978-0073380308 Chapter 10 Solution Manual Part 11

Dynamics 2e 2159

we are viewing the bar

AB

in the plane of the angle

ˇ

, the spring force

Fs

has a component

.Fs/k

in that

plane and a component

.Fs/?

perpendicular to that plane. Fortunately, the work of

Fs

is readily computed

using its potential energy function.

With all of this in mind, we can now write the kinetic energy of the bar AB as

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2160 Solutions Manual

To obtain the components of

E!AB

in the

xy´

frame, we can dot

E!AB

with each of the unit vectors in the

xy´

system in turn. We already know that O|is given by Eq. (5) and we can find O{using

O{DO

K⇥OuA=B

ˇˇˇO

K⇥OuA=B ˇˇˇDO

K⇥1

L⇣lO

IdO

JChO

K⌘

ˇˇˇO

K⇥1

L⇣lO

IdO

JChO

K⌘ˇˇˇDdO

IlO

J

pd2Cl2:(14)

The unit vector O

kcan be found using its definition

O

kDO{⇥O|Dhl O

ICdh O

JCd2Cl2O

K

Lpd2Cl2:(15)

Now, we can determine the components of the angular velocity in the xy´ frame using

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2162 Solutions Manual

constraint force system in the first FBD don’t do any work, the equivalent constraint force system in the

second FBD can’t do any work either.

With all of this in mind, we can now write the kinetic energy of the system as

TD1

2mAB v2

GC1

2IGx!2

ABx C1

2IGy !2

ABy C1

2IG´!2

AB´ C1

2mAv2

AC1

2mBv2

B;(2)

where IGx DIG´ D1

12 mAB L2and IGy D0.

Force Laws.

Placing the datum line for zero gravitational potential energy at

ZD0

, we can conclude that

in ¡

V2D0: (3)

In ¿, we have that

V1DmAB gh1

2CmAgh1:(4)

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Dynamics 2e 2163

The unit vector O

kcan be found using its definition

O

kDO{⇥O|Dhl O

ICdh O

JCd2Cl2O

K

Lpd2Cl2:(15)

Now, we can determine the components of the angular velocity in the xy´ frame using

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July 2, 2012

2164 Solutions Manual

Problem 10.47

When

h1D0:6

m,

`1D0:5

m, and

dD0:9

m, collar

A

is

moving downward at speed

vA1 D3m=s

. The uniform thin

bar

AB

has mass

mAB D4kg

and length

L

and is attached

to collars

A

and

B

with ball-and-socket joints. Collars

A

and

B

slide smoothly on rods

CD

and

EF

, respectively, and collar

B

is attached to the stop at

E

by a linear elastic spring with

constant

kD200 N=m

and unstretched length

2`1

. Neglect the

dimensions of the collars and assume that the angular velocity of

the rod AB is such that AB does not spin about its axis.

Assuming that the mass of collar

A

is

mAD0:5 kg

and the

mass of collar

B

is

mBD0:5 kg

, determine the speed of the

collar Awhen it reaches D.

Solution

Referring to the FBD at the right, we see that the constraint forces

AX

and

AY

on the bar at

A

and

BY

and

BZ

on the bar at

B

do

not do any work since they are perpendicular to the motion of

those points. In addition, the weight force

mBg

does no work

since the motion of

B

is perpendicular to that force. Therefore,

the only forces that do work on the bar as it falls is the weight

force

mAB g

acting at its mass center and the weight force

mAg

acting at A.

Balance Principles.

Since all forces that do work are conser–

vative, we can say that energy is conserved and can apply the

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permission of McGraw-Hill, is prohibited.

of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the

permission of McGraw-Hill, is prohibited.

2166 Solutions Manual

To obtain the components of

E!AB

in the

xy´

frame, we can dot

E!AB

with each of the unit vectors in the

xy´

system in turn. We already know that O|is given by Eq. (??) and we can find O{using

O{DO

K⇥OuA=B

ˇˇˇO

K⇥OuA=B ˇˇˇDO

K⇥1

L⇣lO

IdO

JChO

K⌘

ˇˇˇO

K⇥1

L⇣lO

IdO

JChO

K⌘ˇˇˇDdO

IlO

J

pd2Cl2:(14)

The unit vector O

kcan be found using its definition

ICdh O

JCd2Cl2O

K

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permission of McGraw-Hill, is prohibited.

Dynamics 2e 2167

2168 Solutions Manual

Computation. Substituting Eqs. (??)–(??) into Eqs. (1)–(3), we obtain

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permission of McGraw-Hill, is prohibited.