Question in "A First Course in Mechanics" by Mary

shamus · **Posted:** Sun, 9 Dec 2007 16:42:39 UTC

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 $z=a^{2}/r$ , the axis

pointing vertically downwards and $r,\theta,z$ 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 $a\omega$ 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:

$(1+\frac{a^{4}}{z^{4}})\dot{z}^{2}=-\omega^{2}(z-a)(z+a-\frac{2g}{\omega^{2}})$

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

tashirosgt · **Posted:** Sun, 9 Dec 2007 19:39:56 UTC

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?

shamus · **Posted:** Mon, 10 Dec 2007 00:17:09 UTC

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 $a\omega^{2}=g$ , otherwise the particle rises or falls according to the sign of $a\omega^{2}-g$ .

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.

Opalg · **Posted:** Tue, 11 Dec 2007 18:21:45 UTC

Quote:

A smooth surface of revolution is hyperbolic with equation $z=a^{2}/r$ , the axis

pointing vertically downwards and $r,\:\theta,\:z$ 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 $a\omega$ 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, $r^2\dot{\theta} = a^2\omega$ . But $a^{2}/r = z$ , so we can write this as $r\dot{\theta} = a^2\omega/r = z\omega$ .

Differentiate rz = a^2

to get $\dot{r}z + r\dot{z} = 0$ , from which $\dot{r} = -r\dot{z}/z = -a^2\dot{z}/z^2$ , and so $\dot{r}^2 = a^4\dot{z}^2/z^4$ .

By conservation of energy, $\frac12m(\dot{r}^2 + \dot{z}^2 + r^2\dot{\theta}^2) - mgz = \frac12ma^2\omega^2 - mga$ . Rearrange this and substitute $r\dot{\theta} = z\omega$ to get
. . . . . . . . . . . . . . . . $\begin{array}{rcl}\dot{r}^2 + \dot{z}^2 &=& 2g(z-a) - \omega^2(z^2-a^2) \\{} &=& 2g(z-a) - \omega^2(z-a)(z+a) \\ &=& (z-a)(2g - \omega^2(z+a)).\end{array}$
Now substitute the value for $\dot{r}^2$ from the previous paragraph, and you get $a^4\dot{z}^2/z^4 + \dot{z}^2 = (z-a)(2g - \omega^2(z+a))$ . Tidy this up a bit and you have the answer given by the book.

leedavies · **Posted:** Fri, 14 Dec 2007 01:22:45 UTC

Cheers Opalg

leedavies · **Posted:** Sun, 16 Dec 2007 03:58:07 UTC

Hi Opalg,

I've finally returned to dynamics, working my way through your answer. Just a quick question, where did you get $r^{2}\dot{\theta}=a^{2}\omega$ from?

I get $r^{2}\dot{\theta}=\frac{a^{3}\omega}{z}$ since $r\dot{\theta}=a\omega$ and $r = \frac{a^2}{z}$

Cheers,

Lee

Opalg · **Posted:** Sun, 16 Dec 2007 10:36:42 UTC

leedavies wrote:

Just a quick question, where did you get $r^{2}\dot{\theta}=a^{2}\omega$ 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 $\vec{a} = (\ddot{r}-r\dot{\theta}^2)\vec{r} + (r\ddot{\theta}+2\dot{r}\dot{\theta})\vec{\theta} + \ddot{z}\vec{z}$ . In this problem, there is no force acting in the θ direction, and so $r\ddot{\theta}+2\dot{r}\dot{\theta} = 0$ . If you multiply that by r and integrate then you get $r^2\dot{\theta} =$ const. That is where I got the equation $r^2\dot{\theta} = a^2\omega$ from. (I then carelessly tried to justify this more briefly by claiming it came from angular momentum.)

leedavies · **Posted:** Sun, 16 Dec 2007 23:41:28 UTC

Sorry, I've just noticed I already had the answer $r^{2} \dot{\theta}$

Quote:

I get $r^{2} \dot{\theta} = \frac{a^{3}\omega}{z}$ since $r\dot{\theta}=a\omega$ and $r = \frac {a^{2}}{z}$

.

I forgot at t=0

,

, therefore $r=\frac{a^{2}}{a}=a$ and hence $r^{2} \dot{\theta} = a^{2}\omega$

Cheers,

Lee

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Question in "A First Course in Mechanics" by Mary

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