Solutions - Physics

dx!
dt !
"! # " t #
i
mi
f xi
g
dx!
i
gt " #
! dt : Integrable (3)
f
Physics 301: Classical Mechanics " x!
i#
m
Problem
i
Set 2 Oct. 12, 2011
c) Fx " , x # ! f " x # g " x # ! m x!! : Not integrable (4)
! !
i
i i i i i
2-2. Using spherical coordinates, we can write the force applied to the particle as
F! F e & F e & F e (1)
r
r
$ $ % %
But since the particle is constrained to move on the surface of a sphere, there must exist a
reaction force ' e that acts on the particle. Therefore, the total force acting on the particle is
F r r
Ftotal ! F$ e$ & F% e % ! m!!r (2)
The position vector of the particle is
r ! Re (3)
where R is the radius of the sphere and is constant. The acceleration of the particle is
r
30 CHAPTER 2
r
a! !! r ! R!! e
(4)
We must now express e!!
r
in terms of e
r
, e ! , and e " . Because the unit vectors in rectangular
coordinates, e
1
, e2
, e
3
, do not change with time, it is convenient to make the calculation in
terms of these quantities. Using Fig. F-3, Appendix F, we see that
29
er
$ e1 sin ! cos "% e2 sin ! sin " % e3
cos ! #
&
e!
$ e
1
cos ! cos "% e2 cos ! sin " ' e3
sin ! &
(5)
&
&
e e sin " e cos "
" $'
1
%
2
&(
Then
) ! !
* )
!
!
*
e!
$ e '" sin ! sin "% ! cos ! cos " % e ! cos ! sin "% " sin ! cos " ' e ! ! sin !
r
1 2
$ e ! " sin ! % e ! !
" !
3
(6)
Similarly,
e! $' e ! ! % e ! " cos ! (7)
! r "
e! $' e ! " sin ! ' e ! " cos ! (8)
" r
!
And, further,
)
! 2 2 2 2
" sin ! ! ! * )!! ! ! " sin ! cos ! * ) 2 ! !"!
cos ! " sin ! *
!! e $' e % % e ' % e % !! (9)
r r ! "
which is the only second time derivative needed.
The total force acting on the particle is
and the components are
F
!
F
"
F $ mr!! $ mR!!e (10)
total
$ mR '
)!! ! ! " 2 sin ! cos ! *
) 2! !"! cos ! !! " sin !*
$ mR %
r
(11)

which 2-4. gives One of the the correct balls’ result, 2
height as can in be (4) described for the limit by k y, % y0.
' v t & gt . The amount of time it
0 0
2
takes to rise and fall to its initial height is therefore given by 2v0
g . If the time it takes to cycle
2-10. the ball through The differential the juggler’s equation hands we is are ( % asked 0.9 s , to then solve there is Equation must be 3 (2.22), balls which in the air is x"" during $% k"x
. that
Using time (. the A single given ball values, must the stay plots in are the shown air for at in least the figure. 3(, so the Of course, condition the is reader 2v0
g will ) 3(
not , or be able
to distinguish & 1between the results shown here and the analytical results. The reader will have to
v 0
) 13.2 m * s .
take the word of the author that the graphs were obtained using numerical methods on a
%8
computer. 2-5. The results obtained were at most within 10 of the analytical solution.
flightpath v vs t
10
v (m/s)
5
N
e r
0 5 10 15 20 25 30
plane
mg
t (s)
point of maximum
accelerationx vs t
a) From the force diagram we have & m %! mv 2 R"
2
m% '! mv R"
50
r
N g e . The acceleration that the pilot feels is
N g e , which has a maximum magnitude at the bottom of the maneuver.
b) If the acceleration felt by the pilot must be less than 9g, then we have
& 1
! 3* 330 mt *(s)
s "
2
v
R ) %
8g
8* 9.8 m * s
A circle smaller than this will result in pilot blackout.
2-6.
x (m)
v (m/s)
100
0
0 5 10 15 20 25 30
10
5
v vs x
& 2
0
0 20 40 60 80
x (m)
r
! 12.5 km
(1)
100
2-11. The equation of motion is
2
Let the origin of our coordinate system be
dx
at the tail
2
m $% kmv end # mgof the cattle (or the closest cow/bull). (1)
2
dt
a) The bales are moving initially at the speed of the plane when dropped. Describe one of
This NEWTONIAN
these
equation MECHANICS—SINGLE
bales by
can
the parametric
be solved exactly PARTICLE
equations
in the same way as in problem 2-12 and we find 37
x% x0 ' v0t
(1)
2
1 ! g#
kv "
log
0
x $ % 2 &
(2)
2k ' g#
kv (
where the origin is taken to be the point at which v $ v0
so that the initial condition is
xv ) $ v 0 * $ 0 . Thus, the distance from the point v $ v0
to the point v $ v1
is
2
1 ! g#
kv "
0
sv ) 0
+ v1*
$ log % 2 &
2k ' g#
kv1
(
(3)
2-12. The equation of motion for the upward motion is
Using the relation
we can rewrite (1) as
2
dx
2
$# mkv # mg (1)
2
m
dt
2
d x dv dv dx dv
v
2
dt dt dx dt dx
$ $ $ (2)

d
max
2
v0
%
g $
" 1'
sin #
2-15.
mg sin θ
mg
θ
The equation of motion along the plane is
Rewriting this equation in the form
dv
m mg km
dt
2
% sin 6 & v (1)
1 dv
% dt
(2)
42 k g
2
CHAPTER 2
sin 6 & v
k
We know that the velocity of the particle continues to increase with time (i.e., dv dt ! 0 ), so that
2
" g k# sin v
find
$ ! . Therefore, we must use Eq. (E.5a), Appendix E, to perform the integration. We
1 1
v
tanh
k g * g +
sin $ * sin $ +
k
, k -
% 1 & ' ( )
The initial condition v(t = 0) = 0 implies C = 0. Therefore,
" gk #
g
v ( sin $ tanh sin $ t
k
t
C
(3)
( dx
dt
(4)
We can integrate this equation to obtain the displacement x as a function of time:
Using Eq. (E.17a), Appendix E, we obtain
" gk #
g
x( sin $ tanh sin $
k
. t dt
x (
" gk sin $ t#
g ln cosh
sin $
k
gk sin $
) C/ (5)
The initial condition x(t = 0) = 0 implies C/ = 0. Therefore, the relation between d and t is
From this equation, we can easily find
" gk $ t#
1 d ( ln cosh sin
(6)
k
" e dk
#
% 1
cosh
t ( (7)
gk sin $
2-16. The only force which is applied to the article is the component of the gravitational force
along the slope: mg sin 0. So the acceleration is g sin 0. Therefore the velocity and displacement
along the slope for upward motion are described by:
" 0#
where the initial conditions vt " ( 0# ( v0
and " 0#
v( v0 % g sin t
(1)
1
2
x( v0t % " g sin 0 # t
(2)
2
xt( ( 0 have been used.
At the highest position the velocity becomes zero, so the time required to reach the highest

" #"
2
2!gs VB
VB
#
2!
gs
V # 15.6 m/sec
B
2-25.
a) At A, the forces on the ball are:
N
The track counters the gravitational force and provides centripetal acceleration
Get v by conservation of energy:
So
b) At B the forces are:
mg
2
N" mg # mv R
E # T $ U # 0 $ mgh
top top top
1
E T U mv
2
2
A
#
A
$
A
# $
E # E % v # 2gh
top
A
N # mg $ m2gh R
& 2h'
N # mg( 1 $
*
)
R +
0
N
45˚
mg
2
N mv R mg
Get v by conservation of energy. From a), Etotal
1 2
At B, E# m v $ mgh-
2
# $ cos 45,
2
# mv R $ mg 2
(1)
# mgh .

R 2
R
R 45˚
Rcos 45˚ = R 2
h′
So
Etotal ! TB " UB
Solving for
2
v
becomes:
R
1
R ! " h# or h# ! R $ ' 1 &
%
2
)
(
2 *
mgh mgR $ 1 % 1
! ' 1 & "
)
(
2 * 2
mv
2
Substituting into (1):
1
2 + $ %
- gh & gR' 1& (
, ! v
) 2 * .
/ 0
2
2h
3
N mg + $ %
! - "' & 2(
,
R ) 2 *.
/ 0
c) From b)
B
! 2 + h & R " R 2,
/ 0
2
v g
d) This is a projectile motion problem
1 2 12
v! + 2g h & R " R 2 ,
/ 0
45˚
B
45˚
Put the origin at A.
The equations:
become
Solve (3) for t when y = 0 (ball lands).
A
x! x " v t
0 x0
1
y! y0 " vy0t & gt
2
2
R v
x ! " B
t (2)
2 2
v B
1 2
# (3)
y! h " t & gt
2 2

2
gt ! 2 vBt ! 2h#
" 0
t "
2
2
B
2
B
8
v $ v % gh#
2g
We discard the negative root since it gives a negative time. Substituting into (2):
2
NEWTONIAN MECHANICS—SINGLE PARTICLE R v &
B
2 vB
$ 2vB
% 8gh#
'
71
x " % ( )
2 2 ( 2g
*
)
+
The equilibrium point (where dU d! " 0 ) that we wish to look at is clearly ! = 0. At that point,
Using the previous expressions for v B
and h# yields
we have dU 2 d! 2 " mg # R % b 2$
, which is stable for R&
b 2 and unstable for R' b (2 . We can
12
use the results of Problem 2-46 & 2 3 2 2 '
x" to obtain
, 2 ! 1-R
stability
% h %
(
h
for
!
the
R
case
% 2
R
R
" b 2 , where we will find that
* 2 )
the first non-trivial result is in fourth order and is negative. We therefore + have an equilibrium at
! = 0 which is stable for R& b 2 and unstable for R) b 2 .
e) Ux ( )" mgy( x)
, with y(0)
" h, so U( x ) has the shape of the track.
3 2
2-43. F "% kx + kx *
2-26. All of the kinetic energy of the block goes into compressing the spring, so that
2 2
mv 2" kx 2, or x" v m k ! 23 . m , where x is the maximum
4
compression and the given
values have been substituted. When U # x 1 2 1 x
$ "%
there , is F dx "
2
a rough 2 kx %
floor, 4
k it * exerts a force / mg k
in a direction
that opposes the block’s velocity. It therefore does an amount of work 1 / mgd k
in slowing the
2
To sketch U(x), we note that for small x, U(x) behaves like the parabola kx . For large x, the
block down after traveling across the floor a distance d. After 2 m of floor, 2 the block has energy
mv
2 2 ! /
kmgd
, which now goes into
4
1 x compressing the spring and still overcoming the friction
behavior is determined by % k
2
on the floor, which is kx % 4/
mgx * . Use of the quadratic formula gives
2 2
k
U(x)
2 2
/ mg 0/ mg1
mv 2/
mgd
x "! % 2 3 % !
(1)
k 4 k 5 k k
E 0
Upon substitution of the given values, the result is ! 1.12 m.
E 1
E 2
2-27.
x 4
x 5
x
x 1 x 2
x 3 E 3
= 0
0.6 m
E 4
1 2
To lift a small mass dm of rope onto the table, E" mv an amount + U # x$
2
of work dW " , dm- g , z0
! z-
must be
done on it, where z
0
" 06 . m is the height of the table. The total amount of work that needs to be
For E " E 0
, the motion is unbounded; the particle may be anywhere.
done is the integration over all the small segments of rope, giving
For E " E 1
(at the maxima in U(x)) the particle is at a point of unstable equilibrium. It may
2
z0
( ) ( ) / gz
remain at rest where it is, but if perturbed
0
W " 6
slightly, / dz g z it
0
! will z " move away from the equilibrium. (1)
0 2
dU
When What is we the substitute value of E/ 1" mL We find ", 04 . the kgx -
values , 4 m-
, by we setting obtain W ! 0
dx " . 18 . J .
x = 0, - * are the equilibrium points
3 2
0 " kx % kx *
1 2 1 2 1
U# -* $ " E1
" k* % k*
" k*
2
2 4 4
For E " E 2
, the particle is either bounded and oscillates between % x2
and x 2
; or the particle
comes in from -. to - and returns to -..
x 3

m
1
% m2
m1 % m2
At equilibrium, like in previous problem, we have
72 CHAPTER 2
2 2
32
Gm1m2 m1v1 Gm2 2 r1
2 d
# $ v1
# $ " # ! #
!
2
For E , the particle d is either r at the stable d( mequilibrium % m ) point v x = 0, Gm or ( beyond % m ) x!"
x .
3
! 0
1 1 2 1 1 2
4
The For E result , the will particle be the
4
comes same in if from we consider "# to " the x5
and equilibrium returns. of forces acting on 2nd star.
76 CHAPTER 2
2-44. 2-50.
So one can see that in this problem x = a and x = –a are unstable equilibrium positions, and x = 0
is a stable t
d &
equilibrium mv
0 '
position.
& mv
0 ' mv
0 T θ T Ft
a)
# F$ d # # Ft $ v()
t #
dt ( 2 2 T 2
v ) - (
0 v )
2 2
v
2 F t
( 1* ) (
4Ux
0
1* )
k 4U0
c) Around the 1 *
m
2 origin, F
2 2
0
%
2
+ c , "# kx
+ 2c , m 1 c m 2
a
$ # % ! & m
& 2 ma c
d) To escape to infinity from x = 0, the t m 1 particle g 2 m 2 g 2 2
c & needs to
2 Fget t at least ' to the peak of the potential,
$ xt ()#-
v( t)
dt # ( m0 % * m
2
2 0)
From the figure, the forces acting mv on the 0 masses Fgive + the equations c , of motion
min
2U
0
& Umax & U0 % vmin
&
2 m1 !! x1
! m1g
$ T m
b)
(1)
e) From energy conservation, we v have m x !! m g T %
(2)
2 2 !
2
$ 2 cos
2 2 2
2
where x2
is related to x1
by mvthe Ux
relation 0
mvmin
dx 2U
0
' x (
) & % & v &
2 * 1 #
2
2 a 2 dt m ,
+
a -
We note that, in the ideal case, because & b$
x ' 2
1 2
xthe 2
! initial velocity $ d is the escape velocity found in d), (3)
ideally x is always smaller or equal to a, then from 4 the above expression,
cos ) 1
2* + . At equilibrium, !! x1 ! x!! 2 ! 0 and T ! ' mt
1
g . This 8Ugives as the equilibrium
0
x
a exp ' (
2
* t
( 2 + # 1
values c) From for the a) we coordinates find m dx ma a ) x
*
, , ma - +
-
t& ln ( t)
2
2U . & % x &
0 '
0
x ( 8U0
a
vm
# x4
md ' '
1
8U
( (
0 1
0
* #
,
2 + x10
t! # b$
exp t 1
(4)
4m1 $ m
2 2
a - * )
2
v
,
*
,
+
ma - +
-
2
F 1 *
2
x
c
md
2
x20
!
(5)
NEWTONIAN F
2
Now if 10
4m1 $ m
2 2
m # MECHANICS—SINGLE , then
PARTICLE 75
0
We recognize that our expression x 10
is identical to Equation (2.105), and has the same
requirement that m2 m
1
, 2 for
c
when v!
c 2 , we have t ! the ! 0.55 equilibrium year to exist. When the system is in motion, the
descriptive equations are obtained
10 3
from the force laws:
99c
t
when v = 99% c, we have t ! ! 6.67 years m2&
b$
x1'
10 199 m1( !! x1
$ g) ! ( x!! 2 $ g)
(6)
4x2
2-53.
To examine stability, let us expand the coordinates about their equilibrium values and look at
their 2-51. F is a behavior conservative for small force displacements. when there exists Let a - non-singular
1
. x1 $ x10
and potential -
2
. x2 $ function x2
0
. In the U(x) calculations, satisfying
F(x) = –grad(U(x)). So if F is conservative, its components satisfy the following relations
take terms dv in -
1
and -
2
, and their time derivatives, only up to first order. Equation (3) then
2 dv b
mv0
a) m !" bv # ! " dt # v()
t !
2
becomes dt-2 " $ (m1 m% 2v )-
1
. When %
m written in btv terms / F of these new coordinates, the equation of
0
$ m/
F
x y
&
motion becomes
/ y / x
999m
Now let v(t) = v 0
/1000 , one finds t ! ! 138.7 hours
2 2
32
and so on.
.
vb
0
g&
4m1 $ m2
!!
'
- !$
1
1
4mm m / m d -
(7)
a) In this case all relations above are satisfied, so F is indeed a conservative force.
and % ! d (& b$
x '
b)
t
m & btv0
$ m'
! vdt ! ( )
%
x() t
ln
b * m +
0
v
&
1 2 1 2
2
/ U
bx
F ( , )
x
&# & ayz ) bx ) c % U &# ayzx # # cx ) f1 y z
(1)
/ x
2
where f ( y, z)
1
is a function of only y and z
/ U
F ( , )
y
&# & axz ) bz % U & # ayzx # byz ) f
2 x z
(2)
/ y
We use the value of t found in question a) to find the corresponding distance
'
t

NEWTONIAN MECHANICS—SINGLE PARTICLE 77
where
f ( x, z)
2
is a function of only x and z
! U
F axy by c U ayzx byz f ( , )
z
"% " # # $ "% % #
3 y z
(3)
! z
then from (1), (2), (3) we find that
where C is a arbitrary constant.
bx
2
2
2
U "% axyz % byz % cx % # C
b) Using the same method we find that F in this case is a conservative force, and its potential
is
U " % zexp( % x) % yln
z # C
c) Using the same method we find that F in this case is a conservative force, and its potential
is ( using the result of problem 1-31b):
U " % aln
r
2-54.
a) Terminal velocity means final steady velocity (here we assume that the potato reaches this
velocity before the impact with the Earth) when the total force acting on the potato is zero.
b)
mg = kmv and consequently v" g k " 1000 m/s .
dv F dx dv vdv
" " %( g# kv)
$ " dt " % $ dx " % $
dt m v g # kv
& &
g # kv
x
0
0
v0
x
v g g
" # ln " 679.7 m
k k g # kv
0
max 2
0
where v is the initial velocity of the potato.
0
2-55. Let’s denote vx0
and v
y0
the initial horizontal and vertical velocity of the pumpkin.
Evidently, vx0
" v y 0
in this problem.
dvx
dx dv vx0
% vxf
m " F "% mkv $ % " % dt " $ x " (1)
dt v kv k
x
x x f
x
x
where the suffix f always denote the final value. From the second equality of (1), we have
Combining (1) and (2) we have
dv
% " $ " (2)
x
% kt
dt v
f
xf
vx0e
kvx
x
f
v
k
x0
% kt
" ' 1 %
f
(
e (3)