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4.1. Kinematics of mass – point particle. Analogies 121
latter equation because during one half a period (1/2T ) point M moves behind
the mass center (but R sin ωt > 0), and during the second half a period it moves
before the center of mass (but R sin ωt < 0).
The wheel’s mass center and point M travel the same distance during one
period:
20T =2πR,
where 2πR – length of circle arc. Hence,
T =
2πR
20
,ω=
2π
= 20(rad/s).
T
Thus,
x = x
=20t − R sin ωt.
M
Let’s show in Fig. 4.3 the law of motion of point M along the oy –axis.
x
2
1
Figure 4.3
3
4
2R
t
3
4
2
1
It is seen from the figure that the law of motion is cosinusoidal
curve R cos ωt which is displaced up for R.Hereω =20rad/s — angular
velocity of the motion of position vector of point M along the circle.
Then
y = y
= R − R cos 20t.
M
Answer:
x =20t − R sin ωt =20t −sin 20t(m),
y = R − R cos ωt =1−cos 20t(m).
The second equation in the answer is analogous with the equation of motion
of an oscillator disturbed suddenly by a constant force [1]. Let the vibrations of
a linear oscillator take place along the oy – axis. Then the differential equation
of its motion has the form:
m¨y + cy = Q(t),
where m — mass of weight, c — spring constant, Q(t) — force suddenly applied.
122 Chapter 4
The solution is:
y = A sin pt+ B cos pt+
where p =(c/m)
0,5
— natural frequency of vibrations.
1
p
For zero initial conditions and for Q =const,
t
Q(t)sinp (t −τ)dτ,
0
y = y
(1 − cos pt).
st
Thus, the system vibrates relative to the level of statical deformation with
the swing of y =0...2.
The latter expression has a mathematical analogy with the second equation
in the answer: the plot of y as a function of t will like one shown in Fig. 4.3.
The physical analogy: the harmonic variation of a parameter relative to
some average value takes place in both cases. In the second case (forced vibrations) the revealed swing of a linear oscillator allows us to specify the minimum
safety factor for a dynamic system (which is equal to 2 in the case above).
4.2. Dynamics of mass – point particle
Sample problem 2. A ball of mass 100gr falls through air under the action
of a gravity force. The motion of the ball takes place according to the following
equation:
x =4, 9t − 2, 45(1 − e
where x measures in meters, t in seconds, ox – axis directed vertically downward. Determine force R of air resistance and express it as a function of the
ball’s velocity [9].
Solution. It is needed to find one of acting – upon – the – body force if
the law of the body’s motion is known; i. e. to recover an assumed information.
Newton’s second law allows to relate kinematic characteristics to forces:
ma =F,
−2t
),
where a – acceleration of a particle,F — resultant of all forces applied to the
particle.
In the case of falling ball we project the equation of Newton’s second law
onto the vertical axis ox and obtain
2
d
x
m
dt
2
= mg − R.
4.2. Dynamics of mass – point particle 123
The analysis of an additional information given enables us to state that the
air resistance force is the function of, suppose, a linear velocity of the ball,
dx
R = k
dt
.
Then the differential equation of the ball’s motion assumes the following form.
2
m
d
dt
x
2
= mg − k
dx
dt
,
or
m¨x = mg −k ˙x.
We solve this differential equation for x.
mg
where C
k
¨x =
m
d ˙x
mg
− ˙x
k
mg
−ln |
— constant of integration determined from initial conditions:
1
k
t =0, ˙x(0) = 4, 9 −4.9 e
(
k
=
− ˙x |=
− ˙x),
k
dt,
m
k
t + C
m
0
,
1
=0.
The expression for ˙x was obtained by differentiating the equation of motion.
Whence:
−ln |
mg
k
−ln |
− ˙x| =
˙x = −
k
t − ln |
m
mg
k
mg
− 0 |=0+C
k
−
m
(e
mg
x =
mg
|, −
k
k
t
− 1), ˙x =
t +
k
,C1= −ln |
1
k
t =ln|1 −
m
mg
m
·
k
k
dx
dt
· e
mg
=
k
−
t
m
k ˙x
mg
(1 − e
k
+ C2.
mg
k
|, 1 −
−
|,
k
mg
k
t
m
),
Thus the equation of motion with two unknown quantities k and C
obtained. Comparing it with the given equation
−2t
x =4, 9t − 2, 45(1 − e
),
˙x = e
k
−
t
m
,
is
2
124 Chapter 4
one can see that
Whence
C
= −2, 45,
2
k =
mg
k
mg
4, 9
=4, 9, (
0, 1 · 9, 8
=
4, 9
m
)
k
2
g =2, 45,
=0, 2.
m
k
=2.
The unknown force of air resistance is expressed as follows:
R = k ˙x =0, 2V,
or
R =0, 2(4, 9 − 4, 9e
−2t
)=0, 98(1 − e
−2t
).
Answer.
In addition to that, this problem has a simpler solution. By differentiating
the equation of motion twice we get the acceleration of the body
¨x =9, 8 e
−2t
.
Let’s substitute this expression into the differential equation
¨x = g −
R
m
R = m(9, 8 − 9, 8e
−2t
, 9, 8e
−2t
),orR =0, 98(1 −e
=9, 8 −
R
m
,
−2t
).
Answer.
In this case we have used the given information not in the initial conditions
of the integral equation but immediately in the differential equation. That is why
the second solution is simpler.
4.3. Dynamics of translation motion of a system of rigid bodies
Sample problem 3. Two weights M1and M2of masses m1and m
accordingly are connected by an inextensible rope which passes over pulley A;
the weights slide on the smooth sides of a rectangular wedge (see Fig. 4.4).
The base BC of the wedge is supported by a smooth horizontal plane. Find
the horizontal displacement of the wedge caused by sliding down weight M
for h =0, 1m. Mass m3of the wedge equals 4m1and 16m2(m3=4m1=
=16m
).Masses of the rope and pulley should be neglected [9].
2
Solution. Let’s consider some peculiarities of the problem.
2
1
4.3. Dynamics of translation motion of a system of rigid bodies 125
A
M
1
M
2
0
B
30
C
Figure 4.4
1. The problem is a two-dimensional one, so we introduce a coordinate system XOY.
2. There is no need to consider internal forces, because we are interested in an
integral motion of the system. And the external forces are: three forces of
weight m
g, m1g, m2g, and normal reaction force N exerted by a support
3
plane. All the forces act along axis OY (see Fig 4.5).
N
mg
1
mg m g
2
Figure 4.5
To solve the problem an information predetermined is
d
dt
(m
V
1
1
+ m
V
+ m
2
2
3
e
V
)=
3
F
— Newton’s second law for a system.
HereF
e
=(m3g, m1g, m2g,N) — external forces,V1,V2,V3— velocities
of mass centers of bodies.
Since the unknown displacement occurs along ox – axis, we project the
vector equation onto this axis
d
BecauseF
e
x
=0.
dt
(m
+ m2˙x2+ m3˙x3)=0,
1˙x1
126 Chapter 4
After integration:
m
+ m2˙x2+ m3˙x3= C1.
1˙x1
Initially the system was at rest,
˙x
(0) = ˙x2(0) = ˙x3(0) = 0 = C
d
dt
1
(m
m
+ m2x2+ m3x3)=0,
1x1
+ m2x2+ m3x3= C2.
1x1
1
Determining the constant of integration from the initial conditions we obtain:
+ m2x2+ m3x3= m1x1(0) + m2x2(0) + m3x3(0) = const. (4.1)
m
1x1
We have got one of the forms of the information operator of null action –
principle of conservation of a position of system’s mass center:
(m
+ m2+ m3) · xc= m1x1+ m2x2+ m3x3=const.
1
Then we find coordinates of body’s mass centers after the weight M
has
1
covered distance h downward. The motion of this weight forces the wedge to
move rightwards and the weight M
of the weights are x
(0),x2(0),x3(0).
1
to rise on the wedge. The initial positions
2
x
= x3(0) + l,
3
where l — is the displacement of the wedge,
= x1(0) −
x
1
h
tan 30
+ l, x2= x2(0) − h + l.
0
We substitute the results into (4.1) and obtain:
m
+m
3(x3
h(
l =
tan 30
m1+ m2+ m
(0) −
1(x1
(0) + l)=m1x1(0) + m2x2(0) + m3x3(0).
m
(l −
1
m
1
+ m2)
0
tan 30
h
tan 30
,l=
3
h
+ l)+m2(x2(0) − h + l)
0
)+m2(l − h)+m3l =0,
0
0,5
0.1(3
· 4+1)m
(16 + 4 + 1)m
2
=0, 038(m).
2
Answer.
4.3. Dynamics of translation motion of a system of rigid bodies 127
Sample problem 4. A projectile of mass m1is fired at a horizontal velocity
of V into a box of sand of mass m
, where it embeds itself and moves with the
2
box. The box has rolling supports and is constrained by a spring of stiffness c
connected with the vertical fixed wall (see Fig. 4.6). Determine the maximum
displacement of box with bullet within it if the box was at a rest before the
collision [7].
m
1
V
0
m
2
C
x
Figure 4.6
Solution. This problem deals with the analysis of the motion of a linear
oscillator, i. e. a body which can vibrate under the action of the only linear
restoring force.
It is known that the equation of a linear oscillator motion has the following
form:
where C
c
x = C
and C2— constants of integration determined from the initial condi-
1
1
cos((
m1+ m
0,5
)
t)+C2sin((
2
c
m1+ m
0,5
)
t),
2
tions:
C
2
(
1
(
m1+ m
˙x = −C
=0=C1, ˙x0= C2(
x
0
t =0,x
c
m1+ m
c
=0, ˙x0=0.
0
0,5
)
2
0,5
)
cos((
2
c
m1+ m
sin((
m1+ m
0,5
)
2
c
m1+ m
c
2
,C2=
)
2
0,5
)
0,5
t)+
t),
(
m1+ m
˙x
0
c
.
0,5
)
2
Thus, the equation of motion has the form:
˙x
x =
(
m1+ m
0
c
sin((
0,5
)
2
m1+ m
c
0,5
)
t).
2
128 Chapter 4
It is obvious that the maximum displacement is
˙x
x
max
=
(
c
m1+ m
0
.
0,5
)
2
This problem is reduced to the problem of vibration of the box being hit by
the bullet. To find velocity ˙x
of the box just after the impact we use the known
0
information of the corollary of Newton’s second law for a system. It is linear
momentum principle.
d
dt
(m
V
+ m
1
1
2
e
V
)=
2
F
,
whereF
e
=(m1g, m2g,N2) — external forces.
In projection onto ox –axis:
d
(m
dt
+ m2V2= m1V1(0) + m2V2(0).
m
1V1
But V
Hence ˙x
= V2=˙x0,andV1(0) = V , V2(0) = 0.
1
0
=
m1v
m1+ m
.
2
We put this result into the expression for x
x
=
max
m1+ m
m
1V1
v
1
2
+ m2V2)=0,
max
·
c
m1+ m
:
−0,5
2
.
Answer.
Sample problem 5. The truck B travels at a constant relative speed U
the horizontal platform A moving with constant velocity V
(see Fig. 4.7.). At
0
on
0
some instant the brakes of the truck are applied. Determine the total speed V of
the platform and truck after the truck stopes. Mass of the platform is M,mass
of the truck is m.
Find also distance S travelled by the truck on the platform, and time τ of
its decelerated motion. Let the brakes exert constant force F on the truck [3].
Solution. Since the platform was initially moving at a constant speed, there
were no external forces acting on the system in the direction of motion. Therefore, by linear momentum principle (or Newton’s second law) the projection of
the linear momentum of the system onto the direction of motion conserves. So,
4.3. Dynamics of translation motion of a system of rigid bodies 129
B
r
U
Figure 4.7
0
A
r
V
0
platform A is accelerating due to truck B deceleration. Both bodies are acted on
by resistance force F .
Now the bodies differential equations with respect to the fixed axes can be
written (see Fig. 4.8, 4.9).
B
r
F
Figure 4.8 Figure 4.9
rr
()U+V
r
F
r
V
For truck B:
d
(U + V )=−F. (4.2)
m
dt
For platform A :
dV
M
= F. (4.3)
dt
Here U — variable relative velocity of the truck, V — variable absolute
velocity of the platform.
Let’s use both these equations to determine the relative motion of the truck
m
dU
dt
= −F −
dU
dt
U = U
= −
−
0
dV
dt
F
m
F
m
t −
= −F −m
F
−
M
F
M
F
,
M
,
t. (4.4)
130 Chapter 4
Here the form of (−
velocity due to its proper motion, and the term of (−
F
t) — characterizes decreasing the truck’s relative
m
F
t) - shows change in this
M
relative velocity because of the platform’s acceleration.
Integrating equation (4.4) for the zero initial condition we obtain the relative
path of the truck.
s =
τ
Ud t = U0τ −
0
F
m
2
τ
−
2
M
2
τ
F
.
2
At the instant when the truck stopes its speed U =0. Then from (4.4) we get
τ =
U
0
F
+
m
s =0, 5
=
F
M
mM
M + m
mM
M + m
U
0
·
,
F
2
U
0
·
.
F
The velocity of the platform after the truck stopes can be obtained by the
integration of equation (4.3)
V =
F
M
t + V
,Vτ=
0
F
M
t + V
= V0+
0
mM
M + m
.
U
0
Answer.
One could also use methods of relative motion dynamics to solve this prob-
lem.
4.4. Dynamics of rotation of a system of rigid bodies
Sample problem 6. To determine the moment of inertia J of flywheel A
of radius R with respect to the centroidal axes the following operations are
performed. A thin wire is wrapped around the flywheel, weight B of mass m
is tied to the end of the wire, and the time T1required to the weight to cover the
distance h down is measured. To exclude the influence of friction in bearings the
second experiment with the weight of mass m
h is carried out. Regarding a moment of friction force as a constant quantity
that does not depend upon the mass of the weight, we obtain the moment of
inertia J . (See Fig. 4.10) [9].
, time T2, and the same distance
2
1
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