Science > Physics > Mathematical Problem in Solving "Motion in a central force field"
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Science > Physics |
| User: |
"" |
| Date: |
28 Feb 2006 11:52:16 PM |
| Object: |
Mathematical Problem in Solving "Motion in a central force field" |
Hi there, We got a problem in Dynamics lecture,
Consider motion in a central force field, mr''=F(r,theta)*e_r where r
and e_r are vectors.
In this case, for F(r,theta), textbook derives the formula below,
v^2=(r')^2+(h_0/r)^2 where v is velocity, h_0 is r(cross
product)v=specific angular momentum,
Those processes are not problems, but when considering a special case
for F=F(r) only,
The problem can be described by a potential V, V(r)=-int{F(r)dr}
From energy conservation,
0.5mv^2+V(r)=T_0+V_0=E_0=0.5m{(r'^2+(h_0/r)^2+V(r)}
Transpose and process bilaterial, we get,
dr 2 mh_0^2
(------)^2 = ----(E_0 - V(r) - --------------) = (blablabla)
dt m 2r^2
(The formula is so BEAUTIFUL that I call them (blablabla),
The problem is when we remove square of left-hand side, should
right-hand side be
denoted as a square root? That is,
dr
(------)= + - (blablabla)^0.5
dt
My textbook show us just positive square root and then solve the ODE,
but I don't
know why it neglects the minus sign.
.
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| User: "" |
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| Title: Re: Mathematical Problem in Solving "Motion in a central force field" |
28 Feb 2006 11:57:03 PM |
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In article <1141192336.737963.101700@u72g2000cwu.googlegroups.com>, writes:
Hi there, We got a problem in Dynamics lecture,
Consider motion in a central force field, mr''=F(r,theta)*e_r where r
and e_r are vectors.
In this case, for F(r,theta), textbook derives the formula below,
v^2=(r')^2+(h_0/r)^2 where v is velocity, h_0 is r(cross
product)v=specific angular momentum,
Those processes are not problems, but when considering a special case
for F=F(r) only,
The problem can be described by a potential V, V(r)=-int{F(r)dr}
From energy conservation,
0.5mv^2+V(r)=T_0+V_0=E_0=0.5m{(r'^2+(h_0/r)^2+V(r)}
Transpose and process bilaterial, we get,
dr 2 mh_0^2
(------)^2 = ----(E_0 - V(r) - --------------) = (blablabla)
dt m 2r^2
(The formula is so BEAUTIFUL that I call them (blablabla),
The problem is when we remove square of left-hand side, should
right-hand side be
denoted as a square root? That is,
dr
(------)= + - (blablabla)^0.5
dt
My textbook show us just positive square root and then solve the ODE,
but I don't
know why it neglects the minus sign.
Both are valid. The trajectory consists of branches where dr/dt > 0
(the positive root) and ones where dr/dt <0 which are mirror images of
the first ones.
Mati Meron | "When you argue with a fool,
meron@cars.uchicago.edu | chances are he is doing just the same"
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