angular momentum and kinetic energy

8/15/2019 Angular Momentum and Kinetic Energy

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Vector Dynamics and Vibrations Angular Momentum and Kinetic Energy 1

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CHAPTER 4 : ANGULAR MOMENTUM

4.0 Introduction4.1 Angular Momentum of a Rigid Body in Three Dimensions4.2 Kinetic Energy of a Rigid Body in Three Dimensions4.3 ApplicationsTips and MotivationWorksheetsExercises 4.0

4.0 INTRODUCTION

The two fundamental principles used in the analysis of the plane motion of a rigid

body remain valid in the most general case of the motion of a rigid body in three

dimensions, namely,

F = maG

M G = H G

Free-Body Diagram Equivalent Force-Couple system Kinetic Diagram

(a) (b) (c)Figure 4.1

We note that the system of forces acted on the body, Figure 4.1(a), can be reduced to

an equivalent force-couple system, Figure 4.1(b), which results in the two inertia

forces (also called effective forces) as in Figure 4.1(c).

G

F 1

F 3 F 2

F 4

G

M G F

G

H G maG


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Angular Momentum and Kinetic Energy Vector Dynamics and Vibrations 2

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4.1 ANGULAR MOMENTUM OF A RIGID BODY IN THREE DIMENSIONS

Consider a rigid body having a three-dimensional motion with being the angular

velocity of the body at the instant considered as shown.

The position vector of the particle P of mass

dm can be written as

R = RG + p

Its linear velocity is then given by

v = R = R G + p

where p = p

The linear momentum of the particle P is

dL = dm v = dm R

Angular Momentum of a Rigid Body About Any Given Point O

The angular momentum of the particle P about point O is

dH O = R dL = R dm v = R dm R

= ( RG + p) dm R

= RG dm R + p dm R

and therefore, the total angular momentum of the entire body about point O is

H O = RG dm R + p dm R

where RG dm R = RG dm R = RG mvG

p dm R = H G = the angular momentum about the mass center G

Therefore, the total angular momentum of the entire body about point O is then

H O = RG mvG + H G

which is known as the transfer theorem for angular momentum.

z

x

dm

y

G

RG R

P


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Angular Momentum of a Rigid Body About Its Mass Center G

H G = p dm R = p dm( R G + p )

= p dm R

G + p dm p

= (dm p) R G + p dm p

Note that (dm p) = 0 since it represents the first moment of the mass with respect to

the centroidal plane i.e. the location of the mass center is itself in a coordinate plane.

Therefore,

H G = p dm p = p ( p) dm

Using the local coordinates, we may express p and as

p = x i + y j + z k and = x i + y j + z k

and substituting them into the above expression, we have

p ( p) = ( x i + y j + z k ) [( x i + y j + z k ) ( x i + y j + z k )]

= [( y2 + z

2) x – xy y – xz z ] i +

[ – yx x + ( y2 + z

2) y – yz z ] j +

[ –

zx x –

zy y + ( y2

+ z 2

) z ] k

It follows that

p ( p) dm = dm [( y2 + z

2) x – xy y – xz z ] i +

dm [ – yx x + ( z 2 + x2) y – yz z ] j +

dm [ – zx x – zy y + ( x2 + y

2) z ] k

= [ I x x – I xy y – I xz z ] i +

[ – I yx x + I y y – I yz z ] j +

[ – I zx x – I zy y + I z z ] k


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For simplicity, we may express H G as

H G = H x i + H y j + H z k

where H x = [ I x x – I xy y – I xz z ]

H y = [ – I yx x + I y y – I yz z ]

H z = [ – I zx x – I zy y + I z z ]

We can see that a new value of moments and

products of inertia will be obtained if a

different set of coordinate axes were used.

If the centroidal coordinate axes were chosen such that they coincide with the principal

axes of inertia (i.e. each of the planes is symmetry), then all the products of inertia will

be zero. The above equation reduces to

H G = I x x i + I y y j + I z z k

Example 4.1

A homogeneous disk of mass m =

4.8 kg spins at the constant rate s =

12 rad/s relative to the uniform bentarm OAB as shown. At the sameinstant, the bent arm OAB rotates at

the constant rate p = 6 rad/s aboutthe Y axis. Knowing that the mass ofthe bent arm OAB is 4 kg, determinethe angular momentum of theassembly about point O. Figure E4.1

Solution: H O = RG mvG + H G

Rod OA, p1 = 0.25 K ; 1 = p j = 6 J

vG = (6 J ) (0.25 K ) = 1.5 I (since i = I ) RG mvG = 0.25 K 2(1.5 I ) = 0.75 J kg m

2/s

I x = I y = 121 (2)(0.5

2) = 0.0417 kg m

2 , I z = 0 kg m

2

and I xy = I yz = = I zx = 0 kg m2 (due to symmetry)

O

p

A

B

Y X

Z

s

50 cm

30 cm

50 cm

Z

X

Y

G

H G

x

y z


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H x = I x x – I xy y – I xz z = 0 – 0 – 0 = 0 kg m2/s

H y = – I yx x + I y y – I yz z = 0 + (0.0417)(6) – 0 = 0.2502 kg m2/s

H z = – I zx x – I zy y + I z z = 0 – 0 – 0 = 0 kg m2/s

H G = 0.2502 j = 0.2502 J kg m2/s (since j = J )

H O (rod OA) = 0.75 J + 0.2502 J = 1.0 J kg m2/s

Rod AB, p2 = 0.5 K + 0.25 I ; 2 = 1 = 6 J

vG = (6 J ) ( 0.5 K + 0.25 I ) = 3 I – 1.5 K

RG mvG = (0.5 K + 0.25 I ) 2(3 I – 1.5 K ) = 3.75 J kg m2/s

I y = I z = 121 (2)(0.52) = 0.0417 kg m2 , I x = 0 kg m

2

and I xy = I yz = = I zx = 0 kg m2 (due to symmetry)

H x = I x x –

I xy y –

I xz z = –

0 –

0 = 0 kg m2

/s H y = – I yx x + I y y – I yz z = 0 + 0.0417(6) – 0 = 0.2502 kg m

2/s


H G = 0.2502 j = 0.2502 J kg m2/s

H O (rod AB) = 3.75 J + 0.2502 J = 4.0 J kg m2/s

Disk , p3 = 0.5 K + 0.5 I ; 3 = (6 J + 12i)

vG = (6 J ) (0.5 K + 0.5 I ) = 3 I – 3 K

RG mvG = (0.5 K + 0.5 I ) 4.8(3 I – 3 K ) = 14.4 J k g m2/s

I y = I z = 41 (4.8)(0.3)2 = 0.108 kg m2, I x = 2

1 (4.8)(0.3)2 = 0.216 kg m2

and I xy = I yz = I zx = 0 (due to symmetry)

H x = I x x – I xy y – I xz z = 0.216(12) – 0 – 0 = 2.592 kg m2/s

H y = – I yx x + I y y – I yz z = 0 + 0.108(6) – 0 = 0.648 kg m2/s


H G = 2.592i + 0.648 j = 2.592 I + 0.648 J kg m2/s (since i= I , j = J )

H O (disk) = 14.4 J + 2.592 I + 0.648 J = 2.592 I + 15.048 J kg m2/s

Finally we obtain for the entire assembly,

H O, total = 2.592 I + 20.05 J kg m2/s [ Ans]


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4.2 KINETIC ENERGY OF A RIGID BODY IN SPATIAL MOTION

The kinetic energy of a rigid body having a three-dimensional motion with being the

angular velocity of the body at the instant considered can be derived as follows.

Consider a particle of mass dm on the rigid

body. The kinetic energy of this particle is

given by

dT =2

1 dm (v P •v P )

and from v P = R = R G + p , we now have

dT =2

1 dm( R G + p )•( R G + p )

Expanding the terms and noting that p = p , we obtain

dT =2

1 dm[ R G• R G + 2( R G• p ) + ( p • p )]

Substituting p = p, we have

dT =2

1 dm[( R G• R G) + 2( R G•( p)) + ( p)•( p)]

The total kinetic energy for the entire body is, therefore,

T = dT

=2

1 dm[( R G• R G) + 2( R G•( p)) + ( p)•( p)]

=2

1 dm[( R G• R G) + dm( R G•( p) + 21 dm( p)•( p)

=2

1 m(vG•vG) + 21 dm( p)•( p)

where ( p) = ( x i + y j + z k ) ( x i + y j + z k )

= ( y z –

z y) i + ( z x –

x z ) j + ( x y –

y x) k

( p)•( p) = ( y2 + z 2) x

2 + ( z 2 + x2) y2 + ( x2 + y2) z

2

– 2( xy x y + yz y z + yz y z )

z

x

dm

y

G RG

p

R

P


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dm( p)•( p) = ( y2 + z

2)dm x

2 + ( z

2 + x

2)dm y

2 + ( x

2 + y

2)dm z

2

– 2( xy dm x y + yz dm y z + yz dm y z )

= I x x2 + I y y

2 + I z z 2 – 2( I xy x y + I yz y z + I zx y z )

We obtain finally

T =2

1 m(vG•vG) + 21 ( I x x

2 + I y y

2 + I z z

2 )

– ( I xy x y + I yz y z + I zx y z )

If the axes coincide with the principal axes of inertia, the kinetic energy is merely

T =2

1 m(vG•vG) + 21 ( I x x

2 + I y y2 + I z z

2 )

Alternatively, the kinetic energy of a rigid body moving with the angular velocityand the velocity of its mass center vG can be expressed in general form as

T =2

1 vG• L + 21 • H G where L = mvG

When a rigid body is constrained to rotate in three-dimensional space about a fixed

point O or when there is a point O in the body which momentarily has zero velocity,

the kinetic energy of such body reduces to

T =2

1 • H O


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Example 4.3

A homogeneous disk of mass m =

4.8 kg spins at the constant rate s =

12 rad/s relative to the uniform bentarm OAB as shown. At the sameinstant, the bent arm OAB rotates at

the constant rate p = 6 rad/s aboutthe Y axis. Knowing that the mass ofthe bent arm OAB is 4 kg, determinethe total kinetic energy of theassembly. Figure E4.3

Solution:

T total = T rod OA + T rod AB + T disk where

T =2

1 m(vG•vG) + 21 ( I x x

2 + I y y

2 + I z z

2 ) – ( I xy x y + I yz y z + I zx y z )

From Example E4.1, we have

Rod OA, vG = 1.5 I m/s

T rod OA = 21 (2)(1.5)

2 +

2

1 (0.0417)(6)2 = 3.0 J

Rod AB, vG = 3 I – 1.5 K m/s

T rod AB = 21 (2)[(3)

2 + (1.5)

2] +

2

1 (0.0417)(6)2 = 12.0 J

Disk B, vG = 3 I – 3 K m/s

T disk = 21 (4.8)[(3)2 + ( – 3)2] +

2

1 [(0.216)(12)2 + (0.108)(6)2 ] = 60.7 J

T total = 3 + 12 + 60.7 = 75.7 J [ Ans]

O

p

A

B

Y X

Z

s

50 cm

30 cm

50 cm


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Example 4.4

The assembly which consists of tworectangular plates, each of mass 0.1 kg,

attached to a uniform slender arm ofnegligible mass spins about its axis at a

constant rate of 2 = 4 rad/s. The arm at the

same instant rotates at a constant rate of 1 = 2 rad/s about its Y -axis shaft. For theinstant represented in Figure Q4, determinea) the angular momentum of the assembly

with respect to point A, andb) the kinetic energy of the assembly.

Figure E4.4

Solution:

H A = RG mvG + H G

[Note that the C.G is as shown at point G (neglecting size of rod AB)]

RG = (l +2

a) I = 0.24 I m [note that p = 1 and q = 2 ]

arm = p = pj = 2 J rad/s (since j = J )

vG = arm RG = (2 J ) (0.24 I ) = – 0.48 K m/s

RG mvG = (0.24 I ) (0.2)( – 0.48 K ) Note: total mass is m = 0.2 kg

= 0.02304 J kg m2/s

I x = 121 (0.2)(0.2)

2 = 0.000667 kg m

2 ; I y = 12

1 (0.2)(0.12)2 = 0.00024 kg m

2

I z = 121 (0.2)(0.122+ 0.22) = 0.000907 kg m2 ;

I xy

= I yz

= I zx

= 0 kg m2 (due to plane of symmetry)

plate = p + q = 2 j – 4i [Note that i = I , j = J , and k = K ]

H x = I x x – I xy y – I xz z = 0.000667( – 4) – 0 – 0 = – 0.002668 kg m2/s

H y = – I yx x + I y y – I yz z = – 0 + 0.00024(2) – 0 = 0.00048 kg m2/s

H z = – I zx x – I zy y + I z z = – 0 – 0 + 0.1323(0) = 0 kg m2/s

H G = – 0.002668 I + 0.00048 J kg m2/s

H A = – 0.002668 I + 0.02352 J kg m2/s [ Ans]

T =2

1 m( Gv Gv ) + 21 ( x I

2

x + y I 2

y + z I 2

z ) – ( xy I x y + yz I y z + zx I z x)

T plate = 21 (0.2)( – 0.48)

2 +

2

1 [0.000667( – 4)2 + (0.00024)(2)

2 + (0.000907)(0)

2 ] – [0]

= 0.02304 + 0.005816

A

1

Y

X

Z

l

2

a

B

b

b

Given:

a = 120 mm

b = 100 mm

l = 180 mm

G

z


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T plate = 0.028856 J [ Ans]

Alternatively, since both angular motions 1 and 2 pass through the same point A (i.e. motion

about a fixed point A), we may consequently compute H A using

H A = ( I X X – I XY Y – I XZ Z ) I + ( – I YX X + I Y Y – I YZ Z ) J + ( – I ZX X – I ZY Y + I Z Z ) K

where plate = p + q = 2 j – 4i = 2 J – 4 I => X = – 4 , Y = 2 , Z = 0

I X = I x = 121 (0.2)(0.2)

2 = 0.000667 kg m

2

I Y = I y + m(l +2

a)

2 =

12

1 (0.2)(0.12)2 + 0.2(0.18 + 0.06)

2 = 0.01176 kg m

2

I Z = I z + m(l +2

a)

2 =

12

1 (0.2) (0.122+ 0.2

2) + 0.2(0.18 + 0.06)

2 = 0.012427 kg m

2

I XY = 0 kg m2 (due to symmetrical plane ZX )

I YZ = 0 kg m2 (due to symmetrical plane ZX )

I ZX = 0 kg m2 (due to symmetrical plane XY )

H A = ( I X X – I XY Y – I XZ Z ) I + ( – I YX X + I Y Y – I YZ Z ) J + ( – I ZX X – I ZY Y + I Z Z ) K

= 0.000667( – 4) I + 0.01176(2) J

H A = – 0.002668 I + 0.2352 J kg m2/s [ Ans]

For this type of motion, the kinetic energy is then given by

T =2

1 H A = 21 (2 J – 4 I ) ( – 0.002668 I + 0.2352 J ) = 0.028856 J [ Ans]


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Figure E4.5

Example 4.5

A thin rectangular plate of mass m = 1.2 kg isdesigned to spin at a constant rate of q = 1.5 rad/s

relative to arm AB which at the same instantrotates at the constant rate of p = 3 rad/s relativeto arm OA which itself rotates at the constant rate

of = 2 rad/s. Determine(a) the angular momentum of the plate about

point O, and(b) the kinetic energy of the plate.

Solution:

H O = RG mvG + H G note that point G = B

RG = 0.24 J + 0.36 I ; = + p = – 2 K – 3 J

vG = RG = ( – 2 K – 3 J ) (0.24 J + 0.36 I )

= 0.48 I – 0.72 J + 1.08 K m/s

RG mvG = (0.24 J + 0.36 I ) (1.2)(0.48 I – 0.72 J + 1.08 K )

= 0.311 I – 0.467 J – 0.449 K

I y = 121 (1.2)(0.09)

2 = 0.00081 kg m

2, I z = 12

1 (1.2)(0.12)2 = 0.00144 kg m

2

I x = 121 (1.2)(0.12

2 + 0.09

2) = 0.00225 kg m

2; I xy = I yz = I zx = 0 (symmetry)

plate = + p + q = – 2k – 3 j + 1.5i (i = I , j = J , k = K )

H G = H x i + H y j + H z k

where H x = I x x – I xy y – I xz z = 0.00225(1.5) – 0 – 0 = 0.003375

H y = – I yx x + I y y – I yz z = – 0 + 0.00081( – 3) – 0 = – 0.00243

H z = – I zx x – I zy y + I z z = – 0 – 0 + 0.00144( – 2) = – 0.00288

H O = (0.311 I – 0.467 J – 0.449 K ) + (0.003375 I – 0.00243 J – 0.00288 K )

= 0.314 I – 0.469 J – 0.452 K = 0.314 I – 0.469 J – 0.452 K kg m2/s

T = 21

m(vG vG) + 21

( I x2

x + I y2

y + I z 2

z ) –

( I xy x y + I yz y z + I zx z x)

T plate = 21 (1.2)[(0.48)2 + ( – 0.72)2+ (1.08)2] +

2

1 [(0.00225)(1.5)2 + (0.00081)( – 3)

2 + (0.00144)( – 2)

2 ] – [0]

= 1.1491 + 0.00906 = 1.1582 J [ Ans]

O

360

240

120

q

A

X

Y

Z

p90

x

y

z

B

O

360

240

120

q

AB

X

Y

Z

p 90


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Figure W4.1

Worksheet 4.1 :

A 5.1-kg homogeneous disk of radius 90

mm rotates at the constant rate 2 = 10rad/s with respect to arm OAB, which

itself rotates at the constant rate 1 = 2rad/s about the Y axis. Determine theangular momentum of the disk aboutpoint O and also its kinetic energy.

O2 A

B

Z

X

Y

1

400 mm

90 mm

280 mm


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Figure W4.2

Worksheet 4.2 :

A thin rectangular plate of mass m = 1.2 kg is

designed to spin at a constant rate of q = 1.5 rad/s

relative to arm AB (m = 0.9 kg) which at the sameinstant rotates at the constant rate of p = 3 rad/srelative to arm OA which itself rotates at the

constant rate of = 2 rad/s. Determine(a) the angular momentum of arm AB about

point O, and(b) the kinetic energy of arm AB.

Solution:

O

360

240

120

q

AB

X

Y

Z

p 90

O

360

240

120

q

A

X

Y

Z

p90

x

y

z

B


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Figure W4.3

Worksheet 4.3 :

A thin homogeneous disk of radius r

rotates at the constant rate q rad/swith respect to the fork-ended rod ABCD which itself rotates at theconstant rate p rad/s about the y axis.Determine the angular momentum ofthe disk about the support at B andalso the kinetic energy of the disk.

A B

x

a b

z

q

C Dr


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Exercise 4.0:4-1

End of Chapter 4

angular momentum and kinetic energy

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