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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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33

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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55

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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88

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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99

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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1010

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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1212

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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1313

Complete Motion

The complete motion of 1M is*

( )

( )

( ) 4T

t0#ormm2

tcos2lm

t 21

2

+

=

( ) ( )( ) 4Tt#or

mm22t2lmt

21

2 >+

+=

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1414

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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1515

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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1616

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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1818

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 =

==