Nicolae–Doru Stănescu
Equilibrium without Friction of a Particle on a Mobile
Surface with Bilateral Constraints
In this paper we will study the equilibrium of a particle on a mobile sur-face in the case characterized by bilateral constraints between the par-ticle and the surface, and the absence of friction. Based on our previ-ous work, the conditions for the equilibrium are obtained. We prove that the positions of equilibrium on a mobile surface are no longer the same with those obtained for a fixed surface, the system could have ei-ther oei-ther equilibrium positions, completely different, or some more equilibrium positions, or no equilibrium position.
Keywords: mobile surface, equilibrium position, conditions of equilib-rium
1. Introduction
The dynamics without friction of the particle on a mobile surface with bilateral constraints was discussed in our previous work [4]. The approach was a multi-body type one similar to that presented in [2]. An equivalent approach [3] was also presented. The case of the fixed surface [1] is a particular case.
Let us consider that the particle of mass
m
and acted by the resultant forcek
j
i
F
=
F
X+
F
Y+
F
Z , (1)is situated on the mobile surface of equation
(
X
,
Y
,
Z
,
t
)
=
0
f
, (2)where
i
,j
, andk
are the unit vectors of the axesOX
,OY
, andOZ
, respec-tively,F
X,F
Y, andF
Z are the components of the resultant force on the axesOX
,OY
, andOZ
, respectively, whilet
is the time.In our previous work [4] we obtained the matrix equation of motion
[ ] [ ]
[ ]
{ }
{ }
{ }
[ ]
{ }
−
=
λ
−
q
B
C
F
q
B
B
m
ɺ
ɺ
ɺ
ɺ
ɺ
0
T
, (3)
ANALELE UNIVERSITĂŢII
“EFTIMIE MURGU” REŞIŢA
where
[ ]
=
m
m
m
0
0
0
0
0
0
m
, (4)[ ]
=
∂
∂
∂
∂
∂
∂
Z
f
Y
f
X
f
B
, (5){ }
∂
∂
−
=
t
f
C
, (6){ }
[
]
TZ Y
X
F
F
F
=
F
, (7){ }
[
]
TZ
Y
X
=
q
. (8)2. Equilibrium conditions
The parameter
λ
(Lagrange multiplier) is given by [4][ ][ ] [ ]
[
B
m
−1B
T]
{
{ }
C
−
[ ]
B
{ }
q
−
[ ][ ] { }
B
m
−1F
}
=
λ
ɺ
ɺ
ɺ
(9)Let us assume that the point
P
of coordinatesX
P,Y
P andZ
P situated on the mobile surface of equation (2) is an equilibrium point. It results that the coor-dinatesX
P,Y
P,Z
P, and their derivatives are always known; hence,λ
is always known (according to the formula (9)).Moreover, from the relation (3) we get
[ ]{ } [ ]
m
q
ɺ
ɺ
−
B
Tλ
=
{ }
F
, (10) and, taking into account that{ }
q
ɺ
ɺ
is also known, it results that the equation (10) leads to the expression of the forceF
.The inverse problem, that is, one knows the expression of the force
F
and asks for the equilibrium positions, is more difficult (in fact, it is a difficult problem even in the case of the fixed surface) and must be discussed depending on each particular case.A few examples will clarify the problem.
3. Example 1
The particle is situated on the mobile sphere of equation
(
X
,
Y
,
Z
,
t
) (
=
X
−
a
cos
ω
t
)
2+
Y
2+
Z
2−
R
2=
0
f
(11)and is acted by a force
F
given byi
Denoting
X
f
b
∂
∂
=
1 ,
Y
f
b
∂
∂
=
2 ,
Z
f
b
∂
∂
=
3 , (13)
we obtain the equations of motion
X
F
b
X
m
ɺ
ɺ
−
1λ
=
,m
Y
ɺ
ɺ
−
b
2λ
=
0
,m
Z
ɺ
ɺ
−
b
3λ
=
0
. (14) Taking into account the expression (11), we obtain thatY
=
ct
,Z
=
ct
(at the equilibrium the coordinatesY
andZ
are constant); henceY
ɺ
ɺ
=
0
,Z
ɺ
ɺ
=
0
, and the second and the third equations (14) become0
2
λ
=
b
,b
3λ
=
0
. (15)There are two possible cases. The first one implies
0
3 2
=
b
=
b
, (16)λ
being arbitrary, that is (recall the expressions (13))0
2
2
Y
=
Z
=
, (17)wherefrom
0
=
=
Z
Y
. (18)From the equation (11) we have
(
X
−
a
cos
ω
t
)
2=
R
2, (19)wherefrom
t
a
R
X
=
±
+
cos
ω
, (20)t
a
X
ɺ
=
−
ω
sin
ω
, (21)t
a
X
ɺ
ɺ
=
−
ω
2cos
ω
. (22)Replacing now these expressions in the first relation (14), we get
(
R
a
t
)
F
Xt
ma
ω
ω
−
±
+
ω
λ
=
−
2cos
2
cos
, (23)(
R
a
t
)
t
ma
F
Xω
+
±
ω
ω
−
−
−
λ
cos
2
cos
2
. (24)
One may easily observe that the parameter
λ
and, consequently, the normal reactionN
,t
ma
F
X
f
N
=
−
X−
ω
ω
∂
∂
λ
=
2cos
, (25)have not constant values as in the case of the fixed surface.
The second case is characterized by
λ
=
0
. In this situation, the first equa-tion (14) leads toX
F
X
m
ɺ
ɺ
=
. (26)Again,
Y
=
ct
,Z
=
ct
, and the equation (11) offerst
a
K
X
=
+
cos
ω
, (27)2 2
2
Y
Z
R
K
=
±
−
−
. (28)It results
t
a
X
ɺ
=
−
ω
sin
ω
, (29)t
a
X
ɺ
ɺ
=
−
ω
2cos
ω
(30)and the equilibrium is possible if and only if the force
F
X is given byt
ma
F
X=
−
ω
2cos
ω
. (31)4. Example 2
The particle is situated on the same mobile sphere described by the equation (11), but it is acted by its own weight,
k
F
=
−
mg
. (32)One asks for the equilibrium positions. We obtain the equations of motion
0
1
λ
=
−
b
X
m
ɺ
ɺ
,m
Y
ɺ
ɺ
−
b
2λ
=
0
,m
Z
ɺ
ɺ
−
b
3λ
=
−
mg
. (33)Again,
Y
=
ct
,Z
=
ct
; henceY
ɺ
ɺ
=
0
,Z
ɺ
ɺ
=
0
.The relation
λ
=
0
is not a convenient one due to the third relation (33) which would lead tog
=
0
.From the third relation (33) we get
mg
Z
=
−
λ
−
2
, (34)ct
2
=
=
λ
Z
mg
. (35)
The second relation (33) offers
0
2
λ
Y
=
, (36)wherefrom
0
=
Y
. (37)The equation of the sphere reads now
(
)
2 2 2 2cos
t
R
Z
K
a
X
−
ω
=
−
=
, (38)where
K
is a constant. It resultst
a
K
X
=
±
+
cos
ω
, (39)t
a
X
ɺ
=
−
ω
sin
ω
, (40)t
a
X
ɺ
ɺ
=
−
ω
2cos
ω
(41)and the first equation (33) reads
(
cos
)
0
cos
2
ω
+
±
+
ω
=
ω
K
a
t
Z
g
t
a
, (42)(
)
t
a
t
a
K
g
Z
ω
ω
ω
+
±
−
=
cos
cos
2 . (43)
Since
Z
=
ct
, we obtaina
=
0
, which is absurd.In conclusion, there is no equilibrium position in this case.
5. Example 3
Let us assume that the particle is situated on the surface
(
X
,
Y
,
Z
,
t
)
=
X
sin
(
θ
0cos
t
)
+
Y
cos
(
θ
0cos
t
)
=
0
f
, (44)that is, a vertical plan that contains the axis
OZ
and makes with the axisOX
an anglet
cos
0
θ
=
θ
. (45)The particle is acted by a force
F
.One asks for the conditions in which the equilibrium position is given by the point
P
of coordinates(
t
)
d
X
P=
cos
θ
0cos
,Y
P=
d
sin
(
θ
0cos
t
)
,Z
P=
0
. (46) We have to work with the formula (10).One successively obtains
(
t
)
t
d
X
ɺ
P=
θ
0sin
θ
0cos
sin
, (47)(
t
)
d
t
(
t
)
t
d
X
Psin
cos
0cos
0cos
sin
0cos
22
0
θ
+
θ
θ
θ
−
=
ɺ
ɺ
, (48)(
t
)
t
d
Y
ɺ
P=
−
θ
0sin
cos
θ
0cos
, (49)(
t
)
d
t
(
t
)
t
d
Y
P 2sin
2sin
0cos
0cos
cos
0cos
0
θ
−
θ
θ
θ
−
=
ɺ
ɺ
, (50)0
=
P
Z
ɺ
, (51)0
=
P
Z
ɺ
ɺ
, (52)(
t
)
X
f
cos
sin
θ
0=
∂
∂
, (53)(
t
)
Y
f
cos
cos
θ
0=
∂
∂
, (54)0
=
∂
∂
Z
f
, (55)
(
t
)
t
Y
(
t
)
t
X
t
f
P
P
cos
θ
0cos
sin
+
sin
θ
0cos
sin
−
=
∂
∂
, (56)
[ ]
B
=
[
sin
(
θ
0cos
t
) (
cos
θ
0cos
t
)
0
]
, (57)[ ]
B
ɺ
=
[
−
cos
(
θ
0cos
t
)
sin
t
sin
(
θ
0cos
t
)
sin
t
0
]
, (58){ }
∂
∂
−
=
t
f
{ }
[
]
TP P
P
Y
Z
X
ɺ
ɺ
ɺ
ɺ
=
q
, (60)(
)
(
)
(
)
(
)
(
cos
)
cos
sin
(
cos
)
,
cos
sin
cos
sin
sin
cos
cos
cos
cos
sin
sin
sin
cos
cos
d
d
0 0
2
0 0
0 2
0
t
t
Y
t
t
Y
t
t
Y
t
t
X
t
t
X
t
t
X
t
f
t
P P
P P
P P
θ
+
θ
−
θ
+
θ
−
θ
−
θ
−
=
∂
∂
ɺ
ɺ
(61)
{ }
∂
∂
−
=
t
f
t
d
d
C
ɺ
. (62)From the relation (9) we deduce the parameter
λ
, while the relation (10) of-fers the forceF
.4. Conclusion
The problem of the equilibrium without friction of a particle on a mobile sur-face with bilateral constraints is a difficult one. The determination of the necessary force for a certain equilibrium position reduces at some derivations. The inverse problem, the determination of the equilibrium position maybe solved only by nu-merical methods in the general case. The equilibrium positions in the case of a mobile surface may be completely different comparing to those when the surface is maintained fixed.
References
[1] Pandrea N., Stănescu N.D., Mechanics, Didactical and Pedagogical Pub-lishing House, Bucharest, 2002.
[2] Pandrea N., Stănescu N.D., Dynamics of the Rigid Solid with General Constraints by a Multibody Approach, Wiley, Chichester, UK, 2015.
[3] Udwadia F.E., Kalaba R.E., Analytical Dynamics. A New Approach, Cambridge University Press, Cambridge, 1996.
[4] Stănescu N.D., Motion without Friction of a Particle on a Mobile Surface with Bilateral Constraints, in press.
Address:
• Assoc. Prof. Hab. PhD Eng. PhD Math. Nicolae–Doru Stănescu, Univer-sity of Pitești, Str. Târgul din Vale, nr. 1, 110040, Pitești,
