92 4. TOPIC D-4
z
1
O
x
M
ω
r
α
G
z
y
P
N
2
Φ
C
W
C
90°
N
1
Φ
e
n
W
e
n
Figure 4.31.
4.3 SOLUTION
e reference system Oxyz is attached to the rotating channel (tube) and the axis x coincides with the trajectory of
the relative motion of the particle M. A rotation of this system about axis z
1
is a transfer motion for the particle M.
A relative motion of the particle M is its motion along the tube. As the transfer motion is a rotational motion with
constant angular velocity, a relative motion of the particle M will be described with the following equation:
mw
r
= P
i
+ Φ
e
n
+ Φ
C
.
e forces acting on the particle M: the weight G
, the reaction force of the spring P
and the reaction force normal
to the tube wall N
. e reaction force N
can be resolved into two mutually perpendicular components: N
= N
1
+ N
2
.
93
Additionally, a normal component of the transfer inertia (eective) force Φ
e
n
and a Coriolis inertia force Φ
c
will be
added to the forces acting on the particle M:
Φ
e
n
= –mw
e
n
,
Φ
c
= –mw
C
.
e direction of the Coriolis acceleration w
C
is determined if we assume that the x component of the relative
velocity v
r
is positive. e Coriolis inertia force Φ
c
will be parallel to the axis Oy and it will be perpendicular to the
plane xOy (Figure 4.31).
e magnitudes of the normal component of the transfer inertia (eective) force Φ
e
n
and the Coriolis inertia
force Φ
c
will be determined as:
Φ
e
n
= mw
e
n
= mw
e
2
(r + x sin α ),
Φ
C
= mw
C
= 2mω
e
v
r
sin α,
where ω
e
= ω, v
r
= ||.
e reaction force of the spring P:
P = c(x – l
0
).
e equation of relative motion:
mw
r
= G
+ P
+ N
1
+ N
2
+ Φ
e
n
+ Φ
c
. (4.1)
e dierential equation of the relative motion of the particle M along axis Ox:
m = F
xi
= Φ
e
n
sin α – G cos α – P,
m = mω
2
(r + x sin α) sin α – mg cos α – c(x – l
0
),
+ (
c
– ω
2
sin
2
α) x = ω
2
r sin α – g cos α +
cl
0
. (4.2)
m
m
A general solution of the dierential equation (4.2) consists of the general solution x
*
of the respective homogeneous
dierential equation and a particular solution x
**
of the given inhomogeneous dierential equation (4.2):
x = x
*
+ x
**
.
e characteristic equation of the dierential equation (4.2) will be:
λ
2
+
c
– ω
2
sin
2
α = 0.
m
A solution of this quadratic equation:
λ
1
=
ω
2
sin
2
α –
c
=
π
2
∙ 0.5
2
–
1
= 9.876 i,
m
0.01
λ
2
= –9.876 i.
Hence, the general solution of the respective homogeneous dierential equation:
4.3 SOLUTION
94 4. TOPIC D-4
x
*
= C
1
cos 9.876t + C
2
sin 9.876t.
e particular solution of the inhomogeneous dierential equation (4.2) will be dened as:
= B.
From the dierential equation (4.2):
x
**
= B =
ω
2
r sin α – g cos α +
cl
0
=
π
2
∙ 0.2 ∙ 0.5 – 9.81 ∙ 0.866 +
1 ∙ 0.
= 0.128 m.
c
– ω
2
sin
2
α
1 – π
2
∙0.5
2
e solution of the dierential equation (4.2) of the relative motion of the particle M:
x = C
1
cos 9.876t + C
2
sin 9.876t + 0.128 (m). (4.3)
e velocity of the relative motion of the particle M:
= –9.876C
1
sin 9.876t + 9.876C
2
cos 9.876t (m⁄s). (4.4)
To nd the integral constants C
1
and C
2
we will use the initial conditions.
At t = 0, x
0
= 0.3 m,
0
= 2 m⁄s.
Equations (4.3) and (4.4) at t = 0:
x
0
= C
1
+ 0.128,
0
= 9.876 C
2
.
From here:
C
1
= 0.3 – 0.128 = 0.172,
C
2
=
2
= 0.202.
9.876
en the equation of the relative motion of the particle M:
x = 0.172 cos 9.876t + 0.202 sin 9.876t + 0.128 (m).
e relative velocity of the particle M:
= –1.69 sin 9.876t + 1.99 cos 9.987t (m⁄s).
To determine the components of the reaction force on the tube wall N
1
and N
2
at t = τ = 0.2 s the vector equation
(4.1) will be rewritten in the projections on the axes y and z. As the vector w
r
is perpendicular to both of these axes,
we will have:
0 = N
2
– Φ
C
,
0 = N
1
– G cos 60° – Φ
e
n
cos 30°.
From these equations:
N
2
= Φ
C
= 2mωv
r
sin α,
N
1
= G cos 60° + Φ
e
n
cos 30° = mg cos 60° + mω
2
(r + x sin α) cos 30°.
m
m
0.01
0.01
95
To determine the magnitudes of N
1
and N
2
we will calculate the coordinate x and the x components of the relative
velocity at t = 0.2 s:
x = 0.172 cos 9.876 ∙ 0.2
+ 0.202 sin 9.876 ∙ 0.2 + 0.128
=0.172 cos 113° + 0.202 sin 113° + 0.128
= –0.172 ∙ 0.391 + 0.202 ∙ 0.92 + 0.128 = 0.246 m;
= –1.69 sin 113° + 1.99 cos 113° = –1.69 ∙ 0.92 – 1.99 ∙ 0.391 = –1.55 – 0.78 = –2.33 m⁄s.
Hence, the reaction force components can be determined:
N
1
= 0.01 ∙ 9.81 ∙
1
+ 0.01π
2
(0.2 + 0.246 ∙ 0.5) ∙ 0.866 = 0.077 N,
2
N
2
= 2 ∙ 0.01π ∙ 2.33 ∙ 0.5 = 0.08 N.
en the reaction force on the tube wall:
N = N
2
+ N
2
=0.077
2
+ 0.08
2
= 0.111 N.
e unknown pressure force on the tube wall will be equal to N by magnitude and in the opposite direction to it.
5.1 DETERMINATION OF ABSOLUTE VELOCITY AND ABSOLUTE ACCELERATION OF PARTICLE
1 1
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