# Homework Help: 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.