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Solution to 'An Introduction to Mechanics' by Kleppner and Kolenkow
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100% found this document useful (1 vote)
1K views162 pagesSolutions To Kleppner
Solution to 'An Introduction to Mechanics' by Kleppner and Kolenkow
Uploaded by
Arkish ChakrabortyCopyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF or read online on Scribd
/ 162
PREFACE
This solutions manual was made available when An /ntreduction to Mechanics was
published but it soon went out of print. Nevertheless, a number of copies have been
circulated privately and over the years some errors have been discovered, We have
corrected all the errors that have come to our attention, A word of caution: These
solutions are offen terse, more terse that we would provide today. In the Second Edition
of our book, which is under preparation, we plan to provide solutions in a more
discursive style
nal solutions.
We thank Gilbert Hawkins for preparing the ot
Daniel Kleppner
Robert J. Kolenkow
March, 2010 Chapter 1
a
@) Aro = tars) f scan feck = rheafssk
) KB slay lean fh ecrark = -3f- ap esl
&) APB = cayesy + (3001) 7a) = at
a jk a“
Gy axe = fans 7] = -sfsafeirk
6 12
a) 33) UR
cos (X89 = ABs (A 8) we "2Y+C1NC-3S + CON (-1),
B Varneay® Wear eCayeea
= -0.P06
ts
Ayt Ail = Acose, = Ax
Ay” x? = Acre, = AB
Ag = AR = Acoso, = AY
A= VAx+Ay TAR = A Vaerptey™
Mence st4g2+y? =]
(4) RBH 1B) hence
¢8-8)- CA-B) = (A+B) CAB)
Aah Bre? = AT+QN Tt Be Kea
-2n8 = +2neB = ABO
oy \s
OL Oe
qt
a)
>)
; wr Ob
+
°
= A-B * 0
B =-+, =, ca
C-c’= CAtB)-(A-B) = A? B* =O
[él a aAteee =o
73 = ~ ~_—— ~=a = a
A Sc o= Ry A+ Bte)= AxB+ Axe
Hence 1AXB I= 1AKS)
=
= AB sin Y= AC sin
—a =
sing Sin ¥ ete,
a ALA A
VEZ) 44 4 Be aye eta jy)
a
t & = Mdeose f sciedsind
rie Similarly bo cor Pi esine
* aA
ab = CCN cos CO-%)
Bot B.D = (eas eNcosP) #Csine) Criny)
R-& <0 GYAN Ay Cag =o
Bho GY ny -Odny trys Fo
Hence nys 7 Va ng and ny= Se My
Wie a= Vapendeng = Ingl [ay Cero?
ngs tose and Aa t For (-2i+s]} +3k) = K-08 xe)
os ao
A= Bre
>I
ob
B= Acos@
B-(RAYA
Cm Asine
S=CaxR) ed [73] zs hey, G-7,)- 44 (4-7, )*
Zz ty osm hey, - £977
Be ve h/t,
h Meme he ETT
(T2 +T,) Diebance moved by center is
related te angle of rotation
by xsRE
Mee Yada'Re< Rae
ay
R= 2tsinwt f
Qe ciacout f
Hence
A
Weves 2tew coswt & Bt s= YT
Hence s= Fece Fbrak +
Face TAL ay
T= [ 28 (aace * Perak ) T
Face Shrak
A SOU KIO mi /g® 44% #2RKIO mi/st
T ic minimized by making ay, a6 lage as
possible. Trin = 33-75
(a) Vs tee red = heard
v= Valtecaa 7 VSR m/s
() fe CF- 6) t+ lrSrare) S
= -arth + @ycua &
as YaN*+6 = 24/ m/e
6 => A
de a= -Rwit
re] ~
aca)
alte at) a
sale A /eor d= lade
aw
Vv
a
jerk
e
— dz _ ja, dea
jekt ag st lel ge
*
e
rk = - Rw?
y= RE@-Rsin&
qt RCt-cos 8)
Se R6-Reose &
X= V-Vecos@
4 = Reine oaVeng
> fe Vwsne, 9 = Vw cos &
ve He Geax
Ga (FF) Fe (vs RPO) S
ae m, 242
aa (3- Byte ages ay
(by as H- Sto when HE
Be
Hence ame when en = Ee tt tad
(c) a= # (0-268) +505]
lapl=ia,| becomes li-267] = [sei
For O = {tay °
5+V33 1S:
Be SHS = 2.69 rad = ISH
Let v, be the imitial velocity
x= (y, eos @)t 4
7 Me sin) = Fg V
Let rock hit slpe at t=¢. _ hos.
Find valve of @ which maximizes xtt,). Lp ux
Xo) y(t)
te xed Syed
i Wyre kaxtras |
yc) = Wvesine) #, - bgt? = (tend) x04) - aie
Bt yh) = -tan? xt)
=> x)= YE (costo tan P teon? sine @-)
Maximising x(t) as a fenetion of @ yichl
tan (26) = tan(E-P)
Gy - F/
Fay a ~ Fem Chapter 2
. ze A
a= Eats st fn
@) w= Cade = (HE A. oe? Alms
a = « 24 A . a A
W) Pe (Vat (4- ¢ a2 i}m
) ig k
. 2 46 4\a
pve fe % ° | CR +H )h
1 pe
cs a
Fe¥- te kms
a | —
. I=
T= M% \
Meg -T™ Ma% ne
y= X% since the hength of
AT TS abning cent. [4
Mag = (M.+M.) %, Mag
ss MM, M. *
y= He > «= dae
9 Hene ope mee 2 WR 2 aM
a= (RS )4
T" (SSE) q
10 ize!
Engredients most likely t fall from top
mg-N =m Rw®
N= mg -mRw?
Neo at crvheal cpeed
N
Fe TR ™
mg
Ray ——+
(a) ¥
F-F=mMa Le mM
F=Maa £ Mg
Fe (“eMs) £ a
Bt Fr an=_ums
Hence F= (MMe) amg th
My f
ay ¢ maf
“gle ™,
aa £'=M,@ rt
F's “(a wei)
Bt iy asin > Fis (Mita
Re) Analysis in Prob 2.7 £-
F al
Nsin® = ™y
Nese = my /R
aR /y>
tan &
tan ® =
2
Re we
i
™@
mR wre
RFe G Me /. 32
But GM,
Gw» Me
R=
+
(2.8) C8. 6hyr08)* 77
(enxro*) #72 | = e62 re]
Feeng
Hence a= say.
Yo= ug T, where
Fe jemy in direction shown,
T 4s the
Lime For cleth b be polled out.
fe
aaa
rf |
a’m seg
vie vy~ gt
Bless comer be vest ad te Ee
Eb travels a distance 5, = vet BAY #3
stwie a =
2 ae) Ti
Fer ssc b fiz “a
2 ca)
T C4) (0.5) (32) rae
ais) + er +
+
|
*, Mg Mag Xa TOMGg EM x,
Tl mag = Max
ar’-T=2 (Massless pulley)
vy, ee, tt, + hatha’ Xe
te th, 3
= cons#,
Yenee eH Ma/a
x= (ama) 9
Solving >
uMatM,
\ze] +c"
Ts Xe &b-—>T
A
T= My, . i Tr sled
Mg> T's MX
Tr .
T-2T= 0 (Masslecs pelley) Ort
Constracmd:
Ken kp = comet.
Yence Ke = Xp
Cxpr xa) + Cx p> a= const
Hence Xt Xp = 2xpe 2X
Belving ,
me me Med = MghMeg
A Laymg + Sete + MMe] D
gs MAMeS = ke (MeMg) Meg /0
Oo N, Ng
fe ob x {Mat—> f
Mag
Xa, Fay = feng th of shring eee
x.-% + 27 70°
ref OM fm AM
f.-T = ML Fa Mag
Myy-2T = Mz ¥
Selon Te ae
"f 7
a wan * Hs)
\a1é]
A~mi
He -mg my
x-X= h-y
#-As-§ h ‘
Salvi 4
clvingy 5 Atg g- At
a2 4
15 al @) f N
N= mgqcos &
mg sn@ af s uN = wing cose “eS
Hence, tan O = ng
tam Orgy Ht oe
£
) feese+ Nene - mq=e e! N
Nese -fsin@ = ma 1S. e
Fm uN ab shipping bah
Solvng,
ae (S24) ™g
Tn ad +sine I
N
I
©) Nein = Feoe@ ~ mgz0 ‘e
Neoso +f£sinre =-ma — fear
foun at himed oN
Saloiap, ; ms
cos S tusine
Snax ™ Cae sees te d
2, F F
12 4 4 ewer
2F +N- Mg = Ma I t
2F-N- my 2 ™a | P |
Silvia Mg ¢
1 ye
a= em J ™s [217] Becawe Mg is ig ae
metionless, all bedres i
have the same heritenta/ rt gh
acceleration Q. |
Mag
T= Maa —_>F
M3q-T=0
F'= Maa
F-F'-T=M,@
7
RT,
ae ' |
Eliminating {F' T, a} Palleg halter 7
on M,
M.
an ‘Ma (mamtMa) 9
(= 221 a
FUT =-M,y, [ae T -
Ts Mh 7 y
F’=s M3 Xz F
% T
Moq-T > Med Ir M38
traints:
Cons rain s M,
+ x
hs
x, —xaty = length of string
Wace X40
-MaMag
X=
DM M+ MMg+2MM5* M5)
7 eal ee og
oI
F-r’s (A)a oa
Fis Ems BED Ja mit») mx
a 4
Satving , Foo Gtakfome & a(t Rea t)F
(3! Tand Tand
@) IT gie=W “Ee
Wut” es w
28ine
Tend
Tidy Tons e000 ss aN
—> Triddle
Inward force ™ 2T sin &
Sar tS = Tae
Taos (am\rat am 2 Zot
Mla
Hence T= “Gry
18 . 4
Inward force = Tsin & + Crear) sin
& ar 4S = Tae
Hence N= TA r
Merisontal forees are alse in eguilibrivm.
tra) cos 82 ~ Tros ME -f <0
>
Ff Sar
Assume repe & on verge of cliporng.
Then
feun
ATs ur ae
gt aut
T! id 7 ae
(Epa + abe yey
2
EE = mrust Ore
r
w= TE - EGP TS
2.
Reid T= AE = an Vee m= STR
We want ras small es possible, hence make r=R
Then Trin = Viar/Gp
19 Pm = 22g lem? For avai lable materials. Then
Trin = 5000 seconds,
ae)
Gravitational ferce on m at radius r= GMm />*
Me™, GR)?
r3 GNemr
Fee on m = QE Ne ge” < = Mar
Hence mit = — La r
SHM Pardod = ar Pq = Pik minotes
R
GM
Re Ruse in
ok Ge
os ad fe R
T= avr TA as abeve |
Bat] (See problem 2.26) Mereage
Z—_|™
3 ma, Bis XN
E Fe rece = me * 7
my = 7 a *
=> sum, period Ts a7 ho Fe mar
e \2.28]
Minimum speed:
(Secces as shown)
Neose+ fsin = My
Name -feoe = Mv/R
At lwo, EAM
S.tvng,
- SmO~ pcos @
Vine jr 3 ( tos ane) T
Maximom speed:
Nees @ -Fsrine = My
Nemo +feos6 © mv/R £
An
Setemnp
[Ry ( SO tecose sbesuseese) 1
cos & Tusine
“!
= T=(F-v|d? + (rS+ ars) FS
But reyt > tay,
ar -)
Hence Te-vywt? +aviw &
2 (t) Starts te ckid when Mes 4M oR ayug
a
by2 wie vew'tt= (ag)
Hence f= Tet Vous W2u*
. ==
“ aa a wot S
tena = 22
Gage 408 00™
me me
Fez a De,”
~~ 4
-T= mm, C= Hed — < n° a
“T= ™ ( &- re) ™ wa
uo A-Y.
: 2
-T= ™e L-#,- (4-5) s]
¥ . bw?
Stlving, ~= etn we
("BS ee
lay k,
axe F/R, . .
Ax, = F/ ky
Ox, aKa =F OK + Zk)
Wenee kepe = Flax, a UWne i) Ve Vea)
kee 7
ws Vk Be Ve
* w= lm Che, Tk)
k,
{by F
Fi = k,ax ‘
Fi=kax . &
Fae” TTR 7 (kok) ax
Mence keg = keke and to = Vike
=,
zz]
N
—>v
(a)
mg-N = mv? /R
N cannot be negative
Hence pebble Fhis off if
mE > mg or v> Ve
23 a) ¢
While in contact N
N-mgcos@ =- mv3/RK
; ; mg
With the condition N=0,
woe must have cose, = v*/Rq
There is @ more stringent condition based on f= “NV.
Frmgsin® 3 4m lgcoce -Vv/R)
Sth® S pcos ree v*/Re
bet wos
Use cos@-sir@ = Va cos(ore)
Then
co (e+ JE) ES G
a
o= mCF —rw)
i
= rw
a
?
rt
Jon te antichied’ Ii
rz Acvts pert
rey = ATE
Veoy = -WAtWB
This epve
Te make B50, choose veele-wrce) Then Tare“?
24 (z3e) r oO
(a) ona = rds +atw Tr m
5° ig
J
Merce § Ge=- | 3g y
oe, %
which yields co = w ( aye w (ss, wre
WY -T= mCF -rut)
ri
T# mrwt= mu? Te
re
2.35 iz
fa) -N=- a
Fe m£é =mdt because valld
f= AN
yz
Hence SE YE
+
tt =- (2 Gdt => ia ele a
One ie AGE)
vA
a) gaye +P) ~» exe, +t tn (iw At) EET
a
min -ee-be
£
Sea = - & (a rF=—)
x(e™ Ee = Ret
+
’
vex (= =x)
shire tte
Nees @ -mg=0 a
Nein = mrw*> mr(3Z)
ten o= = (x)*
:
#- =F)
Meten
a (Be) # ote)
rae heaa(a) =
is 0 parabela ef revolution.
26 Chapter 3
2. 4p,
BD Sade, (Gate
M
4% . 4
. M
Density = (ka)(vEa/a)t
aM
T3 art
B3y
w= EU
R, = EGpmy
Ma
Ry ERM tH = RT May
MM Mit Ma,
3
BE] com Fellews the eriginel ” ene
trapectory . CM and beth
masses land simuftaneous/ oe
because imitia/ velocities are herisento/. When m landg
27 ot bo, emis at 2b. Thus 2b= 2c) + m(o)
smtm
Selviny , x= S/o From The launch stotion
At height h speeds v= VWaagh
Nv = (mMtm)v! ws
viz MX isthe new initial
Velecity after monkey is picked up.
Pair rise an additional height
at
By 4 5g GA) (vragh)
Bey Immediately after collision with sandbag , plares
ity is reduced from v, tev,’ MVe /¢ Mtm)
M= ymass of plane
m= mass of sandbag
Retarding free Fx (0,4)(.250) +900 = #00 Mes
ve yla-ak where a= Fp /Cmem)
Steps at t’= Ve /a, . a
Distance of travel =S= yt’ kat” = Vv, /aa,
sm(Be) ve (mim) a = the (BX)
28 + BSEYCEE)(00y*
= 2560 feet
&)
While block #1 is in contack motion of bleck #2 is given by
where w= TR/ing
Mx, + Maxe mn. ( Zz )
= Tt Make ma ~
Ken ™tma Tema \t- “@ coswt
xno t~ £ cosut
Bleck #1 loses contact when xeL, This ocoure af wie Z
After this time the cm moves at a constant velocity
Vem- Vem enn be found fromm #, and x, of £,
= RO em x2!) ma wl
Yeon oe MA) nam, ) SS
mre
Initial velocity v,= Tags
Impulse = mv, = 50 Vaxaexok
= 199 kym/s
Be)
Let met) be the amount
of sand in box car af time
+. am is the amount
veleaced in time 4+,
29 Fea Cotemned +Ayn)(v av)
7 Cem tt) +4) ¥
PoP, = Cmte Av= Fat — te first order
= (mem (2)
dy= _Fdt
Mamce)
Let eo => mlt)= m- 24-6 ~o
ene A=Be 4/a
BAG} A= mass fant length = mp1 - wee
where 2 sess denscty of water
OP, 7 ARL, = Av, at
AP transferred to byvramd is at 45°
as theun and hes megndede Taviat. ft
33 Hence the firce on hydrant is
A
4B - av = = Pp oy
AP transferred
[EET] water speed at height h is
v= W2-39h
Asseme water bounces elasteally
Seem can. Then F = 2av® |
max
whee ave Avy, = (sm/ut),
Hence owe 2 (¢z)v and ro
h= 22 -w
73 9432),
nom OW
Pex (m+an)(veav) *~S
P.-P; = mav + (QM) = mq at | wo .
Hence Mgt + UM veg Fem, “CO
on | MEE + kM = My
34 ve q-kv*
As Vs increases | de decreases and eventually
becomes sero. Wence at terminal velocity
4" kvye eR ov = Vip
rm term
Fore on 2 Square centimefer = AV~
=(OR)v = sxioty
D> Total force = (Seo\ews*) = 25x52 N
When bowl i mering epward, do owe‘
and total force =k 9x0 °N
[e320
™ & vu de =— mq -mbyv
™ a -v Ym = —mg- mb
by = Very
y -*h
v= AA she
Cheese the constant A
soch that vod =o
sb)
v= Yes (-e
35 on Chapter 4
*, = mgt
&= mg (QR) re mv>
weg oe
rf N= ™q, mv 2 amgR.
Hee Ge SmgR= mye
#= 3R
E,= a My" |
zt
B= EkL —+_ Lie
4
B;- 6p Me mf mgbxde @ my bct
Bleck comes fe rect when zmvgs (bong h Lo
My,2
> ie
k+ Mg bh
Loss of mechanical! energy = a Mybe>
- far (4 my?)
VE3] (a) mv= (Mem) V
V- (im) v
36 (6) B= E(Mem) V*
E, = (Mem) £(I-cos ¢) 4.
V2= aq (cos g) u _ nes Aimed}
v= (M22) Tage (cosa)
i]
o> mah m
E,= mg (4-2) t
+E MVi +d my?
Hence
oz EMVR dmv? mh
By conservation ef momentum mm
= ey Ute
my => ve (atm 2g
Teel
-F= m (¥- 76?)
O= VS +2Te ZO
= rO+arw
Bo. _ 2e F
a r
Hence das - adr :
a
comt - 4%
tn ws alee tne OR 6 = Ta a where rlbJ=L, and Ot) =e,
a
we SFr =m Yee or 24 YY de
= dmvict.)- kev te) ne 2 (328 -zz)
Avec) + m&wr more -(amvica) + £m b4w?)
z
a mete ited
Aes
ate Ze 2
Hence W
Ww
rwvicey + drm bzeat)-Comyce,) «kmh uf)
kKeCt,) — KECt)
He) N
N-mgcose = — xpy*
Contact is lest when N=O
™]
From energy considerations
mg R= mg Reese + Amv
& = 2g Ci-cose)
Contact is lect when
~gcese = -agli-cose) SP Fes =2
Hence x= Rr cose) = £
38 1c}
Vertical force on Fing = T-ANcos@-Mg =O in equilibrium.
N-mgcos@ = ~mv7/R
T
By conservation of energy
mgR = Smyv2+ mg R cose
BYE = 2 my (I-cose)
Hence N= mq (Feese-a)
Ring starts te rise when T=0.
Amq (2-Bcos@) cos@ = My
Ms
The maximum value of 2ces@ -Sces*@ occurs af
cos@ = Vz, Hence the ring wil net ruse wrless
@m)% =m
m2 om
Angle is gwen by Bees" -acose+ Smo
=> cose= s+le-s
Bk Ska? S agex, +x,)
Ope jy) = AE Cay) Cc
A amplitude = %-%—% ae — %
39 () Bleck comes te vert when kept F or x= £/k
Hence 4fn/k = %-f/k
nee (bps 1) ee
ra ;
mv, + mV, 2 A2mV
Total energy available = Q+ amvyitim =
‘ v
via Elves) e & 5 orev gS ® A
Gt V= acvien) m
ve VV = Cua Y)
Mence we rewire K(vtevi)+& = ECv3+¥2 +2y,-¥,)
or Um = -E Lyi Y2 -24;-¥,,)
This condition cannet be mtishes because Q > 0
and the right hand side is SO:
let Vg=Vi-V.
V2= vitV2 -2v,v, cose BO
>
ay yz ie
Vevey 2 avey,
40 IO
m= M:
@ (0) 2 av Ve = Tee OT
fa) At veo, amplitude is A, . Hence new
amplitde = A.
(B) B= Xk Camphtde)” = & KAS
Mence Finitas 7 “Fine!
) year ee = eT,
(2) mv=CMem)v! mi ,
Ve (te)v Iw, flv
KAS = & (Mem) v2 « (M,) § Mv?
~ (He) SRAR
PR A
(3) ae= KKAZ- SLAP = ERA? fie #57
= (sna) & kAZ
ACT
bm lids the pan with speed
& Cam)v? = Cam) x x
Max
Am= 7
ve Vagx
41 rt delivers momentum (Amv = Ye Vagx ax
Average firce. dee te collicrens
= F- 4B = Sia B- Bip
= 24 x x
Weight 2f chain en pan = Me x
Reading ef seale = 3Ma y Mg
= 248 (49%) for Oct < PE ig
= M for b> V2ey
1 af Rey +
Ea]
Speed at mpect = Vv, = Fagh
FE abe duration of impact ,
Fo at= mv-o Ee
s= vyat- & SS Cas)® shy
Se vee ~ te % s
oak EE
Hence An= 3%
.
Alternate Metheds
== ime
Dse Xmv,*= Fs
sl FE mm a (iMWdCISe)
P= = mah tet Ces) Ub /Eee
on P= 15 4/52 => Safe te drop
42 413] dp I2 <
= ra “AY, Wy,
os o= [ at + am |
(b)
ey cadasyn® _ Gadire 7 _ 7
ke ae] ¢ [Spee Se [ors
wo 12 An
eI
ys -SMm - GMm = -26Mm
(a) vr a Tatex™
6) kmv2- 2GMrn = dmv? - 26uMm
Ta-Fa® vas
vie vie HOM (- pe) = vie SECM
©) w= M
Se ee
“Re
b= Ramis bd 4
M fre] B > A/x2
@v=-S2-% Seay
= Bx+ A/x tconst
UL) Let const =o
@ of > 2-4
> x- Me
UW) w= Thm whee ke 35 = 2
0 ERY
Be] pe ab = amet = (“Sree )°
ge
= ayas ft-lb /
P= AURE = 475 horsepower
S5o°
ee
hp" f£+Wsind fa
Ss
= Wloogs+sine)
44 Bot f-Wene 4
= w 0.06 —sine) oa
Bs ‘Ww
Pe 5 Yop =ny,
V, = (S08 tsi Jy = ( 2esracas le
‘ 5.65 ~ see “P 0.05-6.0a5
= hs mph
[3] He leaves the ground et tome T.
Yo = [E-3]r, x= 4[f -q\r*= LyT
Fs Pon =f)
Mertop I: poate
mye moh > = Yagh = J@OGN@) Yo “
= a sth s= Left
Fe ee) = mp any = oq (Bi) = 480 |b
PR = @80)(139) _
> “eso > 6! he
METHOD 2.
ce | fe wte mgs) oo (h4)
Ps 7 emer met) a1
z 280 2 (200.5) = 0216s
a Ve 139
F = Cj(a+hsS) _ gy he
e (Fy0 (0-216)
45 va (CB wswt-q)dt = te Ge sit -gt
vie wR - 493) =
sex (OE (4% B smust-gt) te (-E caut-ty8)
°
v,
= lagh = Vaan) = 13.96 44/— ord
Vsing
wo and Be
S215 Ft, solve for
w= 296 Ss”
F,= 742 Ib
Powere Po Fra (Reesut) (4 F smut -3*)
= OF, ( oe Sin wt coswdt -yteerwt)
~ a (F sivohcaaed ~mgtersnt)
= (19,92) (142 Sinust casust -/60 Wt cos wt)
Power is a maximum at wt = 077
P= 7230 Hf = 13 barsepeuer
max
Consider macs \ {
7s
of sand Am.
46 The memrentom of the belt and te transported
Sand dees net change m At.
a Po = (amy = Hat
e- oy
Pe fy Sve
GQ) 4 Ke = & CAn)v>
J
fea
The remaining power is needed! te make op for
Frittisna/ -locses 22 the sand mitral shine
on the belt,
FRI
@) apy tamyy,= Fat
J J d
Fe oby, Ga tem oe dem ay,
y,
ey
é
Fe avy
Fetet = Ave+ AgY
WY Pauer delivered = iy Yer AV3HAgy ve
BK Cayyre potential energy
Potental enegy= (2 aydm= (1p aydye & aa?
B- RAG ve Sp aye
BE - 20D Ba gagh = Zavsergyy,
47 There is no diicrepancy with he gravitational
part of dE Ceconserveative Rrece, no Wise pation
of mechanice/ energy.) Methantal energy i lost
in the “collision” of one part of the vepe uth
the nent ; think of the rope as « chain of
links. a
eaal ve vy -qt —* VeCi-e)
A egele a Yo
cs co
T= 2vCime) 4 AVMs Gi-e)?
g
= at G- eS) [ r+ G)~a) » b= see 7
T= Ge) Dorms > Fe ( ge)
a
E23] v= Fagh —_
CMe vo= My, + mY, ad 3m
4 (Mem) v2 = RMytekmy> }” ot
Solver 2(i-™ nm
cn on By - Rn
z R
(Mom) v= M DOR) y- BY tm
2. 2
= MvS-amv,2 +e y4 +e Va
= 2
Am Vy +ame Yeon, * ™M ye
eR vit ~2AVoVm ~3Vz2 =O
48 3 8h)+ Thy aeiayt
Von ‘o My +1. om ay, = alah
Atence marble
rises ta deighd gin dy
a
mel = 2 mv
He ey = 7 f
4
fey] A-B callievons: =—y,
AWB ciksien: CGR
My, = 2Mv M M
Weavy,
Kn emt eo
E’= £(am)y2= £ (MQ) ~ $e, am — Vv’
EZ the eriginal energy is dizerpated im AB colleecon.
AB-C ceilision +
Ses Sp
2M M
CoeXo onde
SM
amv! = S3mMv"
We yin ay,
Bre k (sm) v"*=2(Lmys)
"35
Hence BEF FH bs diseipeked tm the secenc!
cetliscan | divides between C and the other
cars,
mM
THE) ave mr me! moe O
a :
;
Komyte Smv ebm om OM
7 ”
Kimyits $(kmva) ¥ —>Vv [F27|
m O> O2m
1 Yo
amy
mV, =
”
+ cose
ant + mv" co!
av
1
3BY = mv" sine an
Aessoming the collision 40 be elastic,
O
Hmvp = 5 Gm)? + Lv" had
do
Silving, Vor ¥" (sine +eose)
v2 = ve* (sintoet) moO yw
Hence sintetl= sSin*g +ces*6 + 2sinOcar® = /t2sin Bose
oR Sin & = Reese
tan @ =2 => © SH 63°
50 ae|
Yon Tm
(2) Transfer te the O-v oO
GufM system: Ye be
Jon
WUv-Ve Y= Ve “Oo Te)
v-v
Vee ey, eye ity
Energy in the Co system is &,= &lam)¢-v.)"+ 5(7#) x
can [EMO EP J mV" (2g ema g
3
All the energy in the © system can be vosed £r
the reaction, AL threshold,
7 =
=r Fy 2.2 MeV
B= HH Me
At threchhel/, nevtrons are produced at rest im the
© system so thet they heve speed ve in the L sycdm,
Their energy mL is
B.2 Kms = GE) (Kv?) = (5) tam) v2
~ ar Fey
4, O45 MeV (by uv =v -100v! Tm
4m
Om-r O
™
v’g oO” o> v
E(4m)v*= Emve + 4 llem) v'? + 2.8 (Mev)
Solving» we find
56 (nev) .
Nv -~8 VV ~24V3+ GH
—
oR vs OVE Vuroy?- 24E# (mev)
aa
There are ne real reets wnti/ HNR0y? > A464 (yey)
oR (S60) E(4m) v= 246 4Mey
Fem) V* 2 he Mey
(threshold)
There are to forward velocities
if @v)* = n20y2- EEE (Mev)
on (528) £(4m) VS -P464KMeV
Ctm)v* = 467Mev= E& + 0.27MeV
2
The tee groups of Grward projected neutrons
Yepresent neetrens projecte/ at O° and /FO° in
the © Syskm.
52 = oP i mV. my;~
¢ f= - ve
* + Hy 7
()
‘Yo ont
sy ao
Zn Frame in which wall is Stetierary, we have
-o0
feu ; Fev)
Hence in L frome V'= (VtV)eV= Veav
dviay sve 2Y 2 vv a
dt a By =
dee dv dt eit dve -
dx dt ‘dx Vat ea Sr
Hence dv _ oy
x
An v= -tax + tac ®
ve ¥,4
as 2 a 72
z a2. mv = myn MUL
ro at By * >
x
(cy d:
We = Rene mre [Ss = mvterlts)|
4 mye
=
* °
AKE = EE = Xvi Fm” a
ARE = & VEL bye
me — x
™
=, @ ee
Lab F OMe
ned CM Frame
@) mov-Vv)y= MMe
V.= Ve i V,7 Ve
mem Ss
viv. = M¥e
oe eM wo
VE = V4 Cy) =A VCVV,) cosltr~ 2) %-7
2 vv,
Y= (Se. )(mirmtramm cos ©) 7
Y
= Vo @
\e (=) Crt M742 M cos e)*
ib) KS Ey,” a
K.= Hmvi = 2M enxit
‘+ eye (mi4 m+ amm cos ©)
Fo ge Witmt+amMes@ 2 2mm (incor @)
° CMtrnjJ® Gnem)*
54 Chapter 5
Pe-ay- -Bl*_ ou A a
(a) Fe-sy axi -sy j -gtk
3
= ra Ax? 28g} -acak
GG) Fs -Si = 12Ax_ f 2Ay 4S zaz 6
wagtege |” Tene) ~ Gevtepak
4 xAey ee xe ee
ee
— = Ly au
(Yo Fe -au “Ber tHe F
2
z
= 2hcse Ay Asino A
r? v3
mo , ;
A 4A &
@ Es-abp = ae (els tye Fk)
= Art (xhe gp wal)
Cort E = 0 by deck calevlatien or using the
result that all centre! Aerces are conservative,
=> A
oli = = (-28x-28y) k xo
k
*
xo
power
Ss .
Hence A is met conservative,
2,0
oo
Rmvyt + uleo)- Ems - Ue) = f Fede 20
= sf Cytdx-x*dq)
r
ut
A 4.
Bet Ute ed= 0 bec © ORY =A
=e
yn se that re a
amy = amy2 +A + 2S x "dx - 8 yy
,
we kmyt +A - S/s + Bf |
H:
7 (Ree + BBY
Hence
3) =
(a) Cowl FO Ginat) cot wo becavce the
evel eperater acts enly on Spatial coordinates,
Gy
‘\ The werk areund the
indicated closed path is
=> ime ene thw carved
, part bot net aere over
the straight part, becaure
Phas a different
magietude at 70° and 180"
ey
=
F —_ —_
Me FP dees ne ark because
a =n
Fedé =0. Hence (Fide around
a closed path vanishes and
the fecce F is conservative.
56 (a)
)
te)
u= Axe
ap 2 5%
phe - 3Anty%e
k
te
4
= A Ta-r) i]-o
nao
9
¢
&
1
u
>
m Qe one
p= -A C3x+ £74, 2))
he -aAB Az
f= aq + gece)
BE = -Ay- AGE = -Ag
Hence 9 =O and U= -Alsxe yz)
= fj oy
col FA | 2 &2/- A [Cxe-nyie-J
Yt aye xyz
* ©
Zz exe, A
curt B= (Se Antyte -SeAxy edi
<2
+ (aaa atyre® ~a~Axtgie”") f
z\a
+ CisAxty* eX* -15ax7yt ev dj
<0
z
ws Ax gra” + fey)
se
37 (1) curl = (Ax cos (xy) sin (G2)
—pAx~ cos Cy) sin (Bs) ) f ten
# oo
-4
fa) U= Cxe = const,
The lines of constant potential are
quien by x= Ket here K= const,
te) dP = del + dy f
Along a line U= const. , db=o=Ce “da- Cunt
Hence dy= 2 and
d@= dx (itil) aleng « hne U=consd
58 = Pal Pax ft]
ce) WD
Tuedr = cox ret-c!y eo
By Stokes theerem, § Aedi = [CRA)-ds" | here
d® is an element of the surface bounded by the
closed contour,
If TRA =o
§ Rede =o
Hence A ‘is mathematically analogous to a conservative
force. Since any conservatve force can be
written as the qravient of seme potential,
A= ¥%
{a
R= +2. Mert [swe*( 2)” (6 cose sine) ]
FE
The Feree disappears =
at ©=0° and O= 70° 6S
& / Siem Sem = (5.00) C6) Te FE = Leaxco™®
59 A (ad?) Shay = 2Ad?
War = [Pde = Ad? ae:
and
o wa x
= Fydy = 0
Vg = t aly .
xe 5
= a
Hence Weog 7 Ad’
e464
w- gga CeAx-zay) &
RRS = ChAx-zAy) k
Ay 2avo
= $F. vs = 0 mF. a =A S(exaydydn
= Ad s wads - ahd (7 4gdy = 2Ad7= Ad? = Ad?
60 Chapter 6
B= = Hx; .
a a
ms T+ "5
= =! Lae’
= amr =
fj 7 me Pi
Lie Sip! = ETRE
bis Sexe EZ CR-RIK RP
= SEXP ~ Re zp; = Since = Fj =o
(b>
t= SUKe
=F’
= ws = ~ = =>
Tim £0 xR’ = FCF-RIKE
dod - %. vo .
Tl Sxe - Ree ke =F since ZE =o
i
The fetal angular momentum remams constant,
Tavheal angele momentum = (Myst Ng) a3 ww, co)
Angular momentum at time 2 is given by
(My+M,~ a8) atu, + (Mgtr#) bru,
Note thet un= Wted becaure the cand exerts no
berqve om drum A as if eaves.
61 Hence
Cmgtitg) aFu toy = Cyt Mnat)amentos +(Mar rt) bo,
Se ay Atal waco
. (agent) b*
ee)
The verticle companent ef angular momentum Ls
constant and equal te ite initfe/ velve 0.
GY Angelar mementum of bey
abet pivet = mlv-2RwW)2R
Angular mementm of eng piv #
aoe * CE eMR2)w = -2MR wv
Hence ™
mCv-2Rw) aR = 2mRiw ro
= ashy.
oO Cram R
pivet
«by +. °
Hence.
bang L- “pep ° and wo,
62 jeu The central
qrevitational force exerts 5 aR
M
ne feryes hence L¥ const, %
st
L= (mv, sin@)5R =mvR
M
By concervatien of enegy Fe dmv - ¢4m
= gmy? ~ GMm
Nence vr Sv sine
z.
ve
oR sine =
N, +N, = Mq cos &
F,+£, = Mg sin®
Terque abet CM:
om NiLa-M, £4 + FL +44,
= (N-Na) 4, + (Mgsin) 2,
Retving , NZ EMG (cose - 4 sin &)
Na 4Mg(cs6 + & sing)
For the given nemerical valves ,
N,= 724% jks Ni= /67# ths
63 Ni t+No= Mg
f:rf= MVR
Take torques abot CM:
om Nokon, £-CRE)L TR
= (Noni) & ~ “et
atv = 2 Tmo Myth
Beleing, — Ninside ~ * LMa~ aevay J
= 2 My2h
Nyoteide aiMe+ Beg Zz) J
67] d M\ an
Sr linea) yp oe (BE) xd 4
By te parallel axis Theorem, x4 dm
8t> CA) any adm =(¢)* dra b t
12 poe x
Uptee | [ee a
aE
= SNe <—
\ee Consider the sphere. ————
te bea stack +f disks.
dre dm (R* 22)
= 2.22 =
dM T (CR z2)dz ($3) (eM e)ae
64 te) N, By Ne
N,+W2 = Mg
MN, (4+x) = Ny (-4-x) | ag
We no AE Ud) fab sd
n= Ma ( =)
= sa M
Horreental Force *&~f, = (N,-My) = “AZ x
Mea - “MS,
ae
SUM xm X, coswt where we 14E
6} =T=(F+£)R
= AUN tN) R
(nem FE = ATE EE = Mg
Ne Cigad+ Nz (isa) =z Mq
Mee Natt+fy
ee
Mg
N, (ines Ny Cite) € ( for horizontal equil)
. M. i+
Solving, Mi VR is (7)
nr (34) dt = %,
w= (B)t
os fudt= = (SE) *
me F
Ak t=4, , = “length L hee heen #recuned,
inne = £058) at
BE [2eze
a FR*
wee, af bee
e
w+ SEA e
2LE KE XO og
Hence I, = ae = (e.5y* 400 ky ™
éia| re ZT 72 , beam wi// not rotate:
ae .
art
2 dy.
Reine Mig 4 = Tee orl ig
Magq-T'= Mae
nos ofa hes
4 g MaMa Maq
sar = =o
TS 2r Mat Me
66 VMence, the condition becomes
4+ Ma Ms La
ML = 2 me
oe MatMg
a.
es]
Force ic central, hence angular mementom iz
(a)
conserved. Momentum and cnerpy ef mm change
becavse of applied force = 6 Tu wo
(Tat eo
L=myr = my R
v,
ee
v=
'e
al
(b) Angular mementem is not conserved because Kree ip
alee changes since
met central, Momentum
ST ap mo het energy it conserved decaure
os = v
Tide = (TPH =e
7
Nence Emus & my?
v= y,
1 2
Fei
8
Mg
wis Se as € = Za
a Sule = 7
67 +|w
bo
«€) a= xx =
@ Mas Mg-F Mm
Ra Mima Se
i de °
BS] gaz & °\z2
Take torques aboob pived |
-Mgtsine = LS
&+ (Mat\o Zo mal
=
w= yee pc=emrs (Lme% me2)
=
= M(R%*L4)
Faq ue jaz C See preblem G15)
Using the parallel anis theorem, I= &MR*+ ue?
T= 40 = [RA+*
res
- 4 eo +t at _ ( Bie
os at z male = SS
IA
Hence 267s B+ £7
Le the pied point tm the hid, Ldgh
68 Take torques about pot.
T=3rs6
T= - 5h r+ mg sine
Te ~K( Eye ~ klt9 omy to
Ba (kB + kL?-my£)& #0
os KL kL mg E
=
me
=
w= \ek-ag x EA if springs are strong
(we
.
° >
T=xrs
-mq $e -mglow rs ia
M,
mets mgt 3
60 — ee
Mem) £74 Be
re gme + SMR ems? rm ae (eo
Capem) of
If disk is ena Free bearing it dees net contribute te
the retatienal metion, Nence
x= bmeime* Ts ov [Gow e
(Qim)st
69 Energy Method: = (mg +mMegl)(1-coe o)
+e Lome (amex uly] e
SF
ar e
See Fe const. 3
> o- (mE emge)sing + (5 mle ueieme’)
as above, xf Sth ic free, the term
HZ (EMR S27 chetd mot be ineladed tm F,
ara]
@ TATE
-co = (CEmMRNS
+s
aMR*
o- Ze
UW) New mement of mended oo EMR™*MR?S EMR
=o
1
we &?
At 47> %w Oro, & =H
. . 2
Zim me diately before cothsion L= Té-sMRoe
Mierward L= I'6's 2 mptw/@ where 6’
is the mew amplitede, Because L ie conserved,
+ met
Rmetwe,= Sua pp us,’ ob 6/2 a 5
A
70 FE
h
v= $6
and @ atO=70°
e
M3
F C60") = Mg £ cos (s0°)= Mgt
F(R) = kr (6)*= ML?
Hene 2ML*G* = Mgt
«4,5 ER > EBM, Go tus
N
ez FR= (2 mMR~<
fs RMRX = EMa £ ¢/)
Fer relting wrtheot slipping.
Me
N= Mqcor® 4 _@
Mg sing -f = Ma | Mgsine = ZMa
But FS Aw = My coro
Mas “gM cos
“aug sing = Amygoor@ =b fares gu
7 fa) o= mF = rw)
T-rwr=0
we
r=Ae
wt
Mene xf G=O, rare where "o=rte)
wt
+GBe
() Nema, = (aw) N
&
= amw (nue)
wt
= am wire
©) P= Fy= Tw= (Wr)w
awit
= amu? re
KE = kmv™= bm (F* Crwy)
= Em(rdak e7PE, ptute*)
aut
a mrzwe
KE
i amr a7? 2 p
jeez
@) xy s M+ Re
Reps RS r T
on ath = RX ’ fa Bw
> +
(bk) ™g-7 = me mg Ma
Mqg-T =MA
TR= EMRE
Sb ae(Pet)e A-(BM)2 <> (PH)F
72 Beth drums born through
the same angle and therefre
have the same angular acelera tion,
(The torques about the centers are identical )
XL, + ARO
¥ =A = 2R<
Drom A? Mq-T = MA
TR = RMR
Dem Bi TRa EMR
Mg = %m MA
A= #9
a. Wwing
= 2 yer
To MY
E> mgh = Mgleng-
A= tome 1 bY,
Mg sin q sine
,
h
2] Fz = Mgh 4 a
After Falling « height h, speed is v
es £mv7+ £56*
Hence
Bz tuve ts Rav = = (m+n )ve Wenee va [ Mah 7%
me =
ae
Foe a cnifeem sphere, fas M
Fer a uniform cylinder 2p= M>? EM
Mence the speed of the sphere is always greater
than the speed of the cylinder a# « piven lecatin
and the sphere reaches the betm fired:
e27] F-F = MA
FR-FL = (% MR
= &MRA
Cbermuse AZ Re for vellong
without slipping -)
Hence f* EF (i+ 2b)
Bot FEN meg
Thee (i+ 3) = uMy
Sumy
max (4
CF is considered be be gwen)
Fcor@-f= MA (net needed)
F sin O@tN = Mg
There te he tendency te rotate
when torque = O
Hence FL= FR
74 Alse , fopu = (Mg Fsin@) becavse YoYo is
not retating.
Hence Fb = 4 (Mg-Fsino) R
“ MR FL
APR
629] page ma
(@) The (EMR*)<= sme A
+
oR sine =
Ms
> T- i aE
G) f= Ar 2My%
at re %
-
= AMVs
ah %
By conservation of
energy,
f+ ve
. ta fim, ) ft
eimy+ zGnet
75 —++
fa-MA
FR = I~
Sall rolls without slipping uhen ve Re
+ +
Vee) =v, SH vt é= S Bu
4
V decreases and 6 increases und; / veRe
+ R2 t &
Re< Fue Bar § fet =vevi-t [me
Hence ball rolls without clipping when
+ |
am Usde= vm U Sue
oo
mS fate = Xv,
When rolling begins vav- $y = £¥, . After His
ime, Feo and telling without slipping comtinves,
Simpler Selubion; use conservation of angular
Mementum abevt contact point,
Le MR , Let MVR * Tew
om &
z,> mr Lith, @ MVR= MVR + # MR
ve £y,
ey
Angular mementurn is conserved abevt point of impact.
Tiw, = Ijuy + MVR
For pure relling, wee © omy ~
,
Ne
Tews = Iuwe +MR we —7 ote,
T.- EMR* ( )
SE MRIw, = SMR we + MRM toe
¥
= We
— >
fra -Ty%y
fr= gx, “ \
i CD.
l= w,- xy S Fide h B
t
wer F Shut €
Nene w= w,- (2 EC F2)
F continues to act until the surfaces of contact
attain the same linear velocity: Rey ge re
Wp syn Mo ~ (BE)(FE) x 5,,
ye oe/ + 2) = Tee = a
BF in”
7 (a) Angular mementom along the vertical direction
is conserved.
= TO
fhe tm Rw
Zp
“¢ (atm) “
(bk) £, = BT w3+ mgh
=~ rw? + tm?
Egu Re t
Wence ‘™ = (w3-w?) +agh Ye
% = la [- (she) ]+eqh
6.34
E™ mg Ci-cose) + £m viat
Bt ve RS
bp=RS
x gmk!
Hene F= mg R (i-cose) + & Mm R7G* + om
= mgR(i-coso) + jemR6*
on TE = (mg Rsino) d+ FmRSS
i mgR
Hence S + Emre @=0
78 bee Lone
ence wa TFS
(638)
As bleck recks Wwitheod
Slipping, it meareres off
adistance Re.
Bleck is stable if line of
Mg fall within RO f
center, so ac te produce a
restoring borque.
Unstable if + tang >Re
Fer small ©, tan® So
Wence bleck is onstable if L>ar
v,
\e3e) b
After veleace, the CM moves
ata steady velocity a b
ee ge.
™atm, matey,
Ra Maalod+ mvc) 7 my ) V
‘e
The system retetes and translates. Because Yte)=o
angular velocity is corvectly given by
w= Ye. Ry
Altermate methed: transfer te frame in whith CM
is at rest. System retates about cM, Immediately
79 after release, vst
: 4a
t=(Gtn, )% 7 6 my
v-R = GET) % @
4.7 (wim) L
An (a \L
mer,
wo (Se) [4° +
Te Find T, do not vee my as an origin, because it iz
accelerating. Go te CM frame.
_ maVa~ me LZ
4a + T
M
%
mb
ve GEE)
= 2. - Mam,
T= me (Sha) / (a) (=e a
mete,
6-37[ (a) mv,=Mv-V_
Amven amifek Mv? 9ST, o*
Mat cum ov, tammy £4 Tew
to
ve ie (very) w= ml vgrv,) £ /Z, |
80 2
whee I,= Ta mM (at)? = Sue
Wwe find
Cie Sma) ys + (@ ee? =o
which elds \% =
4 (—=
t= a
(b) Momentum need net be
conserved because forces act
at the pived,
Romy eo em sez at y°
Take angular momentum
abevt the pivets i
my (af) = —™hy Gt\+rw
& 22)"
Hence wom Amd (vu) veut
ee i
and (14 2 22) 3 (eey -(1- BP)y2 =o
- 3Bm/,
> Ve ( na
¥ + 8mm _) Yo
ae w “Key
™ ¥, H5o %m
Ss
™MVg = AMV =m Ve
Amvi= dmv + d(am)v24 £T,w2 Dos am(k>
‘° £ = mL
a
Take angular mementom about Gm
81 mv, (BY sintse = -my(E) sin 450% Zeus
Hence. Vor avnVp
Vie Ve 2, a2
da VA savte Ew
Von 7M TF bw
we ve
>
(a) The bey plank system
translates and rotates abevt M {
its new CM. 4 w
my= (Mtn) V Moe | x v
veo mY Vo
nem
Perition of new CM is at
m(£ +x) amx
ML/a,
Mtm
x=
Take angular mementem about new CM
myx Emxtez+ MOE -x)27 ©
es
by parallel! avis theerem
Lmxter, +m ($x)? ] =
Pe Mls)
mMl® Me M+4m_
4 Cmem) hem
= Mette 7 vex mu dv,
lence = lo = ML* (Mt) = pa
NE (nisin) ML* (rem)
Cammy
a 6p
wo
M+ J U7
() the pent at rest has speed -V due te
vetation.
v
V=Yw y
vo
a a (ir) £ (Mite
Mem us v
(Bam) ©
Yex= (Brome 8) eo L a ae
Mtm
fay
x= +b cospt
X= ~pbsin( pt)
Speed of cm ot x= 45 —Bb
Momentum and energy of whee! are net conserved
in collision with the track, Take angular mementum
abevt peint on track, Torque of spring is sero
at xl, torque dee te collision force ooesht onten
The metion for x>4 is quien by
where B= VEG.
83 = MRv = mR pb
Les MRR + MVR
= amMRiw
Hance to = SE tmmediately after callisrer,
v= Rw = Bb
becaure V= Rw on gear tracks,
Fs EMV + EMR? = MR = MBE kb
Spring ‘is compressed to a length L
ek (4-L)* = ERB
t.
tobe Be, LH LE Seem ull, bec)
——
4
(b) There is ne collision
in the ovtward rip and
energy is conserved . Speed at x= i given by
wk (4-1)3= Fkbe AMV? + BCMA) Rr
= mv>
v= xbT¥m = seb
&
we Ha zgb
For x7, w remains constant. Metion of whee/
is given by x= Asin Rte Geos Bt
= BA cosBt -BB sin Bt
xe eyma
kiss Eph => gro, AWE
84 x= & sinpt
x= Bb cope
Wheel ctops af per = x¢pee Ele £
Uence wheel goer oud fo xe h
(ey
x
Wheel it spinning with om SEE as it hidy track,
and it has peed - 4h.
L; = MRv- MRP
= Reb. MRAL LS
2
° ak
Nenee Lp ame\iw’so ,
wm,
v= Rw and wheel comes to rect,
7
Only nevmal forces act at the Ny
surfaces. (Me Friction) The
position ef dhe Cm and the Ne
angular petition are quien by
a Single coordinate, & \
Location of CM is given by aL
xa hese SS
qzh sine Co The herisenta/ fierce is given by
N= Me = ML C- cose @* -sine s)
Contact ic fect when My=e or 6% = -denoe &
cy
Bneegy it conserved, so we can find &*~ ftom the
energy equation . A simple way te get & 1s te
differentiate the energy epvetion,
B= Myyt EM CG) + aT, So emg y
= kM oe + be) MUS tml sine
= Fumes s Mghsine (>)
32
DO Fferentiating,
O= FmMASSemglese &
62 -Etcwso ()
Sebetituting 689 inte C1) gields
e*= £isine
Substituting thie inte (a) we Find
7
May,> SMgh sin@ +Meb cine
= Eyl sine
Nence Y= Lsin® = By.
The beard leaves the wall when the top hes slipped
Ya of the way down.
86 Chapter 7
z
R a
Ws a
Ld =
q L
(2)
2 enarctan
w= eo st =X o -
B= BR nfo ky G7
wy
Te Tel, = 25, +2,2 x,= MR
By parallel oxis theorem, IgeI,+mr™= Eme*™
T= me (+e) = mean ti+2h)
gene
The spin angular momentum is 5
Ls=Z,W,. Assuming that the Ey
2le\ >= 7
ale is close te herizental,
IE] = ALg along the f axis,
7 8 The tergue dve fo each spring is 2fler,
The tetal torque is 4#£OT hence
uLOT = DLLs
e=- Dts
HLT
The linear equations of motion ave
Tcosp = My =
Trinp = MYL Mx
we
az,
Torque = Toop = dm = = OF = nto
with cosa ond Sing ~p we have
2 2
i (Eye
q
But xe Stlanp ~ filp
Let pox (Saye
Fes
: L
Then p= Bot yPeP
n
or 2
88 The axle everts a terqve T. 2m the milletone
and a vertical ahunward foree F, The milletene exerts
oppesite Force and torque om the axle, The coupling
at the vertveal column must alco exert a downward
force F en the axle,
The axle is in vertical eguilibrivm | and for
vebational eguilibriem RE =T=PO
The epvatians of meron for the milletene are
N= F+W
T =f, 22 kMb rw 2
2
Sulving, F2 W/R= Embwsn
7 &R
For cortact at Th vim, abs OR
2
Then Fe 2Mb os 2
R
7
(a) t
Tr
FON, N,
a9
= gamba —» we hytF =m (9+ Lb or) In the absence f the flywheel, the counterclockwise
forque dee t the normal forces at the wheels
must oppese the cleakwice torque dee te the
redial (transverse) fretion free, Fo mv7r - wed
the Flywheel, the nermal ferces exert ne torque;
se the Flgwheel must exert a clockwise torque
= LF, where + is the sistance From the read te
the canter -f mace. The countercleekwise torque
LE muet be jvet eefficient to make the
Flywheel precess st the rate the car is turning.
xf the Flywheel is mounted with bs angular
memento sileways, as shown
the terque it exerts on the cor
ak
will tend bo balance the torque =>
due be the read . ZF thecar Tread
turne te the right instead of
the left, a Westrated, the torques
dee te the rend and doe te the Flywhee! wi
direction.
beth che ye
(a9) v The torque due to the road
ne is TULF = mv*RE
r Ls “Wa vequive
TLR Using R= Y/R,
Mv?rL = bs = Em HE
oR we 2vML /mR®
90 Coin ‘is accelerating, so we must use CM as eryin for
berque eguation.
N= nermal force = Mg
F= Friction Rree = mv7/R
bys spin angular momentum =I 2M = LMby
R= angular speed with which coin rotates about
the vertical axis = W/R
Ri,coeG = rate of change of angular mementoem
Cin herigental plane)
Torque epvetion
Nhsing -fheoe $ = QL, cos B
Mgbsin gg - MY bosom Ei mbvcarg
Hence tang = 2e
AV= wat
AL, (Lu ain ps)or
=(Lw nt p)w at T] = Teosh Reosp = MaRcosp
THe BE 5 MqResp = Tw? sings
tmprps MP, 3
Ta?* Rw
The 5y
Nes yee aperanrm Ten, althowyh hot gustified here,
(b> the radial free is T sinve | so
2
?
where or it the radius of the path of the cM
Deing Tees = Mg, we find ym ge tans -
Tsine = Mrw
[72] P doving the Mew, the frees
TVs potting ore the impeleive force
ia > Fo and the frictional free
"| £. We atsume F>>F and
f bs negleet F
The ferque aboot cm iy T= Fb.
v
v
(a) The spm angelar mementem if be Te +t e
where ZF, is the mement of inertia abovt the
hevisontal axis = Mb”
FL=1l, = mere
=
tom Fly > Pm ny QFE Ta
92 (L) The gyroscope spprenimotion asrerts tet Hla
spin angular momentum i large compared to
any ether components of ange lar mementem .
The menvent of werbes aboot He vertic
axis ix i> MB/a and the vertée
af
component > t a
.
E
ome ee ame gu
> F << 2v¥/5,
R=EWR
lez om
M }
“ ?
* vez Ls
° R
The torque abst cm is
T= RANSHP -atfFoos
For vertical eyvilibriom, W@W and the herisonte/
acceleration ts gwen by
f= Mv7R
T= 24m (qsin?- VYR Csr)
93 The angelar mementem of the tee wheels is
Le= 2lw =2m£*v/£ = amey
The rate f change of herisente! angular momentum as the
beke turn is
dis Qc Fr = *% (amy) cos P
2
te LE 2M (quan Pe YF cos) = aml cosh
2
qn - “Ee cos Pa Et oP
ten P= ee (it BB)
Fe) Bhs ca) 28 the spin anguhr mementom,
A Tyas parallel te the earthis angelr
wy velocity, De, then Le does net
s change direction as the earth
retabes, and the gyre will
remain statienary.
The spin axis is then ot the hatitude angle A with
respect te the local herizental in the N-S direedisn
Chine Nes)
94 Let the gyre spin axis be ot
angle % with respeet. te Ze. The
tate ef change of epin anguler
momentum in the direction A
© henP 2LE1L, PQ,
Zn addition, iF Ty is the mement of mertie
along the ants a-b, the angelar mementem in
divectien A due to a change in F is Zp,
There is
HEY) + La P= 0
ne torgve along the a-b axis, so
TPL QWP-0
This ia the eguation Lor simple harmenic motion.
The ecreilation Frequency is
Fer Lg Tote, 9 woe
For a thin disk, Z, = 2,
amred/dey =
5 rad
Qe* Feuxiots/oay 1.29 110" rad/s
3
Wo, = Kyo” rotations /min X Se = KIPxro rad/s
95 Hence ww = \2Kwmiexso% 728X102 = 2.77 rad/s
se.
The peried of oscillahcn is To B= 22s
a) Tr mlytes?) = Im ye 2
Zyy = 7 my Te 4me
=e = 7& -
Lye mee ™m tes
eta,
Go -é ™
t= (: 3 :) Y
~ 0 x
a ~ x= 2Ci-cos <)
4 <2 snx
Z2-=8s
To First order in ~< xeR
4 = ax
2-7
Tete -4e — 6
zr’ =m -4¥~ 13-6
-6 be A tlnt
Comparing with part (a) we see that the mementech
thertia vary only by terms of x * . The predvets vary linearly.
96 t Ya
w
A
Ny
aN Ww
The ferce clagram ic in the accelerating System . W'
is the fictitices eee: W'= -MA. The torque aboot
the pivet peint is
Tas - sine wis HE cosew
@) Fer equlibriom, To,
cos & Mg = Sin@ MA
2 tan@ = I/A
(L) Te analyze motion about eguilibrium, intredece a
coordinate system with the 2’ .axis alen the board in
equilibrium. The effective gravitational field a
ye.
For smell displacement, terqve is
Te emg?
Hence Ty P= tuy?
7 Ly2
Veing Ee emits mG) = SmL™
auMeg- + mq’ P =0
@-2¢?%/L =0
tYe
The motion is hyperbelic: F™% ©
where Y= [3755"
L
(22) [ A wu z
SS
Comtder the motion of the trock deer im a system
Sued te the treck. The door appears be Fall” from rect
ha gravitational field =A.
G) By conservation of energy, the kinetie energy of
the deor after rofation through ange P= 7/2
is
; we
RRP aw
Using Ins FM w* we hve () F=mre*
Smut g* = MAW/a =m ( ZA)
P= T3Afe Feima
98 For the pendulum te peink centinvally teward the
earth , its angular acceleration ~’ must be epval and
eppesite te the angular acceleration of its point of
suppert with respect te the center of the earth, ~= “R,
In the accelerating system, the pendulum starts te swing
tn an effectwe Field g's Vgtta® from an inital angular
diuplacement % Farctan #. tf the effective length of
the epivasent simple penivlum is and its mass m,
then the torque is
mgt sin %e and
Tr, = mgtsin® /me* a L
= q/sink/e
“
“
Bt g'sinBoa, so hae
x's a/v
Zn order for the pendulum to point to the centr of dhe earth
we veqvire 4g Steet m/s? or 42 = g6axe?
7 eit
. Visersial = Yoagt %F
t
a> = Fr tobabing system, velocity
instantaneously lies along ¥.
Hence V ed re
pore a 2 A
Ase, Ux¥ = 2rs = ors
Therefore Vo = tee red
FF] The Cortolis farce is
Foor™ -am xv
(a) B= 2MRvsn ar
a= && ved/day _
Ss
7.26K10 rad/s
P.6E X10" S/day
102 v= 60mi/h = 88 ft/sec
sikA= sin 60° = O.9c¢
Foor, = 2M RV SiH
w ———
BRvsnDA
ms z
Fy, = Axtacns” rls x8tltls xoee6 x (Hoo x2000)lbs
32 Fl/5%
= 276 Ibs,
() RxV is directed along He nit vector ofr
east, The
coriolis force an the trai is vitected
along —-€, ie toward the werd The reaction
force om the tracks is directed oppeertely — the tracks
are pushed east ac the train ic pushed weet:
a
ae
v
qe
The apparent acceleration of
gravity is q-9.-8 a where &
is the acceleration of the
local
reference frame.
103 Using the law of cosines,
GE 9i+ ai- RagucosA
= 2*Recosr
47% 92 + Re) cata - 207R, Jo cos? >
let x= Q* Re/g,
£
eo (1=(ax-x?) costa)®
4:
eg, (ie xteos2a -2xcos?a) > =
2
The velecdy dependent fictitious
force is F= -amiixv
The apparent change in gravity is
the normal component of E, /m
@ Fart: RXV points radially ovt
&:
Pht cay pemts radially in
Therefore apparent © decreased,
Sa= - Feet | ~ anv
q ™@ s
am rad
-s
aN = 726x/0” rad/s
24 brs /:
2=
104 V= R00 miles/hr = 294 ft/s
485 -Ax7.a6 x10F rad/s x 294 FE/ 5
4 ——— eee
Batt /st
= -4aax0F
(b) West: sign is reversed, 48 = +138 x70"
G Seeth: Rand 7 are parallel. Ne change ing
(d) North:
Again , 49 =o
2
BE) rn the rebating system Se
there is a radial Rree ovtuard.
AL
2 t
Feng = Mr hy.
w > Rent
The torque equation about the pived y= Lsine
pent @& is
Whsine- Fy lee ® = Mere
MLE + Mg Lemé -M¢2sin& DO =O
where we have used r= Cine
For emall angles sin-6- 86, and
ay -H)E=0 whieh is the epvation for SHM
105 The solution is
O= 8 cos lwtr%)
whee w= V5-3*
SB!
Note thet if Q2> ZF the motion is no longer harmonic,
The pandlum flies cut and escillates abevt some
equilibrium angle. ( The situation is reminiscent of
the conical pendulum, ewample 2.2.)
106 Chapter 9
Bae Leas") + Uer)
gE a ta(ati + artée?)+ BF =o
AGE a rtet) oe FE
du = -ferd
Be te Be) = Fen (an. 27a)
Ae Ae
Ba brte +PS6) =o
(Orb tate) me ( en 7.76)
El
= a
Fey = ~ Ar
wer) EArt a
Yegp OUT oo
AnHdgnes, C= /07 gems , m= 50g
« t
fe ws phy He
@ Upewin’ + Seopa Te
3. au
(b) Leg Minimum at HOS Pe
4] Sw kxpot © 5x03
aN 7 r* min
7 Vein BI em
107 (ec) uer,)= Ucar)
4 Joe bet xref
vir Ba aes TE xl ae
A : .
ES) Sexe - 4 # Cattractve inverse cube force)
w+”
ue =e,
Wage woe Sart
EF £7= aA, Upp =O and radial motion is Ike thet
of a free particle , rarirvt
Late 62
apegt eae
on (ab)E
dee (2A)*
= of dt) dr
(ve = (HF (48 = (2% Ra) + =
Fer Gree mdial motion tab = cont.
Y
o-e,> (38)* $(+-4)
108 7 ot
Le
Yee tur are
--AsB get
ie] uu“
rao
For a stable civeular orbit,
pesential must have a
Minimum ef 1.
Hence utr) >0
28
weny zo => BA = SE
. 5
won) = 7m iOA Les = - eH (28). 6B yo
ry nt* re T og a
=Poner € 3B whieh requires nea,
I
(a) Fee metion in a cireclar orkid the energy is
€= Ekr*+ & mvt
‘
Fer equilibrium, ME = ke D Gm kk ethrake
For Faizgrand k= #N/m Then re 7 a
kr? _
{Ee = VE
109
and v Ub) The blow changes the total energy but not
the angular mementem er the effective potential.
The imitial energy ie
B~ Rkr?+tmvi = ke? ar
°
The Fm! energy is
Eo e+ amit = Jats Heaxe? 137
«>
at
ue Rhyt+ ee, Cem Cinta!)
t= mv,r, -
vag Skrtee mye
Ak Burning pets hep ™ Hi
2.
Rket+amvyt Te =
a
kxaye kxaxex =13
Brte ahr3
rhe BE ritleno
rte 2 Le t Wateee = SL ecse Td
N= 162m and = 245m
110 F= -Kré? in
Wee BK Gare
For eitevlar mobion Fe Uy C6,)
ul os pt 2
eff are
Upp (=O ,
2%
ke £3 > ne (E)”
SAS oN
s fra eK * aid
Be dee tu= FKr,* ‘hE 7 we (8 )
Freyvency of eseillaticn is w= —
et
Ute eK t 25;
+ 3 "222 But Kr t= “e
Uys AKG? + Bee rer
4
utr) = 7s. 7 £2 An) . he a
‘o AS “7 4* ear
we (2 ue ) =
The ‘increase in total energy is AERC Kou), ~ tke) yy
The potential energy is constant during the firing ,
un assuming that the Firing occurs quickly . (Thie ie tye
for the actual, but ned the ive, pebentia/. )
Using Wer V,rav
Ags kK, ~Kj = mC; rav)*> my? = mip av +k mavy
AE is ameximum when AV is parallel te v, and
tohen V; 15 a maximern. The speed is qrestest
at perigee.
Fiving point 8 !
[EE] The inetie! sneray is
Bute) + Emit+ oe
Fm R=
'e
= ~ Ste «= my,*
At the top of the trajectory, r=0 and
— n+
B=-GmMm — + SS
E=-Gim * + Emve sit (fe)*
Fyuating the initial and fina/ energies, we have
2
~GMIO + my = - GM tems si (fz)
fe
v2Re ~ . Ke Rev _- a [Re \>
vie & She ~ AR + do Bee sin CE)
uz v2 R,
we have one se
am =! 0 thet
~4 =- fe sm. Re )*
h=- few gem (Re)
let x= Re
wAyt = xe & sind
x2-axt sintaceo
xe & Lar VecusiFe*] Cure = peste root)
ke p+ CERFS = [tcosm
Menee ore Re Citecos)
[EZ] The peried of a satelite depends only on 4s
major aris A. Frem €9n. 280,
2, 4
AS {2c4) ‘Cm+m)@ 14
TH T6qenrs TEX BICKO” = 2.4/0
Mem = Ma. = and” Ay
A= Lax (2x0? 2 a ~7 7s = fe
Bemel Yl x and” x &£7Kx00 J S40 m
Frem Fyn. 725
Y= KA CI-E*)
we RGAxe% ) x FCt-00.967)2 J
Dain -e*= O-e)lisve) Face ce) Ler OF)
= £G* 54x20 y. 083) = 18x70%™m
G) perihelion = We 2 le = 04400"m
113 3 Ye — te th
aphelion= Te = Nea = SSx10 Mm
(bY From By. DIP AGM
where po is the ele mass = "Lemed
At perihelion | -£= myn
wy? he = tm" 6m
v= (“258 “eSt) "
mien
Simee 4, 7 EM
~! F0\ %
= (4GM\ "a _ ( #xee7x16" x2 x10 = €
y (#4) TTB yze” J SF X10 m/s
JRO t. - ~
Fra civeslay orbit order an attractive force
Yee, we have
Fe
my" L
"
a
eR Amv = £5 =-tucry
G) The betel energy
Es Ktu = -kU+ey eye -k = -k
K=-E
Le beantion of cirschr orbit, dean
”
14 GY Work dene by Friehen is awe -arrf- af
wee (Ear ete
~a2rr rr
ar = ab / 4 = ka €
oe
©) In citevlar orbit | Fe -K
4ak= -4e= + awrf
Friction results im the satellite speeding up /
a re 2
7 Kepler's third law states T= Semam) c
2 AP
ue SA
e . te = €
A major axis = Tt Us, = [PeIHeae = 3.6977 x/0'm
T= U9 min = 714x710" seconds
Q = 667xKI0"
> m= 73 x10" by
71% Tn general, F= Emv* - Sm
: , mv? GM
Fic a cirevlar orbit , “PH = SME
Summ GM 2 g
-+ ies m = -t =
and FSe-Q~ 3 RE = aRr,w =
1s where We mga weight at earths surface,
@) a= -sR.w ($e )\e-e RW
e,7 -e Rw
‘ 2
Aes (Pte) RW FRW HS KEHOE SII xP
e
= 2.865x/0° Tevles
(bY The transfer orbit is an ellipse with perigee at
A and apogee ct 8. The Semimajor anie is
Ar grn.
The energy is cm 1
Bw - Show . - Sar (3.5
‘e Ra Re
Els. & * T =- (sé = -+wR,
WR, Se = - WA Rem, < @
Re Re
The mitial speed at A a given by
Amys Ks B= + $Romg
v= 172 TR
The fined peed at A is gwen by
Kira &
Kwmvt = ~dmghet MME = (4-4) mgRe
v= 78 Veg
16 Hence
aye (V% -V%) Vag = -10R%79%10" m/s = 850m
At B the final speed is
eo ey = 5 RG
tia] speed ab @ ie
, ’ a -
mute wi-u, = -gmpRet “hese = (FZ) mg Reni mgh
we (4-18) te = (0.8-0,408) x7 9x70 = 730 m/s.
17 Chapter 10
ZH 4 arr
= + &, Sintwt Jt
Using sin*x = ie Crseaze)
ger
= af C1 coszut) dd salted simauea)|
Fe Te F > Sin 20 (447) = sin quit, tar) = sinawt,
= ar Ltt T- 8, +24, (sim awt-sincawt, i =
= +f Vins cast dd at chrwt-d Cate)
ter %
= op & sin fut | =o
[ Q= 7, w @ atradfeycle x acycles/sec = 4 rad/s
yo Spree = ons 5
be Ym = 00s ky s'
wo = Thm
k= mut= 0.3x(u7)*= 475 Vm
Fer a damped oscillator
x= x, et Sin wt
Zero crossings eceur at tt=0, Wi27, +6. nT
118