Mass, String, and a Ring

1. Dec 3, 2013

Astrum

1. The problem statement, all variables and given/known data
A mass $m$ whirls around on a string which passes through a ring. Neglect gravity. Initially the mass is at distance $r_0$ from the center and is revolving at angular velocity $\omega _0$. The string is pulled with constant velocity $v$ starting at $t=0$ so that the radial distance to the mass decreases. Draw a force diagram and obtain a differential equation for $\omega$

2. Relevant equations

3. The attempt at a solution
So, the force diagram has only one force, which is the tension on the string acting on the mass. Because the string is shortening by $v = \dot{r}$, I set up the equation $T = F_r = m(-r\omega ^2 +r\omega +2v\omega)$

If we divide out $m$ we get $\mathbf a = -r\omega ^2 +r\omega +2v\omega$, I'm not exactly sure what we're suppose to get on the other side.

I'm asking for a hint, or maybe a word of advice. The book says the answer should be simple and easy to integrate.

2. Dec 3, 2013

tiny-tim

Hi Astrum!

Since you don't know what F is, what is the point of your radial equation?

Try the tangential equation.

3. Dec 3, 2013

voko

The dimension of $r\omega^2$ and $v\omega$ is $m \cdot s^{-2}$, while the dimension of $r\omega$ is $m \cdot s^{-1}$. They cannot be added together.

Is the force-based approach mandatory here? Conservation of angular momentum seems far more suitable here.

4. Dec 3, 2013

haruspex

That equation is dimensionally incorrect. The mrω term has units of momentum.
And check the sign on $v = \dot r$.

What is conserved as the string is pulled in?

5. Dec 3, 2013

Astrum

Yes, this needs to be worked out through using force, rather than conservation of angular momentum.

I'm not sure what you mean by tangential, are you talking about $\hat{\theta}$?

$a_r =( \ddot{r}-r\omega ^2 )\hat{r}$ and $a_{\theta} = (r\dot{\omega}+\dot{r}\omega )\hat{\theta}$, these each can be rewritten as $a_r = -r\omega ^2 \hat{r}$ and $a_{\theta} = (r\dot{\omega}-v\omega)\hat{\theta}$. Are you saying to rewrite these in terms of tangential velocity? I'm not sure how that helps us.

6. Dec 3, 2013

voko

One of the terms has a factor of 2 missing.

Tangential velocity is directly related to angular velocity.

7. Dec 3, 2013

Astrum

Sorry, I always seem to miss terms when typing them =/

I realize that, $r\omega = v$. What does this do for us though? All it does is eliminate the very thing we're interested in finding.

I don't really understand the direction you're pointing me in.

Because there's no force in the radial direction, we can say: $0 = r\dot{\omega}-2v\omega$, if we split it up, we get:

$$\int \frac{d\omega}{\omega} = -\int \frac{2v}{r} dt$$

Last edited: Dec 3, 2013
8. Dec 3, 2013

voko

Since there is no tangential force, what can be said about the tangential acceleration?

9. Dec 3, 2013

voko

Oops, I missed your edit. Use the fact that $dr = -v dt$.

10. Dec 3, 2013

voko

Oh, and that minus sign in front of the RHS integral is definitely wrong.

11. Dec 3, 2013

Astrum

I tried to solve it with respect to $t$, because that's what we're after, why would be integrate $r$?

Doing it the way you said, we find $\omega = r + C$, and that just isn't right.

My attempt is here: Solving this, we end up with $$\omega (t) = ce^{\frac{2v}{r}t}$$, solving for initial conditions: $\omega (0) = \omega _0$, therefore $\omega (t) = \omega _0 e^{\frac{2v}{r}t}$, this doesn't pass the hint given in the text though, so I'm still missing something.

The hint given says "if $vt=\frac{r_0}{2}$ then $\omega = 4 \omega _0$. plugging it in we get $\omega= \omega _0 e^{\frac{r_0}{r}}$"

Judging by how simple this turned out to be, I'm sure there's some little detail I just am not seeing.

Edit: sorry if I'm being sloppy, I'm doing half my work on paper and the other half in the posts, so I may have made an algebra error.

12. Dec 3, 2013

voko

I do not see how you end up with either solution.

Method 1. $$0 = \int\limits_{\omega_0}^{\omega} \frac {d\omega} {\omega} + 2 \int\limits_{r_0}^{r} \frac {dr} r = \ln \frac {\omega} {\omega_0} + 2 \ln \frac r {r_0} = \ln \frac {\omega} {\omega_0} + \ln \left( \frac r {r_0} \right)^2 = \ln \frac {\omega r^2} {\omega_0 r_0^2} \\ \omega r^2 = \omega_0 r_0^2$$ which is conservation of angular momentum, as was to be expected. That cannot possibly give you $\omega = r + C$.

Method 2, where explicit dependence on $t$ is retained, is similar, keeping in mind that $r = r_0 - vt$.

13. Dec 3, 2013

Astrum

I see, that makes a lot of sense. I was under the impression that I didn't need to place limits of integration on there. I'll be sure not to make that mistake again.

Thanks for the help, I made that one way harder than it needed to be.

14. Dec 3, 2013

voko

Personally, I prefer using limits of integration from the beginning and avoid having indeterminate constants of integration. But that is not essential here. You would just get $$\ln \omega + 2 \ln r = C \\ \ln \omega r^2 = C \\ \omega r^2 = k = e^C$$ which is conservation of angular momentum again.

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