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 Post subject: Question in "A First Course in Mechanics" by MaryPosted: Sun, 9 Dec 2007 16:42:39 UTC
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Joined: Thu, 29 Mar 2007 12:41:51 UTC
Posts: 12
Location: Wigan, England
Hi,

I'm stuck on a question from "A First Course in Mechanics" by Mary Lunn. Although the book provides the answer, it doesn't show how it's derived. The question is:

Quote:
A smooth surface of revolution is hyperbolic with equation , the axis pointing vertically downwards and being cylindrical polar co-ordinates. A small particle of mass slides on the interior of this surface.

The particle is set in motion with horizontal velocity along the surface at a depth below the origin. Derive a equation involving z and \dot{z} only.

The answer in the back of the book is

Quote:

As a backup does anybody have Dr Lunn's email address

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 Post subject: Posted: Sun, 9 Dec 2007 19:39:56 UTC
 Member of the 'S.O.S. Math' Hall of Fame

Joined: Tue, 20 Nov 2007 04:36:12 UTC
Posts: 826
Location: Las Cruces
I'm curious if you were finally able to work this problem. Does the magnitude of the component of the velocity vector in the xy-plane remain constant?

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 Post subject: Posted: Mon, 10 Dec 2007 00:17:09 UTC
 Member

Joined: Thu, 29 Mar 2007 12:41:51 UTC
Posts: 12
Location: Wigan, England
Quote:
I'm curious if you were finally able to work this problem. Does the magnitude of the component of the velocity vector in the xy-plane remain constant?

circlar motion occurs if , otherwise the particle rises or falls according to the sign of .

I did track down Dr Mary Lunn who is a Fellow in Mathematics at St. Hugh's College, University of Oxford. However I think itâ€™s bad form to badger the poor woman by email for full proofs to questions within the book. I can recommend the book to anyone doing mechanics/dynamics.

Shamus.

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 Post subject: Re: Question in "A First Course in Mechanics" by MPosted: Tue, 11 Dec 2007 18:21:45 UTC
 Member of the 'S.O.S. Math' Hall of Fame

Joined: Sat, 7 Jan 2006 18:29:24 UTC
Posts: 1401
Location: Leeds, UK
Quote:
A smooth surface of revolution is hyperbolic with equation , the axis pointing vertically downwards and being cylindrical polar co-ordinates. A small particle of mass slides on the interior of this surface.

The particle is set in motion with horizontal velocity along the surface at a depth below the origin. Derive a equation involving z and \dot{z} only.

This is not as hard as it looks at first sight, provided that you use the conservation laws.

By conservation of angular momentum, . But , so we can write this as .

Differentiate to get , from which , and so .

By conservation of energy, . Rearrange this and substitute to get
. . . . . . . . . . . . . . . .
Now substitute the value for from the previous paragraph, and you get . Tidy this up a bit and you have the answer given by the book.

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 Post subject: Posted: Fri, 14 Dec 2007 01:22:45 UTC
 Senior Member

Joined: Thu, 28 Oct 2004 00:52:49 UTC
Posts: 119
Location: Bolton, England
Cheers Opalg

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 Post subject: Posted: Sun, 16 Dec 2007 03:58:07 UTC
 Senior Member

Joined: Thu, 28 Oct 2004 00:52:49 UTC
Posts: 119
Location: Bolton, England
Hi Opalg,

I've finally returned to dynamics, working my way through your answer. Just a quick question, where did you get from?

I get since and

Cheers,

Lee

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 Post subject: Posted: Sun, 16 Dec 2007 10:36:42 UTC
 Member of the 'S.O.S. Math' Hall of Fame

Joined: Sat, 7 Jan 2006 18:29:24 UTC
Posts: 1401
Location: Leeds, UK
leedavies wrote:
Just a quick question, where did you get from?

I said it was by conservation of angular momentum. But that is probably nonsense, because we are not dealing with the motion of a rigid body.

The real reason is that the components of acceleration in cylindrical polar coordinates are given by . In this problem, there is no force acting in the θ direction, and so . If you multiply that by r and integrate then you get  const. That is where I got the equation from. (I then carelessly tried to justify this more briefly by claiming it came from angular momentum.)

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 Post subject: Posted: Sun, 16 Dec 2007 23:41:28 UTC
 Senior Member

Joined: Thu, 28 Oct 2004 00:52:49 UTC
Posts: 119
Location: Bolton, England

Quote:
I get since and
.

I forgot at , , therefore and hence

Cheers,

Lee

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