2.29 and 3.7, 3.14, 3.18, 3.20.pdf
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Problem 2.29
The KEY
To apply Newtons Law in polar form all the
forces must be taken from an inertial frame
You MUST see the motion of the car from thegroun an N!T from the re"ol"ing platform
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Problem 2.29 contd.
So conitione#
$% Motion will N!T be raial but a cur"e oneon the re"ol"ing platform
&% 's friction acts opposite the motion# it acts in the irection of resultant of (ran (
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Contd.
(orce is pro"ie by friction ha"ing ma,- "alue mg
mgF,slidingAtmgF =
22222
0
2422 gmvm4rm =+
+ence#
mv2rv4gm
FrFF
0
22
0
22
r
+=
+=
0
slidslid4
22
0
22
slid
v
rt;
v4gr =
=
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Solution
m1 m2
O
2/
x
2m&is release at t = 0
1an you physically imagine the motion3
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Proposition
m1 m2
O
2/
xm2is released at t = 0
$- m&will e,ecute S+M aboutx=l 2mk=
&- m$will N!T mo"e until m&crosses l mark
't what t# this will happen33
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Fixing Time Scale
m1 m2
O
2/
x
4e nee to look at the motion in two time omains*
4Tt)2
4Ttto0t)1
== Case I
Case II
k
m
24
124
Tt 2
===
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Fixing CMs
Case I m1 m2
O
2/
x
( )tcos22lxlxatm;0xatm 22211 ===(measured from the wall)
( )( )
( )
( )21
2
21
21
mm2
tcos2lm
mm
tcos22
lm0m
t
+
=
+
+=
4Tt0
Note it is function of time
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CM ater T!"
The position of the 1M for instants t>T/4is gi"en by*
( )
( )
( )
+
+=
+
+
+
=
+
=
2t2
mm2
lm
4
Tt
mm2
lm
mm
lm
4
Tt"4
Tt
21
2
21
2
21
2
C!
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Complete Motion
The complete motion of 1M is*
( )
( )
( ) 4T
t0#ormm2
tcos2lm
t 21
2
+
=
( ) ( )( ) 4Tt#or
mm22t2lmt
21
2 >+
+=
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Problem 3.#"
.t eals with conser"ation of linear momentum*
$
m
!
1,2,%, 4&&.'
( )$v'm!v(;0( #i +==
7el of men wrt groun
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Problem 3.#" contd.
a% Let flat car attain a "elocity " when all men :umpeout simultaneously
1onser"ing momentum#( )$v'm!v(;0( #i +==
'm!
'm$"
+=
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Problem 3.#" contd.
b% Let "i;$an "ibe the "elocities of the flat car before an
after the it).man :umps off an conser"e momentum
( )[ ] ( )[ ] ( )m$vvmi'!vmi1'! ii1i ++=++
( ) ( )mi1'!
m$v
mi1'!
m$vv i1ii ++
=++
+=
( ) v
mi1'!
1m$vv
iii = ++==
/ (inal "elocity
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Problem 3.#" contd.
c% The case b% yiel larger final "elocity
( ) ++= i mi1'!1
m$v
+a"e a look at the
++
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Problem 3.#$
E=uation of motion* mgdt
dm$dt
dvmF r*l ==
2kmv
dt
dvm
dt
dmv
dt
dvmmg +=+=
.constk
gvtkvg
dt
dv 2 ===
KEY* .nformation about $r*l
's rainrop gains mass from clou at rest# $r*l=+v
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Problem 3.2%dt
dm$
dt
dvmF r*l=
E=uation of motion of the rocket*
mgmvm$dt
dv
m =
( ) ( ) .const
g$v,t*1g$
1v t =
==
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