rigid body rotation [Archive]

ganesh

May15-04, 02:44 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>hi,\nLets say there is a rigid rod of length l and mass m. Its there\nin say outer space (no gravity, and no friction of any sort). A\nforce F is applied at one end of the rod. Question is will it\nrotate?? I\'m basically looking at some v. fundamental reasoning\nstarting from the v. basic, say newton\'s laws.\n\nMy own interpretation is as follows (not v. fundamental).\nCenter of mass will translate with acceleration\n= a = F/m where m = mass of rod.\n\nNow, comes to rotation around the center of mass.\n\nLet me assume a "fictitious" force F\' on the other end in the opp direction\n\nNow translation of CM = a = (F - F\')/m\nRotation around CM : angular acc = {(F + F\')l/2}/I\nwhere I = moment of inertia and l = length of the rod\n\nTaking the limits F\' -> 0, I get\ntranslation of CM = a = F/m\nand angular acceleration = {F.l/2}/I\n\nBy the above argument I can see it should rotate and translate.\nBut can there be a better argument from more fundamental principles (say\ndirectly from newton\'s three laws) which prove beyond doubt that it\nwill rotate??\n\nWhile trying to tackle the above problem, the question also arises\nas to what is a "rigid body"?? If the interatomic forces can\nbe approximately (to first order) be modelled as a spring with\nforce constant "k" between the atoms/molecules; then is a rigid\nbody necessarily be one in the\ntruest sense when k -> infinity? (All this assuming purely newtonian\nmechanics).\n\nganesh\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>hi,
Lets say there is a rigid rod of length l and mass m. Its there
in say outer space (no gravity, and no friction of any sort). A
force F is applied at one end of the rod. Question is will it
rotate?? I'm basically looking at some v. fundamental reasoning
starting from the v. basic, say newton's laws.

My own interpretation is as follows (not v. fundamental).
Center of mass will translate with acceleration
= a = F/m where m = mass of rod.

Now, comes to rotation around the center of mass.

Let me assume a "fictitious" force F' on the other end in the opp direction

Now translation of CM = a = (F - F')/m
Rotation around CM : angular acc = {(F + F')l/2}/I
where I = moment of inertia and l = length of the rod

Taking the limits F' -> 0, I get
translation of CM = a = F/m
and angular acceleration = {F.l/2}/I

By the above argument I can see it should rotate and translate.
But can there be a better argument from more fundamental principles (say
directly from newton's three laws) which prove beyond doubt that it
will rotate??

While trying to tackle the above problem, the question also arises
as to what is a "rigid body"?? If the interatomic forces can
be approximately (to first order) be modelled as a spring with
force constant "k" between the atoms/molecules; then is a rigid
body necessarily be one in the
truest sense when k -> infinity? (All this assuming purely newtonian
mechanics).

ganesh

Stephen Riley

May17-04, 02:02 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>In message <510dfbbf.0405140735.68f9e36e@posting.google.com >, ganesh\n<gans1973@rediffmail.com> writes\n>hi,\n> Lets say there is a rigid rod of length l and mass m. Its there\n>in say outer space (no gravity, and no friction of any sort). A\n>force F is applied at one end of the rod. Question is will it\n>rotate?? I\'m basically looking at some v. fundamental reasoning\n>starting from the v. basic, say newton\'s laws.\n\nThe conclusion that an initially stationary non-rotating object will\ntranslate and rotate given a force or impulse follows easily from\nconservation of linear and angular momentum, IMO. Try working out how\neither could be conserved if say a particle hits a rod (and is of course\nslowed down, therefore changing the particle\'s momentum and angular\nmomentum measured about the rod). What does the rod need to do to\nconserve these quantities?\n\n>\n>My own interpretation is as follows (not v. fundamental).\n>Center of mass will translate with acceleration\n>= a = F/m where m = mass of rod.\n>\n>Now, comes to rotation around the center of mass.\n>\n>Let me assume a "fictitious" force F\' on the other end in the opp direction\n>\n>Now translation of CM = a = (F - F\')/m\n>Rotation around CM : angular acc = {(F + F\')l/2}/I\n>where I = moment of inertia and l = length of the rod\n>\n>Taking the limits F\' -> 0, I get\n>translation of CM = a = F/m\n>and angular acceleration = {F.l/2}/I\n>\n>By the above argument I can see it should rotate and translate.\n>But can there be a better argument from more fundamental principles (say\n>directly from newton\'s three laws) which prove beyond doubt that it\n>will rotate??\n>\n>While trying to tackle the above problem, the question also arises\n>as to what is a "rigid body"?? If the interatomic forces can\n>be approximately (to first order) be modelled as a spring with\n>force constant "k" between the atoms/molecules; then is a rigid\n>body necessarily be one in the\n>truest sense when k -> infinity? (All this assuming purely newtonian\n>mechanics).\n\nA rigid body is an idealisation, but Newton\'s laws still work. Force\nisn\'t useful in the limit because that idealisation implies infinite\nrigidity and zero time to operate. Forces need time to change\nvelocities, so instead of force an impulse (the integral of force *\ntime) can be used, derived from Newton\'s second law :\n\nF = dp/dt, so dp = F(t) dt\n\nIf both sides are integrated, and the integral on the rhs defined as\nimpulse, then impulse is change in momentum.\n\nhttp://musr.physics.ubc.ca/~jess/hr/skept/Mechanics/node4.html\n\n--\nStephen Riley\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>In message <510dfbbf.0405140735.68f9e36e@posting.google.com>, ganesh
<gans1973@rediffmail.com> writes
>hi,
> Lets say there is a rigid rod of length l and mass m. Its there
>in say outer space (no gravity, and no friction of any sort). A
>force F is applied at one end of the rod. Question is will it
>rotate?? I'm basically looking at some v. fundamental reasoning
>starting from the v. basic, say newton's laws.

The conclusion that an initially stationary non-rotating object will
translate and rotate given a force or impulse follows easily from
conservation of linear and angular momentum, IMO. Try working out how
either could be conserved if say a particle hits a rod (and is of course
slowed down, therefore changing the particle's momentum and angular
momentum measured about the rod). What does the rod need to do to
conserve these quantities?

>
>My own interpretation is as follows (not v. fundamental).
>Center of mass will translate with acceleration
>= a = F/m where m = mass of rod.
>
>Now, comes to rotation around the center of mass.
>
>Let me assume a "fictitious" force F' on the other end in the opp direction
>
>Now translation of CM = a = (F - F')/m
>Rotation around CM : angular acc = {(F + F')l/2}/I
>where I = moment of inertia and l = length of the rod
>
>Taking the limits F' -> 0, I get
>translation of CM = a = F/m
>and angular acceleration = {F.l/2}/I
>
>By the above argument I can see it should rotate and translate.
>But can there be a better argument from more fundamental principles (say
>directly from newton's three laws) which prove beyond doubt that it
>will rotate??
>
>While trying to tackle the above problem, the question also arises
>as to what is a "rigid body"?? If the interatomic forces can
>be approximately (to first order) be modelled as a spring with
>force constant "k" between the atoms/molecules; then is a rigid
>body necessarily be one in the
>truest sense when k -> infinity? (All this assuming purely newtonian
>mechanics).

A rigid body is an idealisation, but Newton's laws still work. Force
isn't useful in the limit because that idealisation implies infinite
rigidity and zero time to operate. Forces need time to change
velocities, so instead of force an impulse (the integral of force *
time) can be used, derived from Newton's second law :

F = dp/dt,[/itex] so [itex]dp = F(t) dt

If both sides are integrated, and the integral on the rhs defined as
impulse, then impulse is change in momentum.

http://musr.physics.ubc.ca/~jess/hr/skept/Mechanics/node4.html

--
Stephen Riley

John C. Polasek

May17-04, 02:02 AM

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no, location=no,scrollbars=yes,resizable=yes,status=no ,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>On Sat, 15 May 2004 07:44:45 +0000 (UTC), gans1973@rediffmail.com\n(ganesh) wrote:\n\n>hi,\n> Lets say there is a rigid rod of length l and mass m. Its there\n>in say outer space (no gravity, and no friction of any sort). A\n>force F is applied at one end of the rod. Question is will it\n>rotate?? I\'m basically looking at some v. fundamental reasoning\n>starting from the v. basic, say newton\'s laws.\n>\n>My own interpretation is as follows (not v. fundamental).\n>Center of mass will translate with acceleration\n>= a = F/m where m = mass of rod.\n>\n>Now, comes to rotation around the center of mass.\n>\n>Let me assume a "fictitious" force F\' on the other end in the opp direction\n>\n>Now translation of CM = a = (F - F\')/m\n>Rotation around CM : angular acc = {(F + F\')l/2}/I\n>where I = moment of inertia and l = length of the rod\n>\n>Taking the limits F\' -> 0, I get\n>translation of CM = a = F/m\n>and angular acceleration = {F.l/2}/I\n>\n>By the above argument I can see it should rotate and translate.\n>But can there be a better argument from more fundamental principles (say\n>directly from newton\'s three laws) which prove beyond doubt that it\n>will rotate??\n>\n>While trying to tackle the above problem, the question also arises\n>as to what is a "rigid body"?? If the interatomic forces can\n>be approximately (to first order) be modelled as a spring with\n>force constant "k" between the atoms/molecules; then is a rigid\n>body necessarily be one in the\n>truest sense when k -> infinity? (All this assuming purely newtonian\n>mechanics).\n>\n>ganesh\nDraw line L. Put vector F at this end, and F\' = 0 at the other. Being\na rigid body we can make the leap to a triangle of force, being the\nline from F at this end to F\' = 0 at the other, i.e. the force tapers\nto zero.\nFor translation note the average force is F/2 at the middle. The\nresulting acceleration is .5*F/m.\nFor rotation, slide the triangle up by F/2 till it\'s balanced in the\nmiddle. The length on this end is +F/2. The length of vector at the\nother end is -F/2 (from triangle end to the basis).\nThe moment of inertia I = M*L^2/12 for both halves of L.\nAcceleration is FL/I = FL*12/ML^2\n\nMr. Dual Space\n(If you have something to say, write an equation.\nIf you have nothing to say, write an essay).\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>On Sat, 15 May 2004 07:44:45 +0000 (UTC), gans1973@rediffmail.com
(ganesh) wrote:

>hi,
> Lets say there is a rigid rod of length l and mass m. Its there
>in say outer space (no gravity, and no friction of any sort). A
>force F is applied at one end of the rod. Question is will it
>rotate?? I'm basically looking at some v. fundamental reasoning
>starting from the v. basic, say newton's laws.
>
>My own interpretation is as follows (not v. fundamental).
>Center of mass will translate with acceleration
>= a = F/m where m = mass of rod.
>
>Now, comes to rotation around the center of mass.
>
>Let me assume a "fictitious" force F' on the other end in the opp direction
>
>Now translation of CM = a = (F - F')/m
>Rotation around CM : angular acc = {(F + F')l/2}/I
>where I = moment of inertia and l = length of the rod
>
>Taking the limits F' -> 0, I get
>translation of CM = a = F/m
>and angular acceleration = {F.l/2}/I
>
>By the above argument I can see it should rotate and translate.
>But can there be a better argument from more fundamental principles (say
>directly from newton's three laws) which prove beyond doubt that it
>will rotate??
>
>While trying to tackle the above problem, the question also arises
>as to what is a "rigid body"?? If the interatomic forces can
>be approximately (to first order) be modelled as a spring with
>force constant "k" between the atoms/molecules; then is a rigid
>body necessarily be one in the
>truest sense when k -> infinity? (All this assuming purely newtonian
>mechanics).
>
>ganesh
Draw line L. Put vector F at this end, and F' = at the other. Being
a rigid body we can make the leap to a triangle of force, being the
line from F at this end to F' = at the other, i.e. the force tapers
to zero.
For translation note the average force is F/2 at the middle. The
resulting acceleration is .5*F/m.
For rotation, slide the triangle up by F/2 till it's balanced in the
middle. The length on this end is +F/2. The length of vector at the
other end is -F/2 (from triangle end to the basis).
The moment of inertia I = M*L^2/12 for both halves of L.
Acceleration is FL/I = FL*12/ML^2

Mr. Dual Space
(If you have something to say, write an equation.
If you have nothing to say, write an essay).