C (0.13216 vz - vy¿)j¿ + 0.99123vz k¿ D * (1.5j¿ - 0.2k¿)

0 = -1.48684vz - 0.026433 vz + 0.2 vy¿

vy¿ = 7.5664 vz

Thus,

v = -7.4342 vz j¿ + 0.99123 vz k¿

T1 + ©U1 - 2 = T2

0 + 5(2p) (2) = 0 +

1

1

(0.100)(-7.4342 vz)2 + (13.55)(0.99123 vz)2

2

2

vz = 2 58 rad>s

Ans.

944

1.5 m

B

0.2 m

y

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z

21–26. The 5-kg disk is connected to the 3-kg slender rod.

If the assembly is attached to a ball-and-socket joint at A

and the 5-N # m couple moment gives it an angular velocity

about the z axis of vz = 2 rad>s, determine the magnitude

of the angular momentum of the assembly about A.

Mϭ5Nиm

A

x

Ix = Iz =

Ix =

1

1

(5)(0.2)2 + 5(1.5)2 + (3)(1.5)2 = 13.55

4

3

1

(5)(0.2)2 = 0.100

2

v = -vy j¿ + vz k = -vy¿ j¿ + vz sin 7.595°j¿ + vz cos 7.595°k¿

= (0.13216vz - vy )j¿ + 0.99123 vz k¿

Since points A and C have zero velocity,

vC = vA + v * rC>A

0 = 0 + C (0.13216 vz - vy ) j¿ + 0.99123 vz k¿ D * (1.5j¿ - 0.2k¿)

0 = -1 48684vz - 0.26433vz + 0.2vy

vy¿ = -7.5664 vz

Thus,

v = -7.4342 vz j¿ + 0.99123 vz k¿

Since vz = 2 rad>s

v = -14.868j¿ + 1 9825k¿

So that,

HA = Ix¿ vx i¿ + Iy¿ vy¿ j + Iz¿ vz¿ k¿ = 0 + 0.100(-14.868)j¿ + 13.55(1.9825) k¿

= -1.4868j¿ + 26.862k¿

HA = 2(-1.4868)2 + (26.862)2 = 26.9 kg # m2>s

Ans.

945

1.5 m

B

0.2 m

y

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21–27. The space capsule has a mass of 5 Mg and the

radii of gyration are kx = kz = 1.30 m and ky = 0.45 m .

If it travels with a velocity vG = 5400j + 200k6 m>s ,

compute its angular velocity just after it is struck by a

meteoroid having a mass of 0.80 kg and a velocity

vm = 5-300i + 200j - 150k6 m>s . Assume that the

meteoroid embeds itself into the capsule at point A and

that the capsule initially has no angular velocity.

vG

G

x

Conservation of Angular Momentum: The angular momentum is conserved about

the center of mass of the space capsule G. Neglect the mass of the meteroid after the

impact.

(HG)1 = (HG)2

rGA * mm vm = IG v

(0.8i + 3.2j + 0.9k) * 0.8(-300i + 200j - 150k)

= 5000 A 1.302 B vx i + 5000 A 0.452 B vy j + 5000 A 1.302 B vz k

-528i - 120j + 896k = 8450vx i + 1012.5vy j + 8450 vz k

Equating i, j and k components, we have

-528 = 8450vx

-120 = 1012.5vy

896 = 8450vz

z

vm

vx = -0.06249 rad>s

vy = -0.11852 rad>s

vz = 0.1060 rad>s

Thus,

v = {-0.0625i - 0.119j + 0.106k} rad>s

Ans.

946

A (0.8 m, 3.2 m, 0.9 m)

y

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z

*21–28. Each of the two disks has a weight of 10 lb. The

axle AB weighs 3 lb. If the assembly rotates about the z

axis at vz = 6 rad>s, determine its angular momentum

about the z axis and its kinetic energy. The disks roll

without slipping.

2 ft

2 ft

B

1 ft

A

x

1

6

=

vx

2

vx = 12 rad>s

vA = {-12i} rad>s

Hz = c

vB = {12i} rad>s

1 10

1 10

a

b (1)2 d(12i) + c a

b (1)2 d( -12i)

2 32.2

2 32.2

+ 0 + b 2c

3

1 10

10

1

a

b (1)2 +

(2)2 d(6) +

a

b(4)2(6) r k

4 32.2

32.2

12 32.2

Hz = {16.6k} slug # ft2>s

T =

=

Ans.

1

1

1

I v2 + Iy v2y + Iz v2z

2 x x

2

2

1

1 10

c2 a a

b (1)2 b d(12)2 + 0

2

2 32.2

+

1 10

10

1

3

1

b (1)2 +

(2)2 d +

a

b(4)2 r (6)2

b 2c a

2

4 32.2

32.2

12 32.2

= 72.1 lb # ft

Ans.

947

1 ft

vz ϭ 6 rad/s

y

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z

•21–29. The 10-kg circular disk spins about its axle with a

constant angular velocity of v1 = 15 rad>s. Simultaneously,

arm OB and shaft OA rotate about their axes with constant

angular velocities of v2 = 0 and v3 = 6 rad>s, respectively.

Determine the angular momentum of the disk about point O,

and its kinetic energy.

O

The mass moments of inertia of the disk about the centroidal x¿ , y¿ , and z¿ axes,

Fig. a, are

x

Ix¿ = Iy¿ =

1

1

mr2 = (10) A 0.152 B = 0.05625 kg # m2

4

4

Ix¿y¿ = Iy¿z¿ = Ix¿z¿ = 0

Here, the angular velocity of the disk can be determined from the vector addition of

v1 and v3. Thus,

v = v1 + v2 = [6i + 15k] rad>s

The angular momentum of the disk about its mass center G can be obtained by

applying

Hx = Ix¿vx = 0.05625(6) = 0.3375 kg # m2>s

Hy = Iy¿vy = 0.05625(0) = 0

Hz = Iz¿vz = 0.1125(15) = 1.6875 kg # m2>s

Thus,

HG = [0.3375i + 1.6875k] kg # m2>s

Since the mass center G rotates about the x axis with a constant angular velocity of

v3 = [6i] rad>s, its velocity is

vG = v3 * rC>O = (6i) * (0.6j) = [3.6k] m>s

Since the disk does not rotate about a fixed point O, its angular momentum must be

determined from

HO = rC>O * mvC + HG

= (0.6j) * 10(3.6k) + (0.3375i + 1.6875k)

= [21.9375i + 1.6875k] kg # m2>s

Ans.

The kinetic energy of the disk is therefore

T =

=

1 #

v HO

2

1

(6i + 15k) # (21.9375i + 1.6875k)

2

= 78.5 J

Ans.

948

V2

B

150 mm

y

Due to symmetry, the products of inertia of the disk with respect to its centroidal

planes are equal to zero.

= [21.9i + 1.69k] kg # m2>s

V3

V1

1

1

mr2 = (10) A 0.152 B = 0.1125 kg # m2

2

2

Iz¿ =

A

600 mm

91962_12_s21_p0925-0987

6/8/09

1:19 PM

Page 949

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z

21–30. The 10-kg circular disk spins about its axle with a

constant angular velocity of v1 = 15 rad>s. Simultaneously,

arm OB and shaft OA rotate about their axes with constant

angular velocities of v2 = 10 rad>s and v3 = 6 rad>s,

respectively. Determine the angular momentum of the disk

about point O, and its kinetic energy.

O

The mass moments of inertia of the disk about the centroidal x¿ , y¿ , and z¿ axes.

Fig. a, are

x

Ix¿ = Iy¿ =

Iz¿ =

1

1

mr2 = (10) A 0.152 B = 0.1125 kg # m2

2

2

Ix¿y¿ = Iy¿z¿ = Ix¿z¿ = 0

Here, the angular velocity of the disk can be determined from the vector addition of

v1, v2, and v3. Thus,

v = v1 + v2 + v3 = [6i + 10j + 15k] rad>s

The angular momentum of the disk about its mass center G can be obtained by

applying

Hx = Ix¿vx = 0.05625(6) = 0.3375 kg # m2

Hy = Iy¿vy = 0.05625(10) = 0.5625 kg # m2

Hz = Iz¿vz = 0.1125(15) = 1.6875 kg # m2

Thus,

HG = [0.3375i + 0.5625j + 1.6875k] kg # m2

Since the mass center G rotates about the fixed point O with an angular velocity of

Æ = v2 + v3 = [6i + 10j], its velocity is

vG = Æ * rG>O = (6i + 10j) * (0.6j) = [3.6k] m>s

Since the disk does not rotate about a fixed point O, its angular momentum must be

determined from

HO = rC>O * mvG + HG

= (0.6j) * 10(3.6k) + (0.3375i + 0.5625j + 1.6875k)

= [21.9375i + 0.5625j + 1.6875k] kg # m2>s

Ans.

The kinetic energy of the disk is therefore

=

1 #

v HO

2

1

(6i + 10j + 15k) # (21.9375i + 0.5625j + 1.6875k)

2

= 81.3J

Ans.

949

V2

B

150 mm

y

Due to symmetry, the products of inertia of the disk with respect to its centroidal

planes are equal to zero.

T =

V3

V1

1

1

mr2 = (10) A 0.152 B = 0.05625 kg # m2

4

4

= [21.9i + 0.5625j + 1.69k] kg # m2>s

A

600 mm

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