Two rotating masses balanced by a third mass (rotational dynamics)

[Eirik]

Homework Statement

Hi! This isn't really homework, but I'm practicing for my exam in mechanical physics and I'm really struggling with this one question!

Two masses, m, are rotating around the z axis as shown in the figure on a frictionless surface, but are being pulled down by a third mass, M. Find the differential equation for the movement of the system. The masses should be considered point masses and the system initially has $R=R_0$ and $\omega=\omega_0$ It should be on the following form, where $\alpha$ and $\beta$ are constants you need to find:

Relevant Equations

$\ddot R + \alpha g - \frac{\beta}{R^3}=0$, differential equation I need to find

$a_r = \ddot R -\omega^2R$, acceleration of mass m in the circle, given in task

Here's a diagram of what the system looks like:

So far I have figured out what the initial angular velocity is, if the system is balanced (no movement):

$\sum F_m = m*\frac{v^2}{R_0}-\frac{Mg}{2}=0$
$m \frac{v^2}{R_0}-\frac{Mg}{2}=0$ divide both sides by m
$\omega_0 = \sqrt{\frac{Mg}{2mR_0}}$

One of the hints given for the task, was that we should consider the conservation of angular momentum:
Moment of inerty, start: $I_0=2mR_0^2$
Angular momentum, start: $L_0=I_0\omega_0=2mR_0^2 \sqrt{\frac{Mg}{2mR_0}}$

I am struggling to find the equation for the angular momentum after that, however. I know that the moment of inertia should still be $I=2mR^2$, and that $L=I\omega$. How do I find \omega?

I also have this:
$\sum F_M = Mg-2S=M*\ddot R$
$S=m* (\ddot R -\omega^2R)$ Really not sure if this one is correct

Any help would be very greatly appreciated!

 

[TSny]

Homework Statement::
So far I have figured out what the initial angular velocity is, if the system is balanced (no movement):

If the system were to start out in the balanced state, then it would remain in this state. I don't think they want you to assume the system is in the balanced state initially. I think they want you to express the differential equation for $\ddot R$ in terms of $R_0$ and $\omega_0$.

One of the hints given for the task, was that we should consider the conservation of angular momentum:
Moment of inerty, start: $I_0=2mR_0^2$
Angular momentum, start: $L_0=I_0\omega_0=2mR_0^2 \sqrt{\frac{Mg}{2mR_0}}$

Since the system does not necessarily start in the balanced state, you cannot assume that $\omega_0=\sqrt{\frac{Mg}{2mR_0}}$. Just express $L_0$ in terms of $m$, $R_0$, and $\omega_0$.

I am struggling to find the equation for the angular momentum after that, however. I know that the moment of inertia should still be $I=2mR^2$, and that $L=I\omega$.

Yes.

How do I find $\omega$?

You can use the expression for the angular momentum to find $\omega$ as a function of $R$.

I also have this:
$\sum F_M = Mg-2S=M*\ddot R$

Be careful with signs. The left side expresses the net downward force on $M$. So, the acceleration on the right should be the downward acceleration of $M$. Does the downward acceleration of $M$ equal $\ddot R$ or $-\ddot R$?

$S=m* (\ddot R -\omega^2R)$

You have a sign error in this equation. Does the force $S$ on $m$ act in the radially outward direction or the radially inward direction?

 

[Eirik]

Thank you so much @TSny ! That was really, really helpful!

Then we have
$L_i=L$
$2mR_0^2\omega_0=2mR^2\omega$
Solving for $\omega$ gives us $\omega=\frac{R_0^2\omega_0}{R^2}$

And if we choose the positive direction for the accelaration along -z, it should be $S=m*(\omega^2R-\ddot R)$ instead, as the direction for m's acceleration will be radially inward. If I now substitute $\omega$ in this expression, I get:

$S=m*((\frac{R_0^2\omega_0}{R^2})^2R-\ddot R) = m(\frac{R_0^4\omega_0^2}{R^3}-\ddot R)$

Substitution in $Mg-2S=M\ddot R$ gives us:

$Mg-2(m(\frac{R_0^4\omega_0^2}{R^3}-\ddot R))=M\ddot R$

And after doing some algebra I get:

$\ddot R - \frac{M}{M-2m}g + \frac{\frac{2mR_0^4\omega_0^2}{M-2m}}{R^3}=0$

Meaning that $\alpha = \frac{M}{M-2m}$ and $\beta = \frac{2mR_0^4\omega_0^2}{M-2m}$

Yay! Does that look right?

 

[TSny]

Everything looks good up to here:

Substitution in $Mg - 2S = M \ddot R$

This equation has a sign error. Review the comments near the end of post #2. This will alter your final result. Otherwise, I think you have it.

 

[Eirik]

@TSny Yes, of course! I thought $\ddot R$ would be M's downward acceleration, but R will get smaller over time, so it has to be $Mg-2S=-M\ddot R$

That finally gives me

$\ddot R + \frac{M}{M+2m}g - \frac{\frac{2mR_0^4\omega_0^2}{M+2m}}{R^3}=0$

with $\alpha = \frac{M}{M+2m}$ and $\beta = \frac{2mR_0^4\omega_0^2}{M+2m}$

That also checks out with the signs the task said I was supposed to end up with lol

Thank you again so, so much for all the help!