Dynamics
D
i (동역학
(동역학))
담당교수:: 석종원
담당교수
제6강
1
Ch. 3. KINETICS OF PARTICLES
Angular Impulse and Angular
Momentum
„
Definition and Angular Momentum
U rightUse
right
i ht-hand
h d rule
l ffor cross product
d t
H O = r × mv
= m(v z y − v y z )i + m(vx z − v z x) j + m(v y x − v x y )k
i
HO = m x
vx
j
y
vy
k
z
vz
In the form of matrix
determinant
so that
H x = m(v z y − v y z ) H y = m(v x z − v z x ) H z = m(v y x − v x y )
제6강
2
Ch. 3. KINETICS OF PARTICLES
Angular Impulse and Angular
Momentum
„
Rate of Change of Angular Momentum
∑ M O = r × ∑ F = r × mv
S b tit ti
Substitution
off N
Newton’s
t ’ second
d llaw
∑ F = mv
= r × mv + r × mv = v × mv + r × mv
H
O
zero
∑ MO = H
O
The moment about the fixed point O of
all forces acting on m equals the time
rate of change of angular momentum
of m about O.
∑ M Ox = H
Ox
∑ M Oy = H
Oy
∑ M Oz = H
Oz
제6강
3
Ch. 3. KINETICS OF PARTICLES
Angular Impulse and Angular
Momentum
„
Angular ImpulseImpulse-Momentum Principle
Angular Impulse (N(N-m-s or kgkg-m2/s)
t2
∫ ∑ M dt = H − H = ΔH
t1
O
O2
O1
Total angular impulse
on m about the fixed
point O equals the
corresponding
change in angular
momentum of m
about O.
O
where H O2 = r2 × mv 2 and H O1 = r1 × mv1
Alternatively
t2
H O1 + ∫ ∑ M O dt = H O2
t1
∫ ∑ M dt = ( H ) − ( H ) = m[(v y − v z ) − (v y − v z ) ]
t2
t1
Ox
Ox 2
Ox 1
z
y
2
z
y
1
x-component: Note it’
it’s a vector
제6강
4
Ch. 3. KINETICS OF PARTICLES
Angular Impulse and Angular
Momentum
„
Plane-Motion Applications and
PlaneConservation of Angular Momentum
For plane motion problems
t2
∫ ∑ M dt = H − H
O
t1
O2
O1
t2
or ∫ ∑ Fr sin θ dt = mv2 d 2 − mv1d1
t1
In case when no external moment is applied
ΔH O = 0
or
H O1 = H O2
Principle of conservation of angular momentum
제6강
5
Ch. 3. KINETICS OF PARTICLES
Angular Impulse and Angular
Momentum
„
•
Example
Sample Problem 3/22
The small 2 - kg block slides on a
smooth horizontal
ho i ontal su
surface
face unde
under the
action of the force in the spring and
a force F. The angular momentum of
the block about O varies with time as
shown
h
iin the
th graph
graph.
h. When
Wh
t=6.5 s, it
is known that r=150 mm and β=60
60°°.
Determine F for this instant.
제6강
6
Ch. 3. KINETICS OF PARTICLES
Angular Impulse and Angular
Momentum
„
•
Example
Sample Problem 3/23
A small mass particle is given an
initial velocity v 0 tangent to the
horizontal rim of a smooth
hemispherical bowl at a radius r0 from
the vertical centerline, as shown at
point
i t A . As
A the
th particle
ti l slides
lid
pastt
point B, a distance h below A and a
distance r from the vertical centerline,
its velocity v makes an angle θ with
the horizontal tangent to the bowl
through B. Determine θ.
제6강
7
Ch. 3. KINETICS OF PARTICLES
Special Applications
„
Impact
If impact is not so severe and
the objects are highly elastic
elastic,
the momenta before and after
are conserved
m1v1 + m2 v2 = m1v1' + m2 v2'
Assumptions
1. Any forces acting on the objects are small
compare to the internal forces of contact
contact.
2. There is no appreciable change in position
s of the mass centers
제6강
8
Ch. 3. KINETICS OF PARTICLES
Special Applications
„
Coefficient of Restitution
There are two unknowns
unknowns, v1 and v2 after impact
impact, so that
we need one more equation to solve them!
t0: Time for deformation
t
F dt m [− v ' − (−v )] v − v '
∫
e=
=
=
[
]
m
−
v
−
(
−
v
)
v −v
∫ F dt
r
t0
t0
d
0
1
1
0
0
1
1
0
1
1
0
Similarly,
y, for p
particle 2 we have
t
F dt m (v ' − v ) v ' − v
∫
e=
=
=
m
(
v
−
v
)
v −v
∫ F dt
t0
t0
0
e=
r
d
2
2
0
2
0
2
0
2
0
2
v2' − v1' relative velocity of separation
=
v1 − v2
relative velocityy of approach
pp
제6강
9
Ch. 3. KINETICS OF PARTICLES
Special Applications
„
Energy Loss During Impact
Elastic Impact: no energy loss
Plastic Impact: max energy loss
제6강
10
Ch. 3. KINETICS OF PARTICLES
Special Applications
„
Oblique Central Impact
There are four unknowns
unknowns, normal and tangential
components of both v1 and v2 after impact, so that we
need four equation tin total o solve them!
1 : m1 (v1 ) n + m2 (v2 ) n = m1 (v1' ) n + m2 (v2' ) n
2 : m1 (v1 ) t = m1 (v1' ) t
3 : m2 (v2 )t = m2 (v2' )t
4: e =
제6강
(v2' ) n − (v1' ) n
(v1 ) n − (v2 ) n
11
Ch. 3. KINETICS OF PARTICLES
Special Applications
„
•
Example
Sample Problem 3/24
The ram of a pile driver has a mass of
800 kg and is released
eleased ffrom
om rest
est 2 m
above the top of the 24002400-kg pile
pile.. If
the ram rebounds to a height of 0.1 m
after impact with the pile, calculate (a)
th velocity
the
l it vp’ off the
th pile
il immediately
i
di t l
after impact, (b) the coefficient of
restitution e, and (c) the percentage
loss of energy due to the impact
impact..
제6강
12
Ch. 3. KINETICS OF PARTICLES
Special Applications
„
•
Example
Sample Problem 3/25
A ball is projected onto the heavy
plate with a velocity of 16 m/s at the
30°° angle shown
30
shown.. If the effective
coefficient of restitution is 0 . 5 ,
compute the rebound velocity v’ and
its angle θ’.
’
제6강
13
Ch. 3. KINETICS OF PARTICLES
Special Applications
„
•
Example
Sample Problem 3/26
Spherical particle 1 has a velocity v1=6
m/s in the di
direction
ection shown and collides
with spherical particle 2 of equal mass
and diameter and initially at rest
rest.. If the
coefficient of restitution for these
conditions
diti
iis e = 0 . 6 , determine
d t
i
th
the
resulting motion of each particle
following impact
impact.. Also calculate the
percentage loss of energy due to the
impact.
제6강
14
Ch. 3. KINETICS OF PARTICLES
Special Applications
„
CentralCentral
-Force Motion: Motion of a Single Body
When a particle moves under the influence of a force
directed toward a fixed center of attraction, the motion is
called central
central-force motion.
F =G
mm0
r2
It is merely angular
momentum conservation
equation
mm0
− G 2 = m(r − rθ 2 )
r
0 = m(rθ + 2rθ) *r/m
r 2θ = h, a constant
A = 1 r 2θ = constant
2
제6강
Kepler’’s second law
Kepler
15
Ch. 3. KINETICS OF PARTICLES
Special Applications
„
CentralCentral
-Force Motion: Conic Sections
Shape of the path
Gm
1
= C cos θ + 2 0
r
h
e = r /(d − r cos θ )
1 1
1
= cos(θ ) +
r d
ed
e<1: ellipse
e=1: parabola
e>1: hyperbola
제6강
h 2C
e=
Gm0
Conic section is
formed by the
locus of a point
which moves so
th t th
that
the ratio
ti e off
its distance from
a point (focus)
to a line
(directrix) is
constant
16
Ch. 3. KINETICS OF PARTICLES
Special Applications
„
CentralCentral
-Force Motion: Conic Sections
„ Case 1:
1 ellipse (e<1)
1 1
1
= cos(θ ) +
r d
ed
2a = rmin + rmax =
r=min. when θ=0, r=max. when θ=π
ed
ed
ed
or a =
+
1+ e 1− e
1 − e2
1 1 + e cos θ
2
−
=
,
r
=
a
(
1
−
e
)
),
r
=
a
(
1
+
e
)
),
b
=
a
1
e
min
max
r a (1 − e 2 )
Kepler’’s first law
Kepler
A πab
2πab
τ = = 1 2 or τ =
A 2r θ
h
For e=0, it becomes a circle
a3/ 2
⇒ τ = 2π
R g
d = 1 / C , a = ed /(1 − e 2 ), b = a 1 − e 2 , Gm0 = gR 2
제6강
17
Kepler’’s third law
Kepler
Ch. 3. KINETICS OF PARTICLES
Special Applications
„
CentralCentral
-Force Motion: Conic Sections
„ Case 2: parabola (e=1)
1 1
1
= cos(θ ) +
r d
edd
The radius vector becomes
infinite as θ approaches π, so
2
hC
e=
Gm0
the dimension a is infinite.
1 1
⇒ = (1 + cos θ ) and h 2C = Gm0
r d
제6강
18
Ch. 3. KINETICS OF PARTICLES
Special Applications
„
CentralCentral
-Force Motion: Conic Sections
„ Case 3:
3 hyperbola (e>1)
1 1
1
= cos(θ ) +
r d
edd
r becomes infinite for θ1 and cosθ1 =-1/e
θ1 for cosθ
For branch II, for making r positive, θ → θ − π , − r → r
1
1
1
= cos(θ − π ) +
−r d
ed
-θ1<θ <θ1
⇔
or
Gm
1
= C cos θ + 2 0
h
r
제6강
1
1 cos θ
=− +
r
ed
d
Branch II is impossible
since Gm0/h2>0
19
Ch. 3. KINETICS OF PARTICLES
Special Applications
„
CentralCentral
-Force Motion: Energy Analysis
Total Energy
Ki ti E
Kinetic
Energy
Potential Energy
1
mgR
g 2
2
2 2
E = m(r + r θ ) −
2
r
For θ = 0, r = 0, 1 / r = C + gR 2 / h 2 , rθ = h / r , h 2C = egR 2
2E
g 2R4
2
2
2 2
= h (egR / h ) − 2
m
h
2 Eh 2
⇒ e = + 1+
mg 2 R 4
e < 1,
e = 1,
E is negative
E is zero
hyperbolic orbit e > 1,
E is positive
elliptical orbit
parabolic orbit
제6강
V=0 when r=∞
r=∞
20
Ch. 3. KINETICS OF PARTICLES
Special Applications
„
CentralCentral
-Force Motion: Energy Analysis
1 2 mgR
R2
The expression for v of m
mv −
=E
2
r
2 Eh 2
2
e = + 1+
,
1
/C
=
d
=
a
(
1
-e
) / e for the elliptical orbit
2 4
mg R
gR 2 m
E=−
2a
⎛1 1 ⎞
& v = 2 gR ⎜ − ⎟
⎝ r 2a ⎠
2
2
g 1+ e
g
=R
a 1− e
a
rmax
rmin
g 1− e
g
vA = R
=R
a 1+ e
a
rmin
rmax
vP = R
1 1 + e cos θ
=
,
2
r a (1 − e )
rmin = a (1 − e), rmax = a (1 + e)
⇐
A: Apogee
p g
P: Perigee
21
Ch. 3. KINETICS OF PARTICLES
Special Applications
„
•
Example
Sample Problem 3/27
An artificial satellite is launched from
point B on the equato
equator by its ca
carrier
ie
rocket and inserted into an elliptical
orbit with a perigee altitude 0f 2000
km.. If the apogee altitude is to be
km
4000 km,
k
compute
t (a)
( ) the
th necessary
perigee velocity vP and the
corresponding apogee velocity vA, (b)
the velocity at point C where the
altitude of the satellite is 2500 km,
and (c) the period τ for a complete
orbit.
제6강
22
Ch. 3. KINETICS OF PARTICLES
Special Applications
„
Relative Motion
a A = a B + a rel
Thus, Newton' s second law ∑ F = ma A becomes
∑ F = m(a B + a rel )
Newton’s second law fails for an
accelerating system since
∑ F ≠ ma
제6강
rel
23
Ch. 3. KINETICS OF PARTICLES
Special Applications
„
Relative Motion – D’Alembert
Alembert’’s Principle
-ma is treated as a force:
force
called inertia force
∑ F − ma = 0
The system is in dynamic
equilibrium
ilib i
∑ Fn = ma n : T sin
i θ = mrω 2
∑ Fy = 0 : T cos θ − mg = 0
Using D' Alembert' s Principle
∑ Fn − ma n = 0 : T sin θ − mrω 2 = 0
제6강
24
Ch. 3. KINETICS OF PARTICLES
Special Applications
„
Relative Motion – Constant Velocity, Nonrotating Systems
For a constant velocity, nonrotating system
∑ F = ma rel
dU rel = ma rel ⋅ drrel = mvrel dvrel = d ( 12 mvrel2 )
dU rel = dTrel or U rel = ΔTrel
(dU = ∑ F ⋅ drA ) ≠ (dU rel = ∑ F ⋅ drrel )
(T = 12 mv A2 ) ≠ (Trel = 12 mvrel2 )
(G = mv A ) ≠ (G rel = mv rel )
WorkWork
-Energy
gy
∑ Fdt = d (mv rel ) Impulse
Impulse-Momentum
∑F = G
rel and ∫ ∑ Fdt = ΔG rel
∑ MB = H
Brel
제6강
MomentMoment
-Angular Momentum
25
Ch. 3. KINETICS OF PARTICLES
Special Applications
„
•
Example
Sample Problem 3/28
A simple pendulum of mass m and
length r is mounted on the flatcar,
flatca ,
which has a constant horizontal
acceleration a 0 as shown
shown.. If the
pendulum is rreleased
eleased from rest
relative
l ti
t the
to
th flatcar
fl t
att the
th position
iti
θ=0, determine the expression for the
tension T in the supporting light rod
for any value of θ . Also find T for
θ=π/2 and θ=π .
제6강
26
Ch. 3. KINETICS OF PARTICLES
Special Applications
„
•
Example
Sample Problem 3/29
The flatcar moves with a constant
speed v0 and carries a winch which
produces a constant tension P in the
cable attached to the small carriage
carriage..
The carriage has a mass m and rolls
freelyy on the horizontal surface
starting from rest relative to the flatcar
at x=0, at which instant X=x0=b. Apply
the work
work- energy equation to the
carriage first,
carriage,
first as an observer moving
with the frame of reference of the car
and, second, as an observer on the
ground.. Show the compatibility of the
ground
two expression.
제6강
27