Angular momentum of a mass-rope-mass system

[Santilopez10]

Homework Statement

Two bodies $m_1$ and $m_2$ are connected by an ideal rope, which passes through a whole of an horizontal table free of friction; $m_1$ moves along a circular trajectory of radius $r_0$, $m_2$ hangs below the table:
1) Find an expression for $v_{0_1}$ that $m_1$ must have so that $m_2$ is stationary.
2) Suposse that as $m_1$ moves with $v_{0_1}$, $m_2$ mass duplicates. Find the angular momentum of $m_1$. Is it constant?

Relevant Equations

$\vec a= (\ddot r-r{\dot{\theta}}^2)\hat r +(r \ddot{\theta} + 2\dot r \dot{\theta}) \hat{\theta}$
$\vec L= \vec r \times \vec p$

1) the motion equations for $m_2$ are: $$T-m_2 g=0 \rightarrow T=m_2 g$$
$m_1$: $$T=m_1\frac{v^2}{r_0} \rightarrow \vec {v_0}=\sqrt{\frac{r_0 g m_2}{m_1}}\hat{\theta}$$

2) This is where I am stuck, first I wrote $m_2$ motion equation just like before, but in polar coordinates: $$-T+m_2 g= m_2 a_r \rightarrow a_r= -\frac{T}{m_2}+g$$
$m_1$: $$\hat r: -T=m_1 a_r=m_1(-\frac{T}{m_2}+g) \rightarrow T=g \frac{m_1 m_2}{m_1-m_2}$$
and at last I have a kind of system of equations:
$$\hat{\theta}:r \ddot{\theta}+2\dot r \dot{\theta}=0$$
$$\hat r: -g \frac{m_2}{m_1-m_2}=\ddot r-r {\dot{\theta}}^2$$

This is all I could do, I think I might be missing some assumption or condition, but honestly I am not sure. Any help would be appreciated.

 

[tnich]

You have $m_1$, $r_0$ and $v$. Suppose you write out the definition of angular momentum in terms of $I$ and $\omega$, and then make the appropriate substitutions.

 

[Santilopez10]

You have $m_1$, $r_0$ and $v$. Suppose you write out the definition of angular momentum in terms of $I$ and $\omega$, and then make the appropriate substitutions.

I do not know what $I$ is, if it is the moment of inertia, then it has not been taught yet. We have already developed the equation of angular momentum for central forces: $$\vec L= m r^2 \dot{\theta} \hat k$$
but still I need to find and expression for r and $\omega$= $\dot{\theta}$ as functions of time to use them in my formulas.

 

[tnich]

It seems to me that the radius is given in the definitions at the beginning of the problem statement as $r_0$. To get $\dot\theta$, figure out the time it would take mass $m_1$ to complete one full circle (from part 1 you already know its speed, and you can figure out the circumference), and divide it into the angle it revolves through in one full circle. That gives you the angular rate.

 

[Santilopez10]

It seems to me that the radius is given in the definitions at the beginning of the problem statement as $r_0$. To get $\dot\theta$, figure out the time it would take mass $m_1$ to complete one full circle (from part 1 you already know its speed, and you can figure out the circumference), and divide it into the angle it revolves through in one full circle. That gives you the angular rate.

the system is not stationary, r is not constant. What I thought would be taking the initial angular momentum and because the force is central then angular momentum is conserved, then the initial angular momentum is constant.

 

[tnich]

the system is not stationary, r is not constant.

That is not apparent from the problem statement. Can you tell me why you say that? Perhaps the phrase "$m_2$ mass duplicates" in the problem statement is copied incorrectly? I can't parse it.

 

[Santilopez10]

That is not apparent from the problem statement. Can you tell me why you say that? Perhaps the phrase "$m_2$ mass duplicates" in the problem statement is copied incorrectly? I can't parse it.

$m_2$ or particle 2 mass duplicate, A.K.A $m_2’=2m_2$ so the system is not stationary anymore.

 

[tnich]

So you are saying that mass $m_1$ has velocity $v_0$ at time $t=t_0$. It must also have an initial radius of revolution. Why not call that $r_0$? Certainly you can write an expression for the angular momentum at time $t_0$. Have you studied conservation of angular momentum, yet? Do you think that principle applies in this case?

 

[Santilopez10]

So you are saying that mass $m_1$ has velocity $v_0$ at time $t=t_0$. It must also have an initial radius of revolution. Why not call that $r_0$? Certainly you can write an expression for the angular momentum at time $t_0$. Have you studied conservation of angular momentum, yet? Do you think that principle applies in this case?

Yes, we have studied it. The rate of change of the angular momentum is the torque applied to the system, but torque is 0 in this case as the position and force vectors are parallel with respect to our origin (the hole). Then angular momentum is constant, and with the initial conditions we can get an answer. Is this reasoning okay?

 

[jbriggs444]

Yes, we have studied it. The rate of change of the angular momentum is the torque applied to the system, but torque is 0 in this case as the position and force vectors are parallel with respect to our origin (the hole). Then angular momentum is constant, and with the initial conditions we can get an answer. Is this reasoning okay?

Yes. Your understanding of the problem matches mine. No calculation is required to reach the conclusion that angular momentum is constant.