★★★★★★★ 부탁 드립니다. ★★★★★★★
수업시간에 말씀하신 것과 같이, Solution을 공개하는 이유는 여러분의 효율적인 학습을 위함입니다.
하지만 현재 공개되는 Solution은 정식으로 배포되는 자료가 아니며,
지적재산권으로서 저작권의 보호를 받고 있습니다.
타인의 지적 재산을 존중하여 무단 수정 및 배포 시 발생하는
불이익 및 법적 문제에 대하여서 본인에게 책임이 있음을 유의 하시기 바랍니다.
또한, 가장 민감한 사항일 수 있는 Matlab Code는 삭제하였으니, 이점 유의하시기 바랍니다.
마지막으로, 절대 무단 수정 및 배포를 하지 않으시길 당부 드립니다.
한 학기 동안 열심히 공부하여 좋은 결과를 얻으시길 기원합니다.
Chapter Two
Newtonian Mechanics
SOLUTION
2.1
Newton's second law only applies to accelerations relative to an inertial
frame.
SOLUTION
2.2
The mass of the accelerating part of the system being analyzed, which is
usually taken to be an equivalently placed infinitesimal particle.
SOLUTION
2.3
We choose the variable x to describe the particle's position, so its velocity
and acceleration are given by x and x, respectively. Since the only force
acting on the particle is in the same direction as the acceleration, we can
use Newton's second law to write the scalar equation
mpx = -bx 2
(S2.1)
.
Following the procedure outlined in Appendix C, we substitute v
rearrange Eq. (S2.1) as
b
2
v
mp
=
x and
Since vis the same as dvjdt, we can multiply both sides by dt and integrate:
J
dv = 1--b-dt
v2
mp
=?
~ = __b_t+C
v
mp
where C is a constant of integration. Since we know that the velocity is 1Jo
at a time to, and assuming that to = 0, we see that C = -1/ vo. Thus, the
equation for the particle velocity as a function of time is
1 ] -l
b
v(t) = [ - t + mp
vo
SOLUTION
2.4
The roller coaster car is constrained to always stay on the track. To satisfy
:mch a geometric constraint for a given speed, the track must exert a force
on the cart which is orthogonal to the path. This is analogous to the normal
4
CHAPTER 2
force exerted by any surface on any object resting on that surface, so we call
this constraint force the normal force due to the coaster track, and label it
N.
SOLUTION
2.5
We know that the luge must maintain contact with the wall of the track.
This represents a constraint to the motion of the luge. We can model the
track as a tube of radius r centered around a three-dimensional path s,
shown in Figure 82.1. To describe the position of the luge we need only
know the distance along the path and an angle that orients the luge along
the wall. This implies that the luge has two degrees of freedom.
. e
~I
I
I
I
0~---------•
I
I
,
I
I
I
Figure S2.1 Diagram of luge track depicting distance along the path s and angle from the
vertical axis (), orienting the luge in three dimensions.
SoLUTION
2.6
Each of the masses, individually moving in the plane, would have two degrees
of freedom. If it weren't for the bar connecting them, the system of two
masses in the plane would have four degrees of freedom. Introducing the
constraint that the distance between masses remains constant removes one
degree of freedom. The barbell, modeled as two point masses connected
by a rod, has three degrees of freedom. We could describe the position of
both masses by using the position of one mass, and the angle of the bar
connecting them - thereby using three scalar variables; i.e. if we know the
coordinates of mass 1 (x1, Yih, then we could use the following equation of
constraint to determine the location of mass 2:
5
NEWTONIAN MECHANICS
where e is the angle of the bar, measured from the horizontal and L is the
length of the rod (see Figure 82.2).
0
L
-
-
-
-
-
-
-
~
Fg
Figure 82.2 Planar sketch and FBD of barbell for Problems 2.6 and 2.8.
SOLUTION 2. 7
Every point in a plane has 2 degrees of freedom, so, if there were no constraints, three points (such as the endpoints of the three links in the robotic
arm described in Cartesian coordinates) would have 6 degrees of freedom.
We've shown, however, that this system has only 3 degrees of freedom, and
so there must be 3 equations of constraint.
Let the three links have lengths lr, l2 and l3, respectively. The endpoint
of the first link must be a distance lr from 0, so we can write
xi+ YI = li ·
The endpoint of the second link must be a distance l2 from the endpoint of
the first link ( ( x 1, yl) I), so we can write
(x2- X1) 2 + (Y2- yl) 2 = l~.
Similarly, the endpoint of the third link must be a distance l3 from the
endpoint of the second link, which can be expressed as
We see that we have now used all of the information provided by the system (the three link endpoint positions and the three link lengths) to relate
the positions of the arm links. As predicted, we now have ~) equations of
constraint of the form:
2
.Tl
+ Yl2 = z21
(.1:2- :rl) 2 + (Y2- Yl) 2 = l~
(:r:3- x2) 2 + (y3- Y2) 2 = z§.
6
CHAPTER 2
SOLUTION
2.8
Figure 82.2 shows the free body diagram for the barbell modeled in Problem
2.6. Each mass is acted on by gravity (F 9 ), and has a force (T) constraining
it to stay on the end of the connecting rod.
SOLUTION
2.9
Figure 82.3 Free body diagram for luge and rider from Problem 2.5.
Figure 82.3 shows the free body diagram of the luge and rider modeled
in Problem 2.5. The luge and rider (approximated as a point mass) are acted
upon by gravity (F 9 ) and a normal force (N) due to the track, which points
in the orthogonal direction to the plane tangent to the track directly beneath
the luge.
SOLUTION
2.10
I
[~
T
w-----.
Fv
OL------- ..
Figure 82.4 Free body diagram for rocket for Problem 2.10.
Figure 82.4 shows the free body diagram of a rocket during its ascent
through the atmosphere. The rocket is acted on by gravity, F 9 , thrust, T,
(which acts in the direction of the rockets flight), drag, Fn, (which acts
in the direction opposite of the thrust) and the force due to the wind, W,
which we have modeled here as contributing in the direction opposite the
rocket's flight (in which the maximum surface area is presented). Note that
we could have also modeled the force due to the wind as pointing in any
arbitrary direction given by whatever the specific conditions happen to be
during the rocket's flight.
7
NEWTONIAN MECHANICS
SOLUTION 2.11
The skier can be modeled as a point mass sliding on an inclined surface with
angle a from the horizontal. In this problem it is convenient to choose a
reference frame aligned with the incline (as shown in Figure S2.5).
I
I
I
I
0 ''
N
Figure S2.5 Free body diagram for skier on inclined plane.
We include all forces acting on the mass in the free body diagram
including the drag force with magnitude IIF D II = 0.5CDApv 2 , the force due
to friction F f = J-LN, gravity and the normal force N. We define x to be
the skier's position along the slope, and x the velocity. Summing only the
forces (and components of forces) acting in the direction of the skier's path
(i.e., along the slope) we get
Fp = mpgsina- J-LN-
1
2cDApx 2 .
Using Newton's second law, this sum of forces is equal to the skier's mass
multiplied by the acceleration in this direction (x) so we get the equation of
motion
..
.
CD A
2mp
·2
x = gsma- J-Lgcosa- - - px
If we sum the forces in the orthogonal direction (i.e., the direction of
the normal force), and use the fact that the skier is constrained to stay on
the surface of the slope, so that acceleration is zero in that direction, we can
find the magnitude of the normal force, N = mg cos a.
By inspection, we can see that the acceleration experienced by a heavier skier will be greater. This is because the gravity force depends upon
mp, but the drag does not. So, in the equations of motion, dividing by
mp cancels the mp in the gravity term, but not in the drag term, thereby
8
CHAPTER 2
making acceleration dependent on mass. We can also compare maximum
speeds. If the skier reaches their maximum speed then they will no longer
be accelerating and we will have x = 0. Solving the equation of motion for
the velocity when acceleration is zero (called the terminal velocity) gives us
v*
=
2mpg
CDAP (sino:- f.-lCOso:)
which shows that the heavier skier should be able to ski faster. However,
if the skier is just big, and not heavy, A will be larger resulting in a lower
terminal velocity.
SOLUTION
2.12
Figure 82.6 Free body diagram for Problem 2.12.
We draw a free body diagram as in Figure S2.6. This is a single
degree of freedom problem in the single vertical dimension, allowing us to
solve the scalar Newton's law in that direction. The spring force is given by
F8 = k(y- lo) for mass position y and spring rest length and constant lo
and k, respectively. The force due to gravity is F 9 = mpg. We can now set
up Newtons Second Law to find the equation of motion:
mjj = ky - mg
'*
..
k
y= - y - g
m
SOLUTION
2.13
We will use the same reference frame and free body diagram as in Tutorial
2.3, Figure 2.7(c). The nonlinear spring force is F.s = -k(x-xo)-c(x-:r 0 ) 3 .
The forces on the mass in the vertical direction sum to zero as the mass is
constrained to the frictionless surface.
a. Summing the forces in the
we get
:1:
direction and using Newton's second law
..
.f.s
J:=-.
mp
(S2.2)
9
NEWTONIAN MECHANICS
If the spring is linear (i.e. c
of motion is
= 0) then the one-dimensional equation
k
(82.3)
m
To put this second order O.D.E. into first order form, we begin by
making a change of variables so that z = x ~ xo. Then the equation of
~-(x ~ x 0 ).
X=
motion is
z = ~ .km z, the simple harmonic oscillator.
Let Z
= (
Zl )
~
where z1 = z and z2 = i1. Then we can write (82.3) as a system of
two first-order O.D.E.s by
z=
(
!
21 )
(
Z2
~~z1
)
The same can be done for the nonlinear spring to get
b. Now that we have the dynamics of our system expressed as a set of
first-order ordinary differential equations, we're ready to use MATLAB 's
ode45 to integrate the equations of motion and find a trajectory for
a particular set of initial conditions. First we must tell MATLAB what
the differential equations are. Listing 2.2 is a function in MATLAB that
will be used by ode45 when we perform the integration.
One way to run ode45 and plot the trajectories for the mass attached
to a linear and nonlinear spring is to create a script-a separate . m
file with all of your code. The code in Listing 2.1 runs ode45 for both
cases and plots the position for each case versus time, as in Figure
S2.7.
Listing 2.1 Script for Problem 2.13
ElJ!C .:--~
-Z,·.:·)'e!:"
= 1;
c_lin
m
{kLY
=
0;
k = 1; ~;_N/m
xO = 0.25;
,,.
nt:eqL--af·e rJ!.!C."J.' p_Z
} _i near
[t_lin, z_lin] = ode45(©springs, ...
[020], [.40], [],m,k,c_lin,xO);
,.,
I j
[t_nonlin,z_nonlin]
[020], [.40],
nd
it
i nq
ode45(@springs,.
[],m,k,c_nonlin,xO);
in a
fer
10
CHAPTER 2
Linear and Nonlinear spring Trajectories
0.5
--,-----·,--,--;========:::;]
,-------.--~-~-
--Linear Spring
Nonlinear Spring
0.45
0.4
0.35
g
0.3
"'
·o
0
§ 0.25
0...
0.2
0.15
0.1
0.05
0
2
4
8
6
10
Time (s)
12
14
16
18
20
Figure 82.7 Trajectories for mass sliding on frictionless surface attached to linear spring
(blue) and nonlinear spring (red).
hold on
pI o t ( t _non l in , z _non l in ( : , 1 ) , ' -- ' , ' Col or ' , ones ( 3 , 1 ) * 0 . 4)
2"
hold off
ylim([0.05,0.5])
%add plot labels, leger1d aJJJ title
y.
set (gca, 'FontName',' Times',' Font Size' , 14)
xlabei('Time (s)')
ylabel ('Position (m) ')
title('Linear and Nonlinear spring Trajectories')
legend('Linear Spring','Nonlinear Spring')
grid on
Listing 2.2 Integrator function for Problem 2.13
function zdot = springs(t,z,m,k,c,xO)
zdot (1, 1)
zdot (2, 1)
SOLUTION
z ( 2) ;
-(k/m)*(z(1)-x0)
-
(c/m)*(z(1)-x0)-3;
2.14
As shown in Tutorial 2.4, the equation of motion of a simple harmonic
osc:illa tor is
i + 2(wz +
= 0,
w6z
with solutions of the form z ( t)
>. 2
=
e>-t, w hic:h leads us to the equation
+ 2(wo>. + w2 = 0.
11
NEWTONIAN MECHANICS
In the critically damped case, ( = 1, so
Al = ->-2 = -wo.
When a solution has repeated roots, the general form of the solution becomes
z (t) = Ae.\t
+ Bte.\t
so for the critically damped case we have
z(t)
+ Bte-wot
-Awoe-wot + Be-wot- Btwoe-wot'
= Ae-wot
i(t) =
where the second expression is just the time derivative of the first. If at time
= 0 we have the state z(O), i(O), we can write
t
z(O) =A
i ( 0) = - Awo + B .
Therefore, the general solution to the critically damped case is
z(t)
r
SOLUTION
=
+ (i(O) + z(O)wo)te-wot
z(O)e-wot
2.15
See Tutorial 2.4
SOLUTION
2.16
Using Newton's second law, we write the equation of motion of the rock as
..
.2
x=g-cx
where x is the distance from the top of the cliff (so, when the rock is at the
top of the cliff, x = 0). This can be rewritten as
d±
dt
d±
2
= g - ex ::::}
=dt=?
--.--=-2
g- ex
1
d±
2 = dt
egje-x
-
Integrating both sides:
-1
II
c.
1fg;
c
- tanh- 1
g
I
d:r •' = dt::::}
g c - :r:2 .
(v
i: )
r::T:.
gjc
=t
i; =
+ k1
=?
{2_tanh ( yegt + ki)
y-;;
12
CHAPTER 2
where k1 is a constant of integration. Since the rock is dropped from rest,
we know that at t = 0, x = 0, which means that k1 must also equal 0.
Integrating the equation one more time:
j x=If j tanh ( vcgt)
j dx =If j tanh ( vcgt) dt
=?
=?
1
ln (cosh ( ygct)) + k2
c
where the second step is due to x = dx j dt and k2 is another constant of
integration. Since we took x to be the distance from the top of the cliff, at
t = 0, x = 0, which means that k2 is again zero. Solving the final equation
algebraically for t give us
x
=-
1
t = - - cosh- 1 (ecx)
ylgC
which means that for the rock to travel y meters (the height of the cliff), it
would need
cosh- 1 (ecx) seconds. Returning to the equation for x, we
plug in this time to find that at the point of impact:
}gc
The terminal velocity is defined as the velocity at which a falling object
no longer accelerates (i.e., x = 0). Returning to our initial equation of
motion:
..
.2
0 = g- CVterm
2
X= g- CX =?
/9TC.
which means that Vterm =
Plugging 0.99 of this value into our solution for the velocity at impact gives us
0.99/f
=If
tanh (cosh- 1 (ecx))
so for the rock to achieve 99% of its terminal velocity as it hits the ground,
the cliff height would have to be
:r
SOLUTION
1
c
= -ln (cosh (tanh- 1 (0.99)))
2.17
a. We are given the equation of motion
k
x =a--x
m,
13
NEWTONIAN MECHANICS
To make the integration easier, we will use a change of variables by
introducing:
k
y =a--x
m
Differentiating y gives us:
.
k .
y=--x
m
..
k ..
y=--X
m
-mjkjj, so we can now replace both sides
Note that this makes x =
of our equation of motion with terms in y, producing:
m ..
--y=y
k
Thus, our new equation of motion is jj = -w6y for wo =
will guess a solution of the form
y
Jk"{Tn.
We
= A cos(wot) + B sin(wot)
where A and B are constants (you can plug this solution into the differential equation to make sure it works). Returning to our definition
of y, we can now write
am
x= k - Acos(wot)- Bsin(wot)
x = Awo sin(wot) - Bwo cos(wot).
Assuming that our initial conditions are x(O) = xo and x(O) = xo, we
can solve for A and B to find that B = -xo/wo and A= amjk- xo.
Putting it all together:
x (t ) = -am
k
+ ( xo-
xo sin(wot)
-am) cos(wot ) +k
wo
b. As usual, to use ode45 we need to turn our equation of motion into a
system of first order ODEs. We do so by defining z = [x xjT so that
. [a-x.!:i_x] = [a-z(2)
]
.l:i_z(l)
z =
rn
m
Listing 2.3 contains the code used to numerically integrate this system
and compare the result to the cmalytical solution from part (a). Note
that we have made use of MATLAB's anonymous function capability to
define the integrator function within the script. You can read more
about anonymous functions in the MATLAB documentation.
Listing 2.3 Script for Problem 2.17
-----------------,
14
CHAPTER 2
C'n
dotz
J.<>
<!l(t,z)
[z(2);
a - k/m*z(1)];
zO = [ 1 , 0. 5] ;
[t,zout] = ode45(dotz,[O 10],z0);
:3().1
x
=
"'
ut
a*m/k + (z0(1) - a*m/k)*cos(sqrt(k/m)*t)
zO (2)/sqrt (k/m)*sin (sqrt (k/m)*t);
+ ...
%plc~c
figure (1)
elf
"'·
""
subplot (2, 1, 1)
p = plot ( t , zou t (: , 1) , ' . ' , t, x, 'k' ) ;
set(p(l), 'Color' ,ones(3,1)*0.4);
set (gca, 'FontName' , 'Times' , 'FontSize' , 14)
1 = legend('Numerical Integration','Analytical Solution');
p = get ( 1 , ' Posit ion ' ) ;
set(l, 'Position', p-[-0.01,0.3,0,0])
ylabel ('Distance (m) ')
grid on
subplot (2, 1, 2)
plot(t,zout(: ,2),'. ','Color' ,ones(3,1)*0.4)
set (gca, 'FontName',' Times',' FontSize' ,14)
xlabel('Time (s)')
ylabel ('Velocity (m/s) ')
grid on
SOLUTION
2.18
If we let c0 = ~poCDA be the coefficient of drag at sea level, then the
coefficient of drag as a function of elevation can be written
C'= coe-yfh.
The new equation of motion becomes
.. = -
y
GM
+coJe-y/h ·2
(Rc + y)2 m ,
y ·
This equation can now be integrated numerically in MATLAB. The first step
is to put it into first-order form. Let ZJ = y and z2 = y. The two first order
equations are
ZJ
= Z2
·, _ _
z2 -
(
GM
Rc + Zl. ) 2
Co
+me
-zJ/h 2
z2 .
15
NEWTONIAN MECHANICS
~
5
0.5
<l)
u
=
~
0
Q
-0.5
2
3
4
5
4,----,----,----,-----,----,----,,----,----,----,,---~
~
2
c
·c
0
0\
.·
c:;
>
.... ·
-2
... ·
-4L---~----~----L----L----~----L---~----~----L---~
0
2
3
4
5
6
7
8
9
10
Time(s)
Figure 82.8 Analytical and numerically integrated solutions to Problem 2.17.
Listing 2.4 shows the integration function used along with ode45 to integrate
this equation of motion in the script in Listing 2.4 to produce the plots m
Figure S2.9.
Listing 2.4 Integration function for Problem 2.18.
function zdot = falling_ode_vary(t,z,G,M,m,Re,c,h)
zdot(1,1)
z(2);
zdot(2,1) = -G•M/(Re + z(1))"2 + c/m•exp(-z(1)/h)•z(2)"2;
Listing 2.5 Script for Problem 2.18.
'"
c = . 05; 1;[-,',f/J''
G = 6.674e-11;
Re = 6378100;
M
5.9742e24;
m = 10;
h
7000;
()!
[t ,z]
.'
\/
ode45(@falling_ode_vary, [to tf], [zinit zdotinit],
[] , G , M, m, Re , c , h) ;
16
CHAPTER 2
Positio11
1000,
I
I
900)
800
-15
l-20
700
·f
600
-25
~ -30
5001
-35
400
~::t- -------------
300[
2000
10
12
14
20
J ; ; ;
0
2
4
6
-~
8
10
12
14
TimC' (s)
Tinw (s)
(a) Particle position
-------------------L
16
'--
"
20
(b) Particle velocity
Figure S2.9 Integration results for Problem 2.18.
figure(i)
elf
y.
plot(t,z(: ,1), 'k', 'Linewidth' ,2)
set (gca, 'FontName',' Times',' FontSize' ,14)
title ('Position',' Interpreter',' Latex')
ylabel ('Height (m)', 'Interpreter',' Latex')
xi abel ('Time (s)', 'Interpreter',' Latex')
grid on
SPlor \/elocicy ::urFe
figure (2)
elf
··.c,
plot(t,z(: ,2), 'k', 'Line\-lidth' ,2)
set (gca, 'FontName',' Times',' Font Size' ,14)
hold on
1••
terminal= -sqrt(9.81*m/c);
plot ([to tf], [terminal terminal],' k--', 'LineWidth' , 2)
hold off
.,_-,
legend(' Velocity',' Theoretical Terminal')
title ('Velocity',' Interpreter',' Latex')
xlabel ('Time (s)', 'Interpreter',' Latex')
ylabel( 'Velocity (m/s)', 'Interpreter',' Latex')
grid on
Integration out to 20 seconds yields trajectories shown in Figures
S2.9(a) and S2.9(b). Note that the particle has a more interesting velocity
profile in an atmosphere with varying density. Rather than reach a terminal velocity, the particle continues accelerating until it reaches a higher
speed than the theoretical maximum velocity calculated at sea level. As the
particle gets closer to sea level the density increases and the effect of drag
becomes more significant.
17
NEWTONIAN MECHANICS
SOLUTION
2.19
The free body diagram for each mass is shown in Figure 82.10.
r r
I
I
mpg mQg
L-------·
1
I
0
Figure 82.10 Free body diagrams for masses P and Q in free fall.
As in the free body diagram in Figure 82.10, the only force acting on
the two masses is gravity. We define yp to be the vertical position of mass
P and use XQ and YQ to describe the horizontal and vertical positions of
mass Q (the horizontal position of mass P does not change, so we do not
need to model it. Applying Newton's second law to mass P, we have
mpyp
= -mpg ==?- YP = -g.
The same procedure tell us that
YQ =-g.
Because there is no force acting on mass Q in the horizontal direction, Newton's second law tells us that
XQ = 0.
Integrating these three equations with respect to time and applying the
initial conditions yp(to)
YQ(to) = h, yp(to) = iJQ(to) = 0, XQ(to) = vo
gives us the system
t2
yp(t) = h- 92
xQ(t) =vat
YQ(t)
=
t2
h -~~-·____
We see that the equations for yp and YQ are identical - this is the basis of
Galileo's famous claim that objects of different masses fall at the same rate.
Because Q's initial velocity is only in the horizontal direction, with no other
forces acting on the masses, they must reach the ground at the same time.
We can solve for this time, finding
t*
=
J2hi9