Finding the initial acceleration rotating masses not about their C.o.M

[Dazed&Confused]

Homework Statement

A disk rotates with constant angular velocity \omega. Two masses, m_A and m_B, slide without friction in a groove passing through the centre of the disk. They are connected by a light string of length l, and are initially held in position by a catch, with mass m_a at a distance r_A from the centre. Neglect gravity. At t=0 the catch is removed and the masses are free to slide. Find \ddot{r_A} immediately after the catch is removed in terms of m_A, m_B, l, r_A, and \omega

The Attempt at a Solution

Since the string is light, the tension on each side is equal.

We have T = m_A\omega^2r_A - m_A\ddot{r_A} = m_B\omega^2(l-r_A) - m_B\ddot{r_B}. If I had another equation in terms of \ddot{r_A} and \ddot{r_B} then I could solve for \ddot{r_A}. There is an angular acceleration of magnitude 2\omega\dot{r_A} but I don't know how to use this. Any help with this would be appreciated.

 

[jbriggs444]

Homework Statement

A disk rotates with constant angular velocity \omega. Two masses, m_A and m_B, slide without friction in a groove passing through the centre of the disk. They are connected by a light string of length l, and are initially held in position by a catch, with mass m_a at a distance r_A from the centre. Neglect gravity. At t=0 the catch is removed and the masses are free to slide. Find \ddot{r_A} immediately after the catch is removed in terms of m_A, m_B, l, r_A, and \omega

The Attempt at a Solution

Since the string is light, the tension on each side is equal.

We have T = m_A\omega^2r_A - m_A\ddot{r_A} = m_B\omega^2(l-r_A) - m_B\ddot{r_B}. If I had another equation in terms of \ddot{r_A} and \ddot{r_B} then I could solve for \ddot{r_A}. There is an angular acceleration of magnitude 2\omega\dot{r_A} but I don't know how to use this. Any help with this would be appreciated.

Can you convince yourself that the string stays taut? If so then the two masses stay at the same position relative to one another. So one can treat them as if they were a single rigid object. Or, if you prefer, this mean that \ddot{r_A} = \ddot{r_B}

 

[Dazed&Confused]

Can you convince yourself that the string stays taut? If so then the two masses stay at the same position relative to one another. So one can treat them as if they were a single rigid object. Or, if you prefer, this mean that \ddot{r_A} = \ddot{r_B}

I don't understand why that would be the case.

 

[verty]

The question is confusing, I think this is what is meant:

Two masses A,B are connected by a string of length l. A third mass C, the catch, is placed on top of the string, being thereby temporarily connected to the string. In this arrangement, and being situated in the frictionless groove of this rotating disk with angular velocity $\omega$, mass A is at a distance $r_A$ from the center and the masses are stationary. The catch is removed, find the radial acceleration of A at this instant.

Suppose that C has mass $m_C$. The point is that we know nothing about C, the mass and position are both variable. All we know is that the position and mass of C is sufficient to root A and B in place. These are the facts you have: A and B were rooted in place and a hypothetical mass C was sufficient to achieve that. Find $\ddot{r_A}$.

 

[verty]

I now understand JBriggs's response, he is right of course. Think again about what he said.

 

[Dazed&Confused]

I now understand JBriggs's response, he is right of course. Think again about what he said.

Does the string stay taut as if it were to go below the m\omega^2r term then there would have to be a radial acceleration outward to compensate?