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Файл:Integrational mechanics. Lecture and exercises
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4.4. Dynamics of rotation of a system of rigid bodies 131
Figure 4.10
Solution. As an information which holds in different experiments one can
regard: a moment of inertia of the flywheel J ,massm of the weight, distance h ,
and time t. It is known that the moment of inertia of a body with respect to some
axes may be considered as the measure of inertia of a body being in a rotational
motion. Using Newton’s second law one can compare the measure of inertia
with characteristics of motion (distance, time).
d
(mV )=F,
dt
where mV — linear momentum of a body.
Based on the analogy between translation and rotation we get differential
equation of rotational motion for a system of particles:
d
(L
)=Moz(Fe),
oz
dt
where L
weight with respect to the oz –axes.
where M
= Jw+ rmV — moment of momentum of the wheel and the
oz
(Fe) — sum of moments of external forces about the oz –axes.
M
oz
d
(Jw + rmV )=mgr − M
dt
— moment of friction forces in the bearings.
fr
fr
,
132 Chapter 4
Expressing linear velocity V of the weight in terms of angular speed w of
the wheel we get a set of equations for two weights. After the integration the set
takes the form:
(J+m
R2)w1= m1gRt1−Mfrt1, (J+m2R2)w2= m2gRt2−Mfrt2.
1
(4.5)
The integration of these equations to obtain an angular displacement is
confronted by certain difficulties.
Let’s use inversion procedure and take the known linear displacement to
find an angular displacement of the wheel and angular velocities at the instant
when they are reaching the bottom.
h
ϕ =
. (4.6)
R
Since the motion in question is caused by a gravity force, let’s assume that
an angular acceleration of the wheel is constant, ε =const.
2
d
dt
ϕ
= ε,
2
dϕ
= w = εt, ϕ=
dt
εt
2
wt
=
2
2
(4.7)
(for zero initial conditions).
From (4.6) and (4.7) we have:
2h
w =
Rt
.
We substitute this expression into the set of equations (4.5) and substitute
the time of reaching the floor for a variable quantity t:
(J + m
R2)
1
2h
= m1gRT1− MfrT1,
RT
1
(J + m
R2)
2
2h
= m2gRT2− MfrT2. (4.8)
RT
2
After eliminating the moment of friction forces and making some transfor-
mations we get
J = R
g
(m2− m1)+(
2h
2
1
T
m
m
1
2
−
2
T
1
1
−
2
2
T
2
1
)
2
T
2
,
4.5. Motion of a body in potential field 133
r
A
or
J = R
g
(−m2+ m1) − (
2h
2
1
2
T
1
m
m
1
2
−
2
T
1
1
−
2
T
2
)
2
T
2
,
Answer.
To make sure that the dimension of the ob-
tained expression satisfies the dimension of a moment of inertia, we may represent the answer in a
consistent form:
2
2
T
T
g
1
J =(m
1
[J]=[kg · m
− m2)R2(
2
m/s
2
(
m
·
2h
s
s
2
− T
2
2
− 1),
2
T
1
4
− 1)] = [kg · m2].
2
Figure 4.11
The dimension of the expression obtained is consistent with the dimension
of a moment of inertia.
4.5. Motion of a body in potential field
gm
Sample problem 7. Weight M of mass m
is hung at fixed point O by string OM of
length l (see Fig 4.12). At the initial instant
the string makes an angle α with the vertical,
and the velocity of the weight is equal to zero.
In its subsequent motion the string meets thin
wire O
which is perpendicular to the plane of
1
O
1
the weight’s motion. The position of the wire is
specified by the polar coordinates: h = OO
and β. Determine the minimum value of the
,
1
M
angle α such that the string will be reeled up
the wire after the collision with it. Find also
Figure 4.12
tension of the string at the instant of the collision. The thickness of the wire should be neglected 9].
Solution. The motion of weight M is caused by a gravity force. So, the
problem is to determine such a minimum initial height that allows the weight
134 Chapter 4
β
r
to be lifted through some height and reel up the wire. Problems dealing with a
change in a position of a body in a potential field are usually solved by the use
of energy methods. First we must determine conditions under which the string
begins to reel. Let’s assume that we are interested in the only revolution about
the wire. It will happen if the velocity of the weight at the instant of the collision
with the wire (at position 1, see Fig. 4.13) will be minimum enough to allow
the weight to be lifted through the height d = r + r cosβ =(l − h)(1 + cos β)
and not to fall on the wire. The latter condition will be satisfied if the tension
force equals zero and the normal component of an inertia force counterbalances
the weight (see Fig. 4.13 and Fig. 4.14).
2
V
Φ
= mg, m
n
2
= mg, mV
r
2
= mgr. (4.9)
2
2
2
Figure 4.13 Figure 4.14
′
hlr −=
d
r
O
1
r
am
=Φ
nn
gm
We find velocity V1of the weight at the instant of the collision from the
equality between the change in a kinetic energy and the work of a gravity force
for the distance (O) and (1):
2
mV
1
− 0=mgl (cos β − cos α). (4.10)
2
We write the similar expression for the distance (1) and (2
mV
2
2
1
mV
2
=
2
2
mV
2
−
2
mV
1
= −mgd,
2
2
2
+ mg (l −h)(1 + cos β).
)
4.5. Motion of a body in potential field 135
We use the expression (4.9).
2
mV
1
= mg [0, 5(l−h)+(l−h)(1+cos β)] = mg (l−h)(1, 5+cosβ). (4.11)
2
From the expressions (4.10), (4.11) we find the angle α.
mg [(l − h)(1, 5+cosβ)] = mgl (cos β −cos α),
cos α =
h
(1, 5+cosβ) − 1, 5,α=argcos[
l
h
(1, 5+cosβ) − 1, 5].
l
Answer.
Let’s determine the cause of the change in the string tension just after the
collision with the wire. At that instant the change in the radius of curvature
happens. And this causes the change in the normal acceleration and the inertia
2
force of the weight Φ
= man= m
n
v
(see Fig. 4.15).
r
Figure 4.15
We discard a constraint and consider a “dynamic equilibrium” of the weight
(i. e. we reveal an “internal null” of the system). So we have:
before the collision
2
mV
1
= T −mg cos β,
l
136 Chapter 4
just after the collision
2
mV
1
∗
− mg cos β.
= T
l − h
Hence
T =
T
∗
=
mV
l
mV
l − h
2
1
+ mg cos β,
2
1
+ mg cos β.
Change in the tension of the string equals
T = T
∗
− T = mV
2
1
l(l −h)
h
,
or
T =2mg [
h
(1, 5+cosβ)],
l
i. e. the tension of the string increases after the collision.
Note that the first part of the problem has been solved on the assumption
that the weight stretches the string when it hits the position (2). But it is also
possible that there are points on the trajectory 2 − 2
in which the string has
a sag. After passing one of such points the weight moves as a free body in a
parabola over the wire. The solution is not given here, we only note that the
minimum value of α in this case equals
h
(1, 5sinα +cosβ) − 1, 5],
l
where 35
0
ϕ 900.
α =argcos[
4.6. Distribution of inertia forces of rigid body being in
general plane motion
Sample problem 8. The end A of uniform thin rod AB of length 2l and
mass m is constrained by block E to move along the horizontal axis at a constant
velocity V . The rod contacts with corner D all time of motion. Determine a
principal vector of inertia forces and a principal moment of inertia forces of the
rod with respect to the centroidal axis (through the mass center C)whichis
4.6. Distribution of inertia forces of rigid body being in general plane motion 137
ϕ
x
y
B
AE H
D
C
•
O
Figure 4.16
perpendicular to the plane of motion. The desired values should be obtained as
functions of angle ϕ (see Fig. 4.16) [9].
Solution. By the definition a principal vector and a principal moment of
inertia forces of a rod being in a general plane motion may be written as follows:
Here a
Φ=−ma
— acceleration of the mass center of the rod; Jcz–= moment of inertia
c
,Mcz= −Jczε.
c
of the rod with respect to perpendicular - to - plane - of - motion central axes;
ε — angular acceleration of the rotation of the rod.
We choose point A as reference point and write the acceleration of the mass
center in form:
a
= aA+ a
C
where a
=0,sinceVA=const, a
A
rotation about the reference point, a
n
CA
n
— normal acceleration of point C in it
CA
τ
— tangential acceleration of point C in
CA
+ a
τ
CA
,
the same motion (see Fig. 4.17).
n
=˙ϕ2· l, a
a
CA
τ
CA
=¨ϕ ·l.
Projections of the principal vector of inertia forces are:
=Φncos ϕ +Φτsin ϕ = m ˙ϕ2l cos ϕ + m ¨ϕl sin ϕ;
Φ
x
Φ
=Φnsin ϕ − Φτcos ϕ = m ˙ϕ2l sin ϕ − m ¨ϕl cos ϕ. (4.12)
y
The principal moment of inertia forces is
1
= −Jcz¨ϕ = −
M
in
m ¨ϕ(2l)
12
2
= −
1
2
ml
¨ϕ. (4.13)
3
138 Chapter 4
Figure 4.17
We have obtained the answer in a general form. However, there are unused
quantities: velocity V of point A and height H. Let’s express angular velocity ˙ϕ
and angular acceleration ¨ϕ in terms of these quantities.
It is known that a rod, as a rigid body, can move along some direction (for
example, along straight line AB) only in such a manner that velocities of all
points in this direction are the same. Velocity of point A is directed along the
axis OX, velocity of point — along the rod. Hence,
V
pr
V
· cos ϕ = VD,VD= V · cos ϕ.
A
AB
A
= pr
AB
V
,
D
Then, one can regard the velocity of point as a measure of the decrease of
distance AD:
H
d
(
= |
V
D
dt
sin ϕ
)| =
H
sin2ϕ
· cos ϕ · ˙ϕ.
From the latter equations we get:
V
2
ϕ, (4.14)
sin
H
· 2sinϕ · cos ϕ · ˙ϕ. (4.15)
¨ϕ =
d ˙ϕ
dt
˙ϕ =
=
V
H
Signs of ¨ϕ and ˙ϕ coincide, therefore the angular acceleration and angular
velocity are in the direction of the increase of ϕ — angle.
4.7. Bearing reactions 139
( )
x
A
α
We substitute expressions for ˙ϕ and ¨ϕ into equations (4.12) and (4.13):
= ml
V
sin4ϕ · cos ϕ + ml
2
H
=3ml
= ml
= −
in
V
H
Φ
y
M
2
V
sin4ϕ · cos ϕ;
2
H
2
l(1 − 3cos2ϕ)sin3ϕ;
2
1
2
ml
3
Φ
x
2
2V
H
2
2V
sin3ϕ · cos ϕ.
2
H
2
sin3ϕ · cos ϕ · sin ϕ =
2
Answer.
4.7. Bearing reactions
Sample problem 9. Because of the fault assembling of the round disk of a
power turbine the plane of the disk makes angle α with axis AB and center C
of mass of the disk does not lie on this axis (see Fig. 4.18). The eccentricity
is OC = a.
hh
B
y
O z
C
′
z
Figure 4.18
Mass of the disk is m, its radius is R, distances AO = OB = h; the angular
velocity of the disk is constant and equal to w. Find side dynamic forces acting
upon bearings A and B [9].
Solution. Because the mass center is out of the axis of rotation, its acceleration does not equal zero. Therefore the motion of a mass center will be
140 Chapter 4
x
ω
ω
r
ω
α
α
described by Newton’s second law. To analyze a rotational motion we will use
moment of momentum principle for a system (an analog of Newton’s second
law for rotational motion).
To simplify the calculations of the products of inertia of the disk we will
make these calculations in rotated coordinate system ox
instead of oxyz.
1y1z1
The coordinate system is rotated with respect to the axis oy which coincides
with the axis oy
(see Fig. 4.19).
1
x
1
1x
o z
1z
, yy 1z
1
a)
R
Ax
x
1
R
1Az
R
1Ax
z b)
1
Figure 4.19
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