Angular momentum and energy conservation.

[golanor]

Homework Statement

A mass m is laying on a frictionless table and is connected to a mass M with a nonelastic string going through the center of the table. At t=0 the m mass is at a r0 distance from the center of the table, and is moving at a v0 velocity in the tangent direction.
Find the r(t) using conservation of angular momentum and energy.

Homework Equations

conservation of angular momentum and energy.

The Attempt at a Solution

Since the net torque on the system is 0, the angular momentum remains the same, so:
m*r0*v0 = m*r*v
the string constraint:
Δy = Δr => Vy=Vr => Ay = Ar
Initial energy - 1/2*m*(v0)^2
Energy after t seconds: 1/2mv^2+MgΔy+1/2M(Vy)^2+1/2m(Vr)^2

this is as far as I've gotten. I tried using a different version with moment of inertia and angular velocity, but i always get stuck with complicated ODEs which, i think, are not what i should get.
I'm probably missing something, can you tell me what?

 

[Simon Bridge]

You want to express all those y's in terms of r ... also - won't mass m have a radial component to it's velocity? Try to work everything in terms of $r$ and $\dot{r}$.

 

[golanor]

That's the string constraint, the thing is that I'm getting a nonlinear ODE.
Vy is the velocity of the M mass and Vr is the radial velocity of the m mass.

 

[Simon Bridge]

If r changes y changes and Δy need not be constant with time... or do you happen to know that the radius changes at constant speed? There are other terms there too - what to do about the non-linearity will become clear once you have expressed everything in terms of r and r-dot.

I suspect (off your work so far) you'll end up with a DE of form: $$\dot{r}^2=f(r)$$ ...

 

[rbrayana123]

Try considering polar co-ordinates. Also make sure you differentiate the vectors properly. If in doubt, look it up. The acceleration is especially non-intuitive.

EDIT: Whoops. Said same thing as above!

 

[rcgldr]

Note that this is an zero-damped system. The tension in the string willl cycle between being lesser and greater than the weight of the suspended mass. This may help you confirm that you've determined the proper r(t), based on intitial conditions.

 

[golanor]

If r changes y changes and Δy need not be constant with time... or do you happen to know that the radius changes at constant speed? There are other terms there too - what to do about the non-linearity will become clear once you have expressed everything in terms of r and r-dot.

I suspect you'll end up with a DE of form: $$\dot{r}^2=f(r)$$ ...

I did get to there, i just thought that i might be missing something, since i don't know how to solve that equation.

 

[Simon Bridge]

What is it that's giving you the problem?
It's first-order in-homogeneous ... admittedly the in-homogeniaty is a pain...

Aside:
I suppose solving this by using conservation of energy and momentum could be the same as using Lagrangian mechanics?
I have a funny feeling that wouldn't count though...

 

[golanor]

It's simply much more complicated than i expected.
Thanks, i guess this was more of a math difficulty than a physical one.

 

[golanor]

It's simply much more complicated than i expected.
Thanks, i guess this was more of a math difficulty than a physical one.

 

[Simon Bridge]

... always assuming you've set up the equations correctly.
I'm kinda surprised not to see a term in $\ddot{r}$ for instance.
Anyway - the first step to solving a problem is to, properly, describe it.
Good luck.

 

[rcgldr]

Just curious, what equations did you finally end up with?

 

[golanor]

Just curious, what equations did you finally end up with?

$\stackrel{mv_{0}}{2}=\stackrel{m(r_{0}v_{0})^{2}}{2r^{2}}+0.5\dot{r^{2}}(m+M) +Mg(r-r_{0})$

 

[Simon Bridge]

Good use of LaTeX - I take it you meant the \stackrel's to be fractions? You'll want \frac instead. You have the dot over the $r^2$ instead of over the $r$. I'm guessing you tried the sigma menu? It's usually easier just to type it out.

Also - did you miss out a square on the RHS? So, I suspect, you intended...

$$\frac{1}{2}mv_0^2=\frac{1}{2}\frac{mr_0^2v_0^2}{r^2}+\frac{1}{2}(m+M)\dot{r}^2+Mg(r-r_0)$$ ... that does look nasty doesn't it?

Tidying up

$$mv_0^2r^2 = mr_0^2v_0^2 + (m+M)r^2\dot{r}^2 + 2Mgr^2(r-r_0)$$

$$\dot{r}=\sqrt{\frac{(mv_0^2-2Mgr_0)r^2 - mr_0^2v_0^2 - 2Mgr^3}{(m+M)r^2}}$$ ... Maple?

 

[golanor]

Thanks, that does look much better. (This is what happens when you do all of your lab reports on MS Word - you forget whatever LaTeX you knew)
I actually just did another derivation over time of the equation and simply left it there, since solving this got me a very complicated equation, and I had what i needed for the second question.
Thanks for all the help!

 

[Simon Bridge]

That must be such a pain! I've been lucky enough to only have to write reports for people who accept true ISO standards so I can write everything in LaTeX. I just use a text editor and a terminal ;)

Still - well done extracting what you needed from all that.