M n I F r 2010 er E ptember 29th, 2010 U your solu n gures ...
Transcript of M n I F r 2010 er E ptember 29th, 2010 U your solu n gures ...
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ME 352 - Ma
Fall Semeste
EXAM 1. O
Use the blanecessary, yorib sheet to
Problem 1 (2Part I. (i) Deabel the loweii) Define vehe direction oiii) Write th
mechanism; aonstraints th
Part II. For a
where 2 5R =
9θ for the Ne
he correction
achine Desig
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PEN BOOK
nk paper pou can use tthe end of p
25 Points). etermine theer pairs and tectors that arof each vect
he vector loand (b) the k
hen write the
a particular p
cm, 2 45θ =
ewton-Raph
ns 9*RΔ and
gn I
K AND CLO
provided forthe given fiproblem 1. A
e mobility othe higher pre suitable fotor on Figureoop equationknown quanconstraint e
planar mecha
5 ,° 1 8 cmR =
son techniqu
9*θΔ (to fou
OSED NOT
r your solugures to shAny work t
of the mechaairs on the fi
for a complete 1. n(s) for thisntities, the uequation(s).
Figure 1. A
anism, the v
IR2√
m, and 1θ =
ue are 9*R =
ur decimal p
1
Name
Lab.
TES.
utions. Writow vectors hat cannot
anism shownfigure. te kinematic
s mechanismnknown var
Planar Mec
ector loop eq
??R R9 1
√√− − =
0 .° The init
6 cm= and θ
laces) that a
e of Student
Div. Numb
We
te on one sand instantbe followed
n in Figure
c analysis of
m and identriables, and
chanism.
quation can
0=
tial estimates
9* 135 .θ = ° C
are to be used
t_________
er________
ednesday, Se
side of the t centers. A
d will assum
1. Clearly n
f the mechan
ify: (a) suitany constra
be written a
s of the posi
Calculate the
d in the next
___________
___________
eptember 29
paper onlyAttach your me to be wro
number each
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table input(sints. If you
as
ition variable
e numerical
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9th, 2010
y. Where one page
ong.
h link and
and show
s) for the identified
es 9R and
values of
M F Pin(iof(iva(ith(iG
ME 352 - Ma
Fall Semeste
Problem 2 (2nput link 2 isi) On Figuref the mechanii) Write theariables, andiii) Calculatehe vector froiv) Determin
Give the mag
achine Desig
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25 Points). Fs 2 12 kω = −e 2, draw andnism. e vector lood any constrae the numeri
om the grounne the angulagnitude and t
gn I
For the mechk rad / s . Thd clearly lab
op equationaints. If you ical values o
nd pivot 4O ar velocity othe direction
hanism in thhe length of tbel all the ve
ns. Clearly iidentified co
of the first-oto point A o
of link 4 andof each vec
Figure 2. A
2
Name
Lab.
he position sthe ground leectors that ar
indicate the onstraints thorder kinemaon link 3). d the velocitytor.
A planar mech
e of Student
Div. Numb
shown in Figength is 2O Ore required f
input, the hen write theatic coeffici y of point A
hanism.
t_________
er________
gure 2, the a4O 10 cm.=
for a comple
known quae constraint eients 4θ ′ and
A relative to
___________
___________
angular veloc ete kinemati
antities, the equations. d 43R′ (wher
the ground p
________
________
city of the
c analysis
unknown
re 43R is
pivot 4O .
M F Pco(i(i
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ME 352 - Ma
Fall Semeste
Problem 3 (2ounterclockwi) List the prii) Show the
Using theiii) The first-iv) The magn
achine Desig
er 2010
25 Points). Fwise with animary instanlocations of
e locations of-order kinemnitudes and d
gn I
For the mechn angular velnt centers andf all instant cf the instant
matic coefficidirections of
Figure 3.
hanism in thlocity 2 9ω =d the secondcenters on Ficenters, deteients of linkf the angular
The Planar M
3
Name
Lab.
he position s9 rad / s. Thedary instant cigure 3. Pleaermine: s 3, 4, and lir velocities o
Mechanism.
e of Student
Div. Numb
shown in Fige mechanismcenters for thase use the K
ink 7. of links 3 and
Drawn full
t_________
er________
gure 3, the inm is drawn fuhe mechanis
Kennedy circ
d 4.
scale.
___________
___________
nput link 2 iull scale in thsm. cle shown be
________
________
is rotating he figure.
elow.
M F Pcocian(iof(iva(i(i(v
ME 352 - Ma
Fall Semeste
Problem 4 (2ounterclockwircular grounnd 3 are 2O Ai) On Figuref the mechanii) Write theariables, andiii) Determiniv) The anguv) The angul
achine Desig
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25 Points). Fwise with annd surface aA AB 8 c= =
e 4, draw andnism. e vector lood any constrane the first-oular velocity lar velocity o
gn I
For the mechn angular velat point C. Tcm. d clearly lab
op equationaints. If you rder kinemaof link 3. Gi
of link 4. Giv
F
hanism in thlocity 2ω =he radius of
bel all the ve
ns. Clearly iidentified co
atic coefficieive the magnve the magn
Figure 4. The4
Name
Lab.
he position s15 rad / s. Thf the ground
ectors that ar
indicate the onstraints th
ents of links nitude and th
nitude and th
e Planar Me
e of Student
Div. Numb
shown in Fighe wheel, lin
d link is 1ρ =
re required f
input, the hen write the3 and 4. he direction
he direction o
echanism.
t_________
er________
gure 4, the innk 4, is in ro10 cm= and
for a comple
known quae constraint e
of this vectoof this vector
___________
___________
nput link 2 iolling contacthe lengths
ete kinemati
antities, the equations.
or. r.
________
________
is rotating ct with the of links 2
c analysis
unknown
S(i
F
Su
T
Nm
OLUTION i) The labels
The Kutzb
or the wheel
ubstituting E
Therefore, the
Note that if thmobility of th
TO PROBL on the link
Fi
bach mobilit
l rolling with
Equation (2)
e mobility of
he wheel is he mechanism
LEM 1. numbers and
gure 1. Num
ty criterion c
hout slipping
n 7=
into Equatio
m
f the mechan
rolling and m is
m =
d the joint ty
mber of links
can be writte
m 3(n=
g on the grou
17, j 8=
on (1) gives
m 3(7 1)= − −
nism is
m
slipping on
3(7 1) 2(7− −
5
ypes in the m
and joint ty
en as
1n 1) 2( j )− − −
und link then
, and
2(8) 1(1)− − =
m 1=
the ground
7) 1(2) 18− =
mechanism a
ypes in the m
21( j )−
n
2j 1=
18 16 1= − −
link then 1j
8 14 2 2− − =
are shown on
mechanism.
7= and 2j
2
n Figure 1.
2.= In this
(1)
(2)
(3a)
(3b)
s case, the
(4)
(i
(i
an
(a(banT
ii) The vecto
iii) Three vec
nd
a) The input b) The knownd θ7.
The unknownThere are
ors for a kine
ctor loop equ
for the mechwn quantities
n variables arfour constra
9 14 9θ = θ −
ematic analy
Figure
uations are n
hanism is ths are the sev
re the three laint equation
0 ,° 2θ
ysis of the me
2. The vecto
necessary for
I2R R√ √
+
33 2
CR R√ √Ι
−
I2R R√
−
e angular poven lengths R
lengths R14, ns, namely
26 2,= θ
6
echanism are
ors for a kine
r the kinema
3 14R R√? ?√ √
− −
2 26
? ?CR R
Ι √+ +
5533C C
R R√ √
+ −
osition of linR2, R3, R33,
R26, and R7,
33 3,θ = θ
e shown in F
ematic analy
atic analysis
09C
R√
=
6 5
? ?0R R
√− =
?07R
√− =
nk 2, that is, θR6, R5, R55
, and the thr
and
Figure 2.
ysis.
of this mech
θ2. 5, and R9, an
ee angles θ3
55θ = θ
hanism, nam
nd the two a
, θ5, and θ6.
5θ + γ
mely
(5a)
(5b)
(5c)
angles θ14,
(6)
7
Since there is rolling contact between link 4 (the pinion) and the ground link 1 (the rack) then the rolling constraint equation can be written as
14 4 4 14R ( )± Δ = ρ Δθ −Δθ (7a)
The positive sign must be used for the mechanism in the given position, that is, the vector 14R is getting longer for a counterclockwise rotation of link 4, or the vector is getting shorter for a clockwise rotation of link 4. Therefore, Equation (7a) can be written as
14 4 4 14R ( )+ Δ = ρ Δθ −Δθ (7b) (iv) From the given VLE, the X and Y components of the error vector ε can be written as
* *2 2 32 32 1 1 XR cos R cos R cosθ − θ − θ = ε (1a)
and * *
2 2 32 32 1 1 YR sin R sin R sinθ − θ − θ = ε (1b) Taking the partial derivates of Equation (1a) with respect to the estimates of the of the position variables gives
*X32*
32cos
R∂ε
= − θ∂
and * *X32 32*
32R sin∂ε
= + θ∂θ
(2a)
Also, taking the partial derivates of Equation (8b) with respect to the estimates of the position variables gives
*Y32*
32sin
R∂ε
= − θ∂
and * *Y32 32*
32R cos∂ε
= − θ∂θ
(2b)
From Cramers rule, the coefficient matrix can be written as
X X* *32 32
Y Y* *32 32
R
R
∂ε ∂ε⎡ ⎤⎢ ⎥∂ ∂θ⎢ ⎥⎢ ⎥∂ε ∂ε⎢ ⎥∂ ∂θ⎢ ⎥⎣ ⎦
(3a)
Substituting Equations (2) into Equation (3a), the coefficient matrix is
* * *32 32 32* * *32 32 32
cos R sin
sin R cos
⎡ ⎤− θ + θ⎢ ⎥⎢ ⎥− θ − θ⎣ ⎦
(3b)
The matrix equation for the corrections to the position variables can be written as
* * * *32 32 32 32 X* * * * Y32 32 32 32
cos R sin R
sin R cos
⎡ ⎤ ⎡ ⎤− θ + θ Δ − ε⎡ ⎤⎢ ⎥ ⎢ ⎥ = ⎢ ⎥− ε⎢ ⎥ ⎢ ⎥ ⎣ ⎦− θ − θ Δθ⎣ ⎦ ⎣ ⎦
(4)
8
Using Cramer’s rule, the corrections can be written from Equation (11) as
* * * ** X 32 32 Y 32 3232
R cos R sinRDET
ε θ + ε θΔ = (5a)
and * *
* 32 Y 32 X32
cos sinDET
θ ε − θ εΔθ = (5b)
The determinant of the coefficient matrix, see Equation (3b), is
* * * * * * *32 32 32 32 32 32 32DET ( cos )( R cos ) ( sin )( R sin ) R= − θ − θ − − θ + θ = (6)
The initial estimates of the position variables are *9 6 cmR = and *
9 135 .θ = ° Substituting this information and the given data into Equations (1), the X and Y components of the error vector are
X 5cos45 6cos135 8cos0 0.2218 cmε = °− °− ° = − (7a) and
Y 5sin 45 6sin135 8sin 0 0.7071cmε = °− °− ° = − (7b) Also, substituting the initial estimates of the position variables into Equation (6), the determinant is
*32DET R 6 cm= = (8)
Substituting Equations (7) and (8) into Equations (5), the corrections can be written as
*32
( 0.2218 cm)(6 cm)cos135 ( 0.7071 cm)(6 cm)sin135R6 cm
− °+ − °Δ = (9a)
and *32
cos135 ( 0.7071cm) sin135 ( 0.2218 cm)6 cm
° − − ° −Δθ = (9b)
Therefore, the corrections that are required for the next iteration are
2 2*32
0.9410 cm 3.0000 cmR 0.3432 cm6 cm−
Δ = = − (10a)
and *32
0.5000 cm 0.1568 cm 0.1095 rad6 cm+
Δθ = = + (10b)
S(i
an
TR
F
(i
an
OLUTION i) The vector
There are
nd the secon
The input is tR43, R45, and
rom the geo
ii) The X and
nd
TO PROBLrs that are su
Figu
two indepen
nd vector loo
he angular pR5 and the a
metry of the
d Y compon
LEM 2 uitable for a v
ure 1. The v
ndent vector
op equation i
position of thangular posit
e mechanism
43R =
nents of Equa
velocity ana
ectors for a v
r loops for th
IR2√
is I
R2√
he input linktion of link 4
43θ =
m, the length
45 2 cR R= = =
ation (1a) are
2 2cosR Rθ −
2 2sinR Rθ −
9
alysis of the m
velocity ana
his mechanis
??R R43 1
√√− −
45
? ?R R5C √
+ −
k 2, that is, θ24, namely, θ4
45 4θ θ= =
of the input
5 10cos30 3
=°
e
43 4cosR Rθ −
43 4 1sinR Rθ −
mechanism
alysis of the m
sm. The first
01√
=
0=
2. The four u4. The constr
link is
cm 5.77 cm=
1 1cos 0R θ =
1 1sin 0θ =
are as shown
mechanism.
t vector loop
unknown varraint equatio
m
n in Figure 1
p equation is
riables are thon is
1.
(1a)
(1b)
he lengths
(2)
(3)
(4a)
(4b)
10
Differentiating Equations (4) with respect to the input position gives
2 2 43 4 43 4 4sin cos sin 0R R Rθ θ θ θ′ ′− − + = (5a) and
2 2 43 4 4 4 4cos sin sin 0R R Rθ θ θ θ′ ′− − = (5b) Writing Equations (5) in matrix form gives
4 43 4 43 2 2
4 43 4 4 2 2
cos sin sinsin cos cos
R R RR R
θ θ θθ θ θ θ
′− + +⎡ ⎤ ⎡ ⎤ ⎡ ⎤=⎢ ⎥ ⎢ ⎥ ⎢ ⎥′− − −⎣ ⎦ ⎣ ⎦⎣ ⎦
(6)
Solving Equation (6) using Cramer’s rule, the first-order kinematic coefficients can be written as
2 43 4 243
sin( )R RR
DETθ θ−′ = (7a)
and 2 4 2
4cos( )R
DETθ θ
θ−′ = (7b)
where the determinant is 43DET R= (8)
Substituting Equation (8) into Equations (7), the first-order kinematic coefficients can be written as
2 43 4 243 2 4 2
43
sin( )sin( )
R RR R
Rθ θ
θ θ−′ = = − (9a)
and 2 4 2
4 4 243
cos( ) cos( )RRθ θ
θ θ θ−′ = = − (9b)
Substituting the given data into Equations (9), the first-order kinematic coefficients for the mechanism are
43 cm10 sin(150 30 ) 5 cm/rad3
R′ = ° − ° = + (10a)
The positive sign indicates that the vector 43R is getting shorter for a clockwise rotation of the input link and
4 cos(150 30 ) 0.5 rad/radθ ′ = ° − ° = − (11) The negative sign indicates that link 4 is rotating counterclockwise (for the input rotating clockwise). (iv) 5 Points. The angular velocity of link 4 is
3 4 2 0.5 ( 12 rad/s) 6 rad/sω θ ω′= = − − = + (12) The angular velocity of link 4 is counterclockwise.
The velocity of point A relative to the ground pivot 4O is
34 43 2 5 cm / rad ( 12 rad/s) 60 cm/sV R ω′= = + − = − (13)
This velocity is acting towards the ground pivot 4O .
S
(ian(i1li2in3li4li5li6li7li8li9li10li
OLUTION The numb
i) 5 points. Tnd 10 secondii) 5 Points. . Connect innes gives the. Connect inntersection o. Connect innes gives the. Connect innes gives the. Connect innes gives the. Connect innes gives the. Connect innes gives the. Connect innes gives the. Connect innes gives the0. Connect ines gives the
TO PROBLber of instant
There are 11dary instant The procedu
nstant centere instant cennstant cente
of these two lnstant centere instant cennstant centere instant cennstant centere instant cennstant centere instant cennstant centere instant cennstant centere instant cennstant centere instant ceninstant centee instant cen
LEM 3 t centers for
1 primary incenters, nam
ure to locate rs I12 and I2nter I15. rs I15 and Ilines gives thrs I12 and I1nter I26. rs I12 and I1nter I27. rs I16 and I1nter I67. rs I14 and I1nter I47. rs I14 and I1nter I46. rs I13 and I1nter I36. rs I13 and I1nter I37. ers I13 and Inter I35.
this mechan
nN =
nstant centersmely, I15, I56,
the seconda5. Connect i
16. Draw a he instant ce6. Connect i
7. Connect i
7. Connect i
7. Connect i
6. Connect i
6. Connect i
7. Connect i
15. Connect
Figure 1. T11
nism is
( 1) 7(72 2
n − −=
s, namely, I I26, I27, I67,
ary instant ceinstant cente
line perpenenter I56. instant cente
instant cente
instant cente
instant cente
instant cente
instant cente
instant cente
instant cent
he Kennedy
1) 212− =
12, I13, I14, I1I47, I46, I36, Ienters (see thers I17 and I
ndicular to th
ers I25 and I
ers I25 and I
ers I56 and I
ers I45 and I
ers I47 and I
ers I34 and I
ers I34 and I
ters I37 and I
y Circle.
16, I17, I23, I2I37, and I35. he Kennedy I57. The inte
he slot thro
I56. The inte
I57. The inte
I57. The inte
I57. The inte
I67. The inte
I46. The inte
I47. The inte
I57. The inte
24, I25, I34, I4 circle) is:
ersection of
ugh the con
ersection of
ersection of
ersection of
ersection of
ersection of
ersection of
ersection of
ersection of
(1)
45, and I57;
these two
ntact. The
these two
these two
these two
these two
these two
these two
these two
these two
T
(i
wm
T
TarT
w
The construct
iii) 5 Points.
where the pomeasurement
Therefore, the
The negative rm). Therefo
The rolling co
where the neg
tion of the se
. The rolling
ositive sign ts and the kn
e first-order
sign indicatore, gear 4 isontact equati
gative sign m
econdary ins
Figure 2. T
g contact equ
must be unown kinema
kinematic c
tes that gear s rotating in tion between
must be used
stant centers
The location
uation betwe
1
4
ρρ
± =
used here beatic coefficie
4 cm1cm
+
oefficient of
4θ′ = −
4 is rotatingthe clockwis gear 3 and g
4
3
ρρ
± =
d here becaus
12
is shown in
of the prima
en gear 4 an
4 2
1 2
θ θθ θ′ ′−
=′ ′−
ecause of thents into Equ
4 1mm 0 1
θ ′ −=
−
f gear 4 is
3 rad/rad−
g in the oppse direction. gear 4 can b
3 2
4 2
θ θθ θ′ ′−
=′ ′−
se of externa
n Figure 2.
ary instant c
nd the fixed g
he internal uation (2a) g
osite directi e written as
al rolling con
centers.
gear 1 can be
rolling contgives
ion to the inp
ntact.
e written as
tact. Substit
put link 2 (t
(2a)
tuting the
(2b)
(2c)
that is, the
(3a)
13
Substituting the measurements and the known kinematic coefficients into Equation (3a) gives
3
4
11cm2 cm 1
θθ′ −
− =′ −
(3b)
Substituting Equation (2c) into Equation (3b), the first-order kinematic coefficient of gear 3 is
3 3 rad/radθ′ = + (3c) The positive sign indicates that gear 3 is rotating in the same direction to the input link 2 (that is, the arm). Therefore, gear 3 is rotating in the counterclockwise direction. Alternative approach. The first-order kinematic coefficient for link j can be written as
12 2
1 2
jj
j j
I II I
θ ′ = (4a)
Therefore, the first-order kinematic coefficient for gear 4 can be written as
12 244
14 24
I II I
θ ′ = (4b)
The distances required to obtain the first-order kinematic coefficient of link 4 are measured as
12 24 3 cmI I = and 14 24 1 cmI I = (5a) Substituting Equation (5a) into Equation (4b), the first-order kinematic coefficient of link 4 is
43 cm 3 rad/rad1 cm
θ ′ = = (5b)
Note that the relative instant center 24I is between the two absolute instant centers 12 14and .I I Therefore, the first-order kinematic coefficient for gear 4 must be negative, that is
4 3 rad/radθ ′ = − (5c) Similarly, the first-order kinematic coefficient for gear 3 can be written as
12 233
13 23
I II I
θ ′ = (6)
Note in this case that the instant centers 12 23 13, andI I I are coincident, see Figure 2. Therefore, the first-order kinematic coefficient of link 3 cannot be obtained from this equation.
Consider the point of contact between gears 3 and 4. Denote this point as A (note that point A on gear 3 is coincident with point A on gear 4 (that is, point A is the instant center 34I ). Therefore, the velocity of point A can be written as
A A3 4V V= (7a)
which can be written as 3 13 34 4 14 34( ) ( )I I I Iω ω= (7b)
14
Dividing both sides of Equation (7b) by the input angular velocity 2ω gives
3 13 34 4 14 34( ) ( )I I I Iθ θ′ ′= (7c) Then rearranging this equation, the first-order kinematic coefficient of link 3 can be written as
14 343 4
13 34
( )I II I
θ θ′ ′= (7d)
The lengths necessary to solve for the first-order kinematic coefficient of link 3 are measured to be
13 34 2 cmI I = and 14 34 2 cmI I = (8a) Substituting Equations (5c) and (8a) into Equation (7d) gives
32 3( )( ) 3 rad/rad2 1
θ ′ = ± − = ± (8b)
Note that the relative instant center 34I is between the absolute instant centers 13 14, and .I I Therefore, the first-order kinematic coefficient for gear 3 has the opposite sign as the first-order kinematic coefficient of gear 4. Therefore, the first-order kinematic coefficient of link 3 must be positive, that is
3 3 rad/radθ ′ = + (9)
The first-order kinematic coefficient for a slider, say link k, can be written as
12 2j jR I I′ = (10a) Therefore, the first-order kinematic coefficient for link 7 can be written as
7 12 27R I I′ = (10b) The distance required for the first-order kinematic coefficient of link 7 is measured as
12 27 4.75 cmI I = (11) Substituting Equation (11) into Equation (10b), the first-order kinematic coefficient for link 7 is
7 4.75 cmR′ = − (12)
The negative sign must be used here because the vector 7R is getting shorter for a positive change in the position of the input link 2, see Figure 4. In other words, link 7 is moving upward for a positive change in the rotation of the input link 2. (iv) 5 points. The angular velocity of gear 3 can be written as
3 3 2ω θ ω′= (13a) Substituting Equation (9) and the input angular velocity into Equation (13) gives
3 ( 3)( 9) 27 rad/sω = + + = + (13b) The positive sign indicates that gear 3 is rotating counterclockwise.
Su
T
The angul
ubstituting E
The negative
lar velocity o
Equation (6)
sign indicat
F
of gear 4 can
and the inpu
es that gear
Figure 4. A s
n be written
ut angular ve
4 (ω =
4 is rotating
suitable choi
15
as
4 4 2ω θ ω′=
elocity into E
( 3)( 9)− + =
g clockwise.
ice of vector
Equation (15
27 rad/s−
rs for the me
5) gives
echanism.
(14a)
(14b)
S
tr
U
T
Su
(i
T
T
an
D
an
T
OLUTION Note that
riangle, that
Using the sum
Therefore, the
ubtracting th
i) 5 points. A
The vector lo
The X and Y
nd
Differentiting
nd
Then writing
TO PROBLt the lengthsis
m of the inter
e triangle O2
he length of
A suitable ch
op equation
components
g Equation (2
Equations (3
LEM 4. s of link 2
rior angles o
2AB is an eq
O2B from th
hoice of vect
Figu
can be writt
s of Equation
2) with respe
3) in matrix
−⎡⎢+⎣
and link 3
2AO
of a triangle,
2 18O AB =
quilateral tria
2O A =
he radius of t
4ρ ρ=
tors for the m
ure 1. The ve
ten as
C2R R
√ √+
n (1) can be
2 2cosR Rθ +
2 2sinR Rθ +
ect to the inp
2 2sinR θ− −
2 2cosR θ+ +
form gives
3 3
3 3
sincos
RR R
θθ
− ++ −
16
are equal, t
2B O BA= =
280 AO B° −
angle and
2AB O B= = =
the ground l
1 2 2 cO Bρ − =
mechanism a
ctors for the
9
? ?03R R
√ √− =
written as
3 3 9cosR Rθ −
3 3 9sin sR Rθ −
put position
3 3 3sinR θ θ ′ +
3 3 3cosR θ θ ′+ −
9 9 3
9 9 9
sincos
RR
θ θθ θ
′⎤ ⎡⎥ ⎢ ′⎦ ⎣
therefore, th
60= °
2 6O BA− =
8 cm
link then the
cm
are shown in
e mechanism
9cos 0θ =
9sin 0θ =
θ2 gives
9 9 9sinR θ θ ′ =
9 9 9cosR θ θ ′− =
3 2
9 2
sincos
RR
θ ′ +⎤ ⎡=⎥ ⎢′ −⎣⎦
he triangle O
60°
radius of lin
n Figure 1.
m.
0=
0=
2
2
nsθθ
⎤⎥⎦
O2AB is an
nk 4 is
isosceles
(1a)
(1b)
(1c)
(1d)
(1)
(2a)
(2b)
(3a)
(3b)
(4)
17
Solving Equation (4) using Cramer’s rule, the first-order kinematic coefficients can be written as
2 9 2 93
sin( )R RDET
θ θθ
− −′ = (5a)
and 2 3 2 3
9sin( )R RDET
θ θθ
− −′ = (5b) where the determinant of the coefficient matrix can be written as
3 9 3 9sin( )DET R R θ θ= − (6) Substituting Equation (6) into Equations (5), the first-order kinematic coefficients can be written as
2 2 93
3 3 9
sin( )sin( )
RR
θ θθ
θ θ− −′ =
− (7a)
and 2 2 3
99 3 9
sin( )sin( )
RR
θ θθ
θ θ− −′ =
− (7b)
Substituting the position data into Equations (7), the first-order kinematic coefficients are
38 cm sin( 60 0 ) 1 rad/rad8 cm sin(60 0 )
θ − − ° − °′ = = +° − °
(8a)
and
98 cm sin( 60 60 ) 1 rad/rad8 cm sin(60 0 )
θ − − ° − °′ = = +° − °
(8b)
The positive sign for the first-order kinematic coefficients in Equations (8) indicate that links 3 and 9 are rotating in the same angular velocity as the input link 2. The answers in Equations (8) also imply that links 3 and 9 have the same angular velocities as the input link 2. Note. For the given input position and link dimensions the answers in Equations (8) should be intuitively obvious. Since the lengths of O2A, AB, and O2B are equal then O2AB is a rigid equilateral triangle rotating about the ground pivot O2. Since the triangle experiences rigid body motion then the angular velocity of this triangle must be same as the angular velocity of link 2. Therefore, the first-order kinematic coefficients of links 3 and 9 must be 1. (ii) 5 points. To determine the angular velocity of link 4, consider the rolling contact equation between links 1 and 4. Recall that the general form of the rolling contact equation between a gear and a pinion can be written as
gear pinion arm
pinion gear arm
ρ θ θρ θ θ
Δ − Δ± =
Δ − Δ (9)
where the positive sign is used for internal contact and the negative sign is used for external contact.
For the given mechanism the gear is link 1, the pinion is link 4, and the arm is link 2. Substituting this notation into Equation (9) gives
1 4 2
4 1 2
ρ θ θρ θ θ
Δ − Δ+ =
Δ − Δ (10)
18
Differentiating Equation (10) with respect to the input position θ2 gives
1 4
4
10 1
ρ θρ
′ −+ =
− (11)
Solving Equation (11) for the first-order kinematic coefficient of link 4 gives
14
41 ρ
θρ
′ = − (12)
Then substituting the radius of the ground and the radius of link 4 into Equation (12) gives
410 cm1 4 rad/rad2 cm
θ ′ = − = − (13)
(iii) 5 points. The angular velocity of link 3 can be written as
3 3 2ω θ ω′= (14a) Substituting the known values into Equation (14a) gives
3 ( 1 rad/rad)( 15 rad/s) 15 rad/sω = + + = + (14b) The positive sign indicates that the angular velocity of link 3 is counterclockwise. (iv) 5 points. The angular velocity of link 4 can be written as
4 4 2ω θ ω′= (15a) Substituting the known values into Equation (15a) gives
4 ( 4 rad/rad)( 15 rad/s) 60 rad/sω = − + = − (15b) The negative sign indicates that the angular velocity of link 4 is clockwise.