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Times-Bold" 14}{SCP_R 1 0 55{COM_R 2 0{TEXT 4 37 "Chain Sliding o
ff the Edge of a Table"}}{COM_R 3 0{TEXT 1 64 "Physics: Classical
 Mechanics\015\015by Darren Redfern and Tony Scott.\015"}}{COM_R 
4 0{TEXT 6 12 "References: "}}{COM_R 5 0{TEXT 1 749 "R.A. Becker,
 \"Introduction to Theoretical Mechanics\", McGraw-Hill,  New Yor
k, pp. 184-190,  (1954).\015\015This particular problem consists 
of a uniform chain of finite length (say L) and linear density (s
ay r), which lies in a heap right on the edge of a smooth table a
nd then starts sliding over the edge.   The question is:    \015 
        \015              What is the acceleration \"a\" of the c
hain during the time it is sliding over the edge?\015\015The mass
 of the chain, which is pulled down by the gravitational force, i
ncreases as the chain is sliding, so the mass in our problem is v
arying in time.   Hence, we can't use the old-fashioned F = ma (f
orce equals mass times acceleration) here. We must use a more gen
eral version of Newton's second law, namely:"}}{OUT_R 6 0 0{DAG ,
3=3n3\`F`(3n4\`diff`,3(3n3\`p`,2n3\`t`pD=3p9*5(3n3\`m`pCj2x0001(3
n3\`v`pCp17}}{COM_R 7 0{TEXT 1 19 "which implies that,"}}{OUT_R 8
 0 0{DAG =3n3\`F`+5*5(3n4\`diff`,3(3n3\`m`,2n3\`t`pEj2x0001(3n3\`
v`pDp11p11*5(3p6,3p13pEp11pAp11p11}}{COM_R 9 0{TEXT 1 108 "We tak
e y=0 as our reference for the table top, and since the chain sta
rts from a rest position t=0 we have:"}}{OUT_R 10 0 0{DAG ,3=3(3n
3\`v`,2=3n3\`t`j2x0000p9=3(3n3\`y`p5p9}}{COM_R 11 0{TEXT 1 75 "\0
15Thus,  the mass m(t) is  r*y(t)  (where the y-axis points downw
ards).   So"}}{OUT_R 12 0 0{DAG =3(3n4\`diff`,3(3n3\`m`,2n3\`t`pA
*5n3\`r`j2x0001(3p2,3(3n3\`y`p9pAp10}}{COM_R 13 0{TEXT 1 23 "whic
h is also equal to:"}}{OUT_R 14 0 0{DAG *5n3\`r`j2x0001(3n3\`v`,2
n3\`t`p3}}{COM_R 15 0{TEXT 1 39 "As well, we can say that F is eq
ual to:"}}{OUT_R 16 0 0{DAG =3*5(3n3\`m`,2n3\`t`j2x0001n3\`g`p8*7
n3\`r`p8(3n3\`y`p5p8pAp8}}{COM_R 17 0{TEXT 1 85 "\015where \"g\" 
is the gravitational constant. Therefore our equation of motion f
or F is:\015"}}{OUT_R 18 0 0{DAG =3*7n3\`r`j2x0001(3n3\`y`,2n3\`t
`p4n3\`g`p4+5*5p2p4(3n3\`v`p9j2x0002p4*7(3n3\`a`p9p4p2p4p6p4p4}}
{COM_R 19 0{TEXT 1 17 "where v(t) equals"}}{OUT_R 20 0 0{DAG (3n4
\`diff`,3(3n3\`y`,2n3\`t`p9}}{COM_R 21 0{TEXT 1 15 "and a(t) equa
ls"}}{OUT_R 22 0 0{DAG =3(3n4\`diff`,3(3p2,3(3n3\`y`,2n3\`t`pDpD(
3p2,3(3n3\`v`pCpD}}{COM_R 23 0{TEXT 1 97 "which is the accelerati
on. Dividing the above by the density \"r\", we get the DE we hav
e to solve."}}{OUT_R 24 0 0{DAG =3*5(3n3\`y`,2n3\`t`j2x0001n3\`g`
p8+5*3(3n4\`diff`,3p2p6j2x0002p8*5p2p8(3p10,3pFp6p8p8}}{COM_R 25 
0{TEXT 1 170 "Thus we obtain a second-order- NON-linear different
ial equation which can be quite daunting for students...  \015How
ever, with Maple, we can get the solution as shown below:"}}
{INP_R 26 0 "> "{TEXT 0 17 "readlib(isolate):"}}{INP_R 27 0 "> "
{TEXT 0 54 "eq := y(t)*g = diff(y(t), t)^2 + y(t)*diff(y(t), t$2)
:"}}{INP_R 28 0 "> "{TEXT 0 23 "ans := dsolve(\", y(t));"}}{OUT_R
 29 0 28{DAG :3n3\`ans`,3=3n3\`t`+5(3n3\`int`,3*7+5*5n3\`y1`j2x00
03n3\`g`j2x0001j2x0002n3\`_C1`p11/3i2x0001p17p11/3p15p17pFp15=3pF
~3j2x0000(3n3\`y`,2p5p15n3\`_C2`p1D=3p5+5(3p9,3+3*7+5*5n3\`y2`p11
p13p15p17p19p11p1Cp11p21p3Ep15p1D=3p3Ep28p15p31p1D}}{COM_R 30 0
{TEXT 1 183 "So, there is a solution of the form t=y(t).  Select 
the solution corresponding to positive time.  Also,  at t=0, v=0 
and y=0: so clearly _C2=0.  Thus,  our solution is obtained from:
\015"}}{INP_R 31 0 "> "{TEXT 0 62 "res := t = int( sqrt(3)*y2/sqr
t(2*y2^3*g+3*_C1), y2=0..y(t) );"}}{OUT_R 32 0 31{DAG :3n3\`res`=
3n3\`t`(3n3\`int`,3*7+5*5n3\`y2`j2x0003n3\`g`j2x0001j2x0002n3\`_C
1`pF/3i2x0001p15pF/3p13p15pDp13=3pD~3j2x0000(3n3\`y`,2p4}}{COM_R 
33 0{TEXT 1 90 "We still have one arbitrary constant left.  Let's
 isolate the expression for the velocity:"}}{COM_R 34 0{TEXT 1 0 
""}}{INP_R 35 0 "> "{TEXT 0 45 "diff(\", t): subs(diff(y(t), t)=v
, y(t)=y, \");"}}{OUT_R 36 0 35{DAG =3j2x0001*9+5*5n3\`y`j2x0003n
3\`g`p1j2x0002n3\`_C1`p8/3i2x0001pDp8/3p1pDp6p1n3\`v`p1}}{COM_R 
37 0{TEXT 1 64 "\015Isolate the velocity itself and see what happ
ens for small \"y\":"}}{INP_R 38 0 "> "{TEXT 0 22 "assign(isolate
(\", v)):"}}{INP_R 39 0 "> "{TEXT 0 2 "v;"}}{OUT_R 40 0 39{DAG +3
*7+5*5n3\`y`j2x0003n3\`g`j2x0001j2x0002n3\`_C1`p6/3pApCp6p11p4i2x
0001/3pAp6}}{INP_R 41 0 "> "{TEXT 0 13 "series(v, y);"}}{OUT_R 42
 0 41{DAG _8n3\`y`*3n3\`_C1`/3j2x0001j2x0002xFFFFFFFF+3*5p4/3i2x0
001p9n3\`g`p7/3p7j2x0003x2(3n3\`O`,2p7x5}}{COM_R 43 0{TEXT 1 109 
"\015This suggests that  _C1 must be zero; but since 1/_C1 also a
ppears above, the case _C1=0 is a special case:\015"}}{INP_R 44 0
 "> "{TEXT 0 17 "subs(_C1=0, res);"}}{OUT_R 45 0 44{DAG =3n3\`t`(
3n3\`int`,3+3*9j2x0002/3j2x0001p9n3\`y2`/3i2x0001p9n3\`g`p11j2x00
03pBpB=3pF~3j2x0000(3n3\`y`,2p1}}{COM_R 46 0{TEXT 1 22 "Here is o
ur solution:\015"}}{INP_R 47 0 "> "{TEXT 0 17 "isolate(\", y(t));
"}}{OUT_R 48 0 47{DAG =3(3n3\`y`,2n3\`t`+3*5p5j2x0002n3\`g`j2x000
1/3pEj2x0006}}{COM_R 49 0{TEXT 1 22 "Here is the velocity:\015"}}
{INP_R 50 0 "> "{TEXT 0 12 "diff(\", t); "}}{OUT_R 51 0 50{DAG =3
(3n4\`diff`,3(3n3\`y`,2n3\`t`pA+3*5pAj2x0001n3\`g`p10/3p10j2x0003
}}{COM_R 52 0{TEXT 1 26 "Here is the acceleration:\015"}}{INP_R 
53 0 "> "{TEXT 0 11 "diff(\", t);"}}{OUT_R 54 0 53{DAG =3(3n4\`di
ff`,3(3p2,3(3n3\`y`,2n3\`t`pDpD+3n3\`g`/3j2x0001j2x0003}}{COM_R 
55 0{TEXT 1 452 "\015which is a constant in time, and which also 
is exactly 1/3 of \"g\".    Since we had v(t=0) = y(t=0) = 0 here
, this is the physical solution.  Armed with hindsight,  one coul
d then see that the  D.E.  resembled  the form:  =>   v^2 = vo^2 
+ 2*a*y  ( a constant acceleration problem ),  where vo=0.  Howev
er,  intuition often fails when you're dealing with a non-linear 
DE.  Nonetheless,  as we have seen,  Maple can readily give us th
e correct solution."}}{COM_R 56 0{TEXT 1 0 ""}}}{END}
