Chapter 21 - adasdas PDF

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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.

SOLUTION Ixx + Iyy + Izz =

Lm

= 2

(y2 + z2)dm +

Lm

(x2 + y2)dm (x2 + z2)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 + I yy + Izz is also indepenent of the orientation of the x, y, z axis. Q.E.D.

1 10 9

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21–2. Determine the moment of inertia of the cone with respect to a vertical y axis passing 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

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

dIy =

=

=

Iy =

L

rpa2 2 x dx. h2

h

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

(4h2 + a2) x4 dx

4h4

dIy =

rpa2 4

4h

(4h2 + a2)

L0

h

x4dx =

r pa2h 20

(4h2 + a2)

However, m =

Lm

h

dm =

r pa2 r pa2h x2 dx = 2 3 h L0

Hence, Iy =

3m (4h2 + a2) 20

Using the parallel axis theorem: Iy = Iy + md2 3h 2 3m (4h2 + a2) = Iy + ma b 20 4 Iy =

3m 2 (h + 4a2) 80

Ans.

Iy' = Iy + md2 =

3m 2 h 2 (h + 4a2) + ma b 4 80

=

m (2h2 + 3a2) 20

Ans.

Ans: 3m 2 ( h + 4a2) 80 m ( 2h2 + 3a2 ) Iy′ = 20 Iy =

11 1 0

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21–3. Determine moment of inertia Iy of the solid formed by revolving the shaded area around the x axis. The density of the material is r = 12 slug/ft3.

y

y2 = x 2 ft x

SOLUTION The mass of the differential element is dm = rdV = r A py2 B dx = rpxdx . dIy =

1 2

dmy2 + dmx2

=

1 4

[rpxdx](x) + (rpxdx)x2

4 ft

1 = rp( x2 + x3) dx 4 Iy =

L

4

dIy = rp

1 ( x2 + x3) dx = 69.33 pr L0 4

= 69.33(p)(12) = 2614 slug # ft2

Ans.

Ans: Iy = 2614 slug # ft 2 1 11 1

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*21–4. z

Determine the moments of inertia Ix and Iy of the paraboloid of revolution.The mass of the paraboloid is 20 slug.

z2  2y

2 ft y

SOLUTION The mass of the differential element is dm = rdV = r A pz2 B dy = 2rpydy. m = 20 =

Lm

dm =

20 = 4rp dIx =

1 4

L0

x

2 ft

2

2rpydy r =

5 p

slug/ft3

dmz2 + dmA y2 B

= 14 [2rpydy](2y) + [2rpydy]y2 = A 5y2 + 10y3B dy Ix =

Iy =

L

dIx =

L0

dIy =

1 2

L

dIy =

2

2 A 5y2 + 10y3 B dy = 53.3 slug # ft

Ans.

dmz2 = 2rpy2 dy = 10y2 dy L0

2

10y2 dy = 26.7 slug # ft2

Ans.

Ans: Ix = 53.3 slug # ft2 Iy = 26.7 slug # ft2 11 1 2

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21–5. 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.

z a a

h

y

SOLUTION x

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

Lm

dm = rh

L0

a

(a - y)dy =

ra2h 2

Using the parallel axis theorem: dIyz = (dIy¿z¿)G + dmyGzG h = 0 + (rhxdy) (y) a b 2 =

=

Iyz =

rh2 xydy 2 rh2 2

(ay - y2) dy

a 2 rh2 ra3h2 m 1 ra h ah b (ah) = = a (ay - y2) dy = 6 2 2 L 6 12 0

Ans.

Ans: Iyz =

1 11 3

m ah 6

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21–6. 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.

z a a

h

y

SOLUTION x

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

Lm

dm = rh

L0

a

(a - y)dy =

ra2h 2

Using the parallel axis theorem: dIxy = (dIx¿y¿)G + dmxGyG x = 0 + (rhxdy)a b (y) 2 =

=

rh2 2

x2ydy

rh2 3 (y - 2ay2 + a 2 y) dy 2 a

Ixy =

rh (y3 - 2ay2 + a 2 y) dy 2 L0

=

2 ra 4h m 1 ra h 2 ba = a2 a = 12 12 2 24

Ans.

Ans: Ixy =

11 1 4

m 2 a 12

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21–7. Determine the product of inertia Ixy of the object formed by revolving the shaded area about the line x = 5 ft. Express the result in terms of the density of the material, r.

y 3 ft

2 ft

y2  3x

SOLUTION L 0

3

L 0

3 ~

dm = r 2p

L0

x 3

3

(5 - x)y dx = r 2p

L0

(5 - x) 23x dx = 38.4rp

3

y dm = r2p

y (5 - x)y dx L0 2 3

(5 - x)(3x) dx L 0 = 40.5rp = rp

Thus, y =

40.5rp 38.4rp

= 1.055 ft

The solid is symmetric about y, thus Ixy¿ = 0 Ixy = Ixy¿ + x ym = 0 + 5(1.055)(38.4rp) Ixy = 636r

Ans.

Ans: I xy = 636r 1 11 5

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*21–8. Determine the moment of inertia Iy of the object formed by revolving the shaded area about the line x = 5 ft. Express the result in terms of the density of the material, r.

y 3 ft

2 ft

y2  3x

SOLUTION

x

3

Iy¿ =

1 1 dm r2 - (m¿)(2)2 2 L0 2 3

3

1 1 dm r2 = r p(5 - x)4 dy 2L 0 L0 2 3 y2 4 1 = rp a 5 - b dy 3 2 L0 = 490.29 r p

m¿ = r p (2)2(3) = 12 r p Iy¿ = 490.29 r p -

1 (12 r p)(2)2 = 466.29 r p 2

Mass of body; m =

L0

3

r p (5 - x)2 dy - m¿

3

L0 = 38.4 r p

r p (5 -

=

y2 2 ) dy - 12 r p 3

Iy = 466.29 r p + (38.4 r p)(5)2 = 1426.29 r p Iy = 4.48(103) r

Ans.

Also, Iy¿ = =

L0 L0

3

r2 dm

3

(5 - x)2 r (2p)(5 - x)y dx 3

(5 - x)3 (3x)1/2 dx L0 = 466.29 r p = 2rp

m =

L0

3

dm L0

3

L0 = 38.4 r p

3

= 2rp = 2rp

(5 - x)y dx (5 - x)(3x)1/2 dx

Iy = 466.29 r p + 38.4 r p(5)2 = 4.48(103)r

Ans. 11 1 6

Ans: Iy = 4.48(103) r Iy = 4.48(103) r

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21–9. Determine the moment of inertia of the cone about the z¿ axis. The weight of the cone is 15 lb, the height is h = 1.5 ft, and the radius is r = 0.5 ft.

r z h z¿

z¿

SOLUTION u = tan-1(

0.5 ) = 18.43° 1.5

Ixx = Iyy = [

1.5 3 m{4(0.5)2 + (1.5)2}] + m[1.5 - ( )]2 80 4

Ixx = Iyy = 1.3875 m Iz =

3 m(0.5)2 = 0.075 m 10

Ixy = Iyz = Izx = 0 Using Eq. 21–5. 2 2 I Iz¿z¿ = u2z¿x Ixx + uz¿y Iyy + uz¿z zz

= 0 + [cos(108.43°)]2(1.3875m) + [cos(18.43°)]2(0.075m) = 0.2062m Iz¿z¿ = 0.2062(

15 ) = 0.0961 slug # ft2 32.2

Ans.

Ans: Iz′ z′ = 0.0961 slug # ft2 1 11 7

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21–10. y

Determine the radii of gyration kx and ky for the solid formed by revolving the shaded area about the y axis. The density of the material is r.

0.25 ft

xy  1 4 ft 0.25 ft

SOLUTION For ky: The mass of the differential element is dm = rdV = r(px2) dy = rp dIy = 21 dmx2 = Iy =

L

dIy = 12 rp

dy y4 L0.25 4

C

1 dy 2 rp y 2

+

dy . y2

1 dy 1 D A y 2B = 2 rp y 4

x

4 ft

1 C r(p)(4)2(0.25) D(4)2 2

= 134.03r

However,

m =

Hence,

ky =

m L

4

dm = rp

L0.25

dy y2

+ r C p(4)2(0.25)D = 24.35r

Iy 134.03r = 2.35 ft = Am A 24.35r

Ans.

For kx : 0.25 ft 6 y … 4 ft dI¿x =

=

1 dmx2 + dmy2 4 1 I dy 2 c rpdy y d a 2 b + a rp y b y 4 y 2

2

1 = rpA 4y 4 + 1 B dy

I¿ x =

L

4

dI¿ x = rp

0.25 L

A 4y1 4 + 1 B dy = 28.53r

I¿¿ x = 41 C rp(4)2(0.25)D (4)2 + C rp(4)2(0.25) D (0.125)2 = 50.46r

Ix = I¿ x + I¿¿ x = 28.53r + 50.46r = 78.99r

Hence,

kx =

Ix = m

78.99r = 1.80 ft 24.35r

Ans.

Ans: ky = 2.35 ft kx = 1.80 ft 11 1 8

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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.

a

a

SOLUTION h

The mass of the differential element is dm = rdV = r A pa2 B dy. dIaa =

1 4

a

dma2 + dmAy2 B

= 41 CrA pa2B dy Da2 + CrA pa2 B dyD y2 = A 14 rpa4 + rpa2y2B dy Iaa =

L

dIaa =

=

L0

h

A 14 rpa4 + rpa2y2 B dy

rpa2h 12

(3a2 + 4h2) h

However, Hence,

r A pa2 B dy = rpa2h m L0 L m = A 3a2 + 4h2 B 12

m = Iaa

dm =

Ans.

Ans: Iaa =

1 11 9

m ( 3a2 + 4h2) 12

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*21–12. Determine the moment of inertia Ix of the composite plate assembly. The plates have a specific weight of 6 lb> ft2.

z

0.25 ft 0.5 ft

SOLUTION

y

0.5 ft x

Horizontial plate: Ixx =

0.5 ft

0.5 ft

1 6(1)(1) ( )(1)2 = 0.0155 12 32.2

Vertical plates: lxy¿ = 0.707,

lxx¿ = 0.707, 1 Ix¿x¿ = ( 3 Iy¿y¿ = (

6(14 )(1 22) 32.2

lxz¿ = 0

1 )( )2 = 0.001372 4

6(14 )(122) 1 6(41)(1 22) 1 1 )( )[( )2 + (1 22)2] + ( )( )2 32.2 12 4 32.2 8

= 0.01235 Using Eq. 21–5, Ixx = (0.707)2(0.001372) + (0.707)2(0.01235) = 0.00686 Thus, Ixx = 0.0155 + 2(0.00686) = 0.0292 slug # ft2

Ans.

Ans: Ixx = 0.0292 slug # ft 2 11 2 0

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21–13. Determine the product of inertia Iyz of the composite 2 plate assembly. The plates have a weight of 6 lb>ft .

z

0.25 ft 0.5 ft 0.5 ft

0.5 ft

SOLUTION

y

0.5 ft x

Due to symmetry, Ans.

Iyz = 0

Ans: Iyz = 0 1 12 1

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21–14. z

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 .

400 mm 200 mm

SOLUTION 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 .

x

y

400 mm

Ixy = ©Ix¿y¿ + mxGyG = C 0 + 8(0.2)(0.2) D + C 0 + 4(0)(0.2)D = 0.32 kg # m2

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.

Ans: Ixy = 0.32 kg # m2 Iyz = 0.08 kg # m2 Ixz = 0 11 2 2

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21–15. Determine the moment of inertia of both the 1.5-kg rod and 4-kg disk about the z¿ axis. 300 mm

z 100 mm z'

SOLUTION Due to symmetry Ixy = Iyz + Izx = 0 1 1 Iy = Ix = B (4)(0.1)2 + 4(0.3)2 R + (1.5)(0.3)2 4 3 = 0.415 kg # m2 Iz =

1 (4)(0.1)2 = 0.02 kg # m2 2

uz = cos (18.43°) = 0.9487,

uy = cos 90° = 0,

ux = cos (90° + 18.43°) = - 0.3162 Iz¿ = Ix ux2 + Iy u2y + Iz uz2 - 2Ixy ux uy - 2Iyz uy uz - 2Izx uz ux = 0.415(-0.3162)2 + 0 + 0.02(0.9487)2 - 0 - 0 - 0 = 0.0595 kg # m2

Ans.

Ans: Iz′ = 0.0595 kg # m2 1 12 3

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*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.

z a

0.3 m

1m

O 0.5 m

SOLUTION x

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 and m3 = 3(0.3) = 0.9 kg . Ixx = c

1 1 (3)( 12 ) + 3( 0.52 ) d + 3 0 + 1.5( 12 ) 4 + c (0.9)( 0.32) + 0.9( 0.152 + 12 ) d 12 12

= 3.427 kg # m2 Iyy = 0 + c

1 1 (1.5) ( 0.52 ) + 1.5 ( 0.252 ) d + c (0.9)( 0.32 ) + 0.9 ( 0.152 + 0.52) d 12 12

= 0.377 kg # m2 Izz = c

1 1 (3)( 12 ) + 3( 0.52 ) d + c (1.5)( 0.52) + 1.5 ( 12 + 0.252 ) d + 12 12

30

+ 0.9( 12 + 0.52 ) 4

= 3.75 kg # m2 Ixy = [0 + 0] + [0 + 1.5(0.25)(-1)] + [0 + 0.9(0.5)(-1)] = -0.825 kg # m2 Iyz = [0 + 0] + [0 + 0] + [0 + 0.9(-1)(0.15)] = -0.135 kg # m2 Izx = [0 + 0] + [0 + 0] + [0 + 0.9(0.15)(0.5)] = 0.0675 kg # m2 The unit vector that defines the direction of the Oa axis is UOa = Thus,

ux =

0.5i - 1j + 0.3k 2 0.5 + (-1) + 0.3 2

0.5 21.34

2

uy = -

=

2

1 21.34

0.5 21.34

i -

uz =

1 21.34

j +

0.3 21.34

0.3 21.34

11 2 4

k

y

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*21–16. Continued

Then IOa = Ixx u2x + Iyy uy2 + Izz u 2z - 2Ixy u xu y - 2Iyz u yu z - 2Izx u zu x = 3.427 a

0.5 21.34

2

b + 0.377a -

-2( - 0.135) a-

1 21.34

ba

1 21.34 0.3 21.34

2

b + 3.75a

0.3 21.34

b - 2(0.0675)a

b

2
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