30 2. TOPIC AK-2
v
C
2y
= – v
C
2r
sin α = – ω
2
R
2
sin α , (2.8b)
v
C
4y
= 0 , (2.8c)
v
C
5y
= v
C
5r
sin β =
1
ω
2
R
2
1
–
r
4
sinβ, (2.8d)
2
R
4
v
C
6y
= v
C
5y
. (2.8e)
Hence, the y component of the momentum of the entire system:
K
y
= – (2m
2
+ m
3
) ω
2
R
2
sin α +
1
(m
5
+ m
6
) ω
2
R
2
1
–
r
4
sin β
2
R
4
or
K
y
= ω
2
R
2
1
(m
5
+ m
6
) 1
–
r
4
sin β – (2m
2
+ m
3
) sin α . (2.9)
2
R
4
e sum of the y components of the impulses of all external forces:
S
ext
=
– Gt cos α +
t
Ndt . (2.10)
iy
0
Substituting the expressions (2.9) and (2.10) into Equation (2.3b) we will obtain:
ω
2
R
2
1
(m
5
+ m
6
) 1
–
r
4
sin β – (2m
2
+ m
3
) sin α=
– Gt cos α +
t
Ndt. (2.11)
2
R
4
0
From here:
t
Ndt = Gt cos α + ω
2
R
2
1
(m
5
+ m
6
) 1
–
r
4
sin β – (2m
2
+ m
3
) sin α. (2.12)
0
2
R
4
Substituting the expression (2.12) in Equation (2.7) we obtain:
mx
̇
+ bx = ω
2
R
2
(2m
2
+ m
3
) (cos α + f sin α ) +
1
(m
5
+ m
6
) 1
–
r
4
(cos β – f sin β) + Gt(sin α
– f cos α) ,
2
R
4
(2.13)
where
ω
2
= ω
0
[1 – exp(–st)].
Dividing both sides of Equation (2.13) into m we obtain:
x
̇
+ ηx = h[1 – exp(–st)] – qt , (2.14)
where η =
b
,
m
h =
ω
0
R
2
(2m
2
+ m
3
)(cos α + f sin α ) + 1
1
(m
5
+ m
6
)1
–
r
4
(cos β – f sin β) ,
2
m
R
4
q = g(f cos α – sin α) .
e solution of Equation (2.14) consists of the general solution of the corresponding rst-order homogeneous equa-
tion x
̇
+ ηx = 0 and the particular solution of this equation:
x = x
1
+ x
2
. (2.15)
31
A general solution of the rst-order homogeneous equation is dened as:
x
1
= Cexp(–ηt).
A particular solution of Equation (2.14):
x
2
= Q
1
exp(–st) + Q
2
t + Q
3
.
Substituting the particular solution x
2
into Equation (2.14), we obtain:
–Q
1
s exp(–st) + Q
2
+ η[Q
1
exp(–st) + Q
2
t + Q
3
] = h[1 – exp(–st)]–qt .
Equating the coecients of the variable parameters exp(–st) and t, as well as the other terms, we will obtain the
following equations:
–Q
1
s + ηQ
1
= –h ,
ηQ
2
= –q ,
Q
2
+ ηQ
3
= h ,
from where:
Q
1
=
h
,
s – η
Q
2
= –
q
,
η
Q
3
=
h
+
q
.
η – η
2
Hence,
x = x
1
+ x
2
= C exp(–ηt) +
h
exp(–st) +
h
+
q
–
q
t . (2.16)
s–η η η
2
η
e constant C we will nd from the initial condition: at t = 0 x
0
= 0 .
Using this condition we can nd the coecient C :
C = –
h
–
1
q
+ h .
s–η
η
η
us, the equation of motion of the object 1 will be expressed as:
x =
h
[exp(–st) – exp(–ηt)] +
1
q
+ h[1 – exp(–ηt)] –
q
t . (2.17)
s–η
η
η
η
e velocity of the object 1 :
v=x
̇
=
h
[ηexp(–st) – sexp(–st)] +
q
+ h exp(–ηt) –
q
=
hs
[exp(–ηt) – exp(–st)] –
q
[1 – exp(–ηt)] . (2.18)
s–η
η
η
s–η
η
2.3 SOLUTION
32 2. TOPIC AK-2
e expression (2.18) reveals that the velocity of the object 1 after certain time interval will be equal to zero and
hence the object will stop. erefore, Equation (2.17) is valid only for this time of interval.
To study the motion of the given mechanical system we will apply the equation of the center of mass motion:
ma
⃗
C
= F
⃗
i
ext
, (2.19)
where F
⃗
i
ext
– is a geometrical sum of all external forces acting on the given system.
e dierential equations of the center of mass motion in algebraic form:
mx
̈
C
= F
ext
= F
ext
, (2.20a)
xi
x
my
̈
C
= F
ext
= F
ext
. (2.20b)
yi
y
e coordinates of the center of mass:
x
C
=
m
1
x
C
1
+ (2m
2
+ m
3
) x
C
2–3
+ m
4
x
C
4
+ m
5
x
C
5
+ m
6
x
C
6
,
(2.21a)
m
y
C =
m
1
y
C
1
+ (2m
2
+ m
3
) y
C
2–3
+ m
4
y
C
4
+ m
5
y
C
5
+ m
6
y
C
6
.
(2.21b)
m
Dierentiating the expressions (2.21a) and (2.21b) twice by time, and considering the following:
x
̈
C
1
= x
̈
, x
̈
C
2–3
= x
̈
C
2
, x
̈
C
6
= x
̈
C
5
, y
̈̈
C
1
= y
̈̈
1
= 0, y
̈̈
C
2–3
= y
̈̈
C
2
, y
̈̈
C
6
= y
̈̈
C
5
,
we will obtain:
x
̈
C
=
m
1
x
̈
+ (2m
2
+ m
3
)x
̈
C
2
+ m
4
x
̈
C
4
+(m
5
+ m
6
)x
̈
C
5
,
(2.22a)
m
y
̈
C
=
(2m
2
+ m
3
)y
̈
C
2
+ m
4
y
̈
C
4
+(m
5
+ m
6
)y
̈̈
C
5
.
(2.22b)
m
Using the x and y components of the velocities of the mass centers provided in Equations 2.25a and 2.25b and 2.8a
– 2.8d, we can determine the x and y components of the mass centers of the objects:
ẍ
C
2
=
dv
C
2
x
= ẍ –
dω
2
R
2
cos α ,
(2.23a)
dt
dt
ẍ
C
4
=
dv
C
4
x
= ẍ ,
(2.23b)
dt
ẍ
C
5
=
dv
C
5
x
= ẍ –
1
dω
2
R
2
1
–
r
4
cos α
,
(2.23c) dt
2 dt
R
4
33
ÿ
C
2
=
dv
C
5
y
= ẍ –
dω
2
R
2
sin α
,
(2.23d)
dt
dt
y
̈̈
C
4
= 0 , (2.23e)
ÿ
C
5
=
dv
C
5
Y
= –
1
R
2
dω
2
1
–
r
4
cos α
,
(2.23f) dt
2 dt
R
4
Substituting the expressions (2.23a – 2.23f) in Equations (2.22a) and (2.22b):
ẍ
C
=
(m
1
x
̈
+ 2m
2
+ m
3
)x
̈
–
dω
2
R
2
sin α + m
4
x
̈
+ (m
5
+ m
6
)x
̈
–
1
dω
2
R
2
1 –
r
4
cos β , (2.24a)
m
y
̈
C
=
–(2m
2
+ m
3
)
dω
2
R
2
sin α + (m
5
+ m
6
) ∙
1
R
2
dω
2
1 –
r
4
sin β , (2.24b)
m
or
ẍ
C
=
mx
̈ –
R
2
dω
2
(2m
2
+ m
3
) cos α +
1
(m
5
+ m
6
)1 –
r
4
cos β ,
m
y
̈
C
=
R
2
dω
2
1
(m
5
+ m
6
) (1 –
r
4
sin β –
(2m
2
+ m
3
) sin α .
m
e x and y components of the resultant vector of all external forces acting on the system:
F
ext
= F
ext
= G sin α –
bx
̇
–
F
fr
, (2.25a)
x
xi
F
ext
= F
ext
= –G cos α + N . (2.25b)
y
yi
Substituting the expressions (2.24a), (2.24b), (2.25a), and (2.25b) into the dierential Equations (2.20a) and (2.20b)
of the motion of the mass center of the system, we will obtain:
mx
̈
– R
2
dω
2
(2m
2
+ m
3
) cos α +
1
(m
5
+ m
6
)1 –
r
4
cos β = G sin α – bx
̇
– fN
dt
2 R
4
or
mx
̈
+ bx
̇
= R
2
dω
2
(2m
2
+ m
3
) cos α +
1
(m
5
+ m
6
)1 –
r
4
cos β + G sin α – fN (2.26a)
dt
2 R
4
and
dω
2
R
2
1
(m
5
+ m
6
)1 –
r
4
sin β – (2m
2
+ m
3
) sin α = – G cos α + N ] . (2.26b)
dt
2 R
4
Considering that the angular acceleration of the wheels 2 are equal:
ε
2
=
dω
2
= ω
0
s exp(–st) ,
dt
dt
dt
dt
dt
dt 2
2
2
R
4
R
4
R
4
R
4
2 dt
2.3 SOLUTION
34 2. TOPIC AK-2
from Equation (2.26b) we can nd the normal reaction force N :
N = ω
0
s exp(–st) R
2
1
(
m
5
+ m
6
)1 –
r
4
sin β – (2m
2
+ m
3
) sin α + G cos α. (2.27)
2 R
4
Substituting the expression (2.27) into Equation (2.26a) we will obtain the dierential equation of the motion of
the object 1:
x
̈̈
+ ηx
̇
= hs exp(–st) – q , (2.28)
where η =
b
,
m
h =
ω
0
R
2
(2m
2
+ m
3
)(cos α + f sin α) +
1
(m
5
+ m
6
)1 –
r
4
(cos β – f sin β) ,
m
q = g(f cos α – sin α) .
e general solution of the dierential equation (2.28) will be:
x = x
1
+ x
2
, (2.29)
where x
1
= C
1
+ C
2
exp(–ηt) – is the general solution of the homogeneous equation:
x
̈̈
+ ηx
̇
= 0 .
e particular solution x
2
will be dened as:
x
2
= Q
1
exp(–st) + Q
2
t .
Substituting the particular solution x
2
into Equation (2.28) we can nd Q
1
and Q
2
:
Q
1
s
2
exp(–st) – ηsQ
1
exp(–st) + ηQ
2
= hs exp(–st) – q .
Equating the coecients of the exp(-st) and the constant terms:
Q
1
(s – η) = h ,
ηQ
2
= –q ,
we will obtain:
Q
1
=
h
,
s – η
Q
2
= –
q
.
η
en the general solution of the dierential equation (2.28) will be:
x = x
1
+ x
2
= C
1
+ C
2
exp(–ηt) +
h
exp(–st) –
q
t . (2.30)
s – η
η
x
̇
=
dx
= –ηC
2
exp(–ηt)
hs
exp(–st) –
q
. (2.31)
dt
s – η
η
2
R
4
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