Unformatted text preview: Equation (2.105), and has the same
c
requirement2that m2 m1t, 2 for the0equilibrium to exist. When the system is in motion, the
when v ! c , we have !
! .55 year
10 ined from the force laws:
3
descriptive equations are obta
t
99c
m2 s
when v = 99% c, we have t !
! 6.67 year& b $ x1 '
m1 x
(x 2 $ g)
(6)
!!
10 199( !!1 $ g ) !
4 x2
253.
To examine stability, let us expand the coordinates about their equilibrium values and look at
2 is abehavior for small when there ex Le 1 . xsingular potential function h ( calculations,
tF51.conservative forcedisplacements.iststanon1 $ x10 and 2 . x2 $ x20 . In tUex) satisfying
heir
F(x) = –grad(U(x)). So if F is conservative, its components satisfy the following relations
take terms in 1 2and 2 , and tb
heir time derivatives, only up to first order. Equation (3) then
mv0
dv
dv
a) m ! " bv # %
! " % dt # v(t) !
)terms of y
becomes 2 " $(m1 m2v 2 1 . When written inbt/Fx$ m Fthese new coordinates, the equation of
dt
m
v0 & /
/y
/x
motion becomes
999m
! 138.7 hou3 s .
r2
Nowso on. t) = v0/1000 , one finds t !
2
2
and let v(
v0 b g 4 m1 $ m2
!! ! $
1
(7)
a) In this case all relations above ar1e satisfied, &sm1F is indeed a conservative force.
4 m1m2 o / m2 ' d
v & Fx & # ' bx 2
/U
& ayz ) bx ) c % U & # ayzx #
# cx ) f1 ( y , z)
2
/x (1) where f1 ( y , z) is a function of only y and z
Fy & # /U
& axz ) bz % U & # ayzx # byz ) f 2 ( x , z)
/y
t b) t x(t ) ! % vdt !
0 m & btv0 $ m '
ln (
)
b*
m+ We use the value of t found in question a) to find the corresponding distance (2) 77 NEWTONIAN MECHANICS—SINGLE PARTICLE where f 2 ( x , z) is a function of only x and z
Fz " % !U
" axy # by # c $ U " % ayzx % byz # f 3 ( y , z)
!z (3) then from (1), (2), (3) we find that
U " % axyz % byz % cx 2 % bx 2
#C
2 where C is a arbitrary constant.
b)
is Using the same method we find that F in this case is a conservative force, and its potential
U " % z exp( % x) % y ln 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 131b): U " % a ln r 254.
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. mg = kmv and consequently v " g k " 1000 m/s . b)
x 0 dv F
dx
dv
vdv
" " %( g # kv) $
" dt " %
$ dx " % &
$
dt m
v
g # kv &
g # kv
0
v0
xmax " 255. g
v0 g
# 2 ln
" 679.7 m
kk
g # kv0 where v0 is the initial velocity of the potato. Let’s denote vx 0 and vy 0 the initial horizontal and vertical velocity of the pumpkin. Evidently, vx 0 " vy 0 in this problem.
m v x 0 % v xf
dvx
dv
dx
" Fx " % mkvx $ %
" % dt " x $ x f "
dt
vx
kvx
k (1) where the suffix f always denote the final value. From the second equality of (1), we have
% dt " dvx
% kt
$ vxf " vx 0 e f
kvx (2) vx 0 '
% kt
1% e f (
k (3) Combining (1) and (2) we have
xf "...
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 Fall '13
 Argryes
 Physics, mechanics, Force, Sin, Trigraph, dt

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