Find velocities when a mass leaves a system of a rod with two masses

[Jeavenasti]

Homework Statement

A rod with mass M (without a fixed axis) is attached to two masses m (they are point masses) located at its ends. The system rotates about the center of mass with velocity $w_o$ k . At the instant in which a mass is in the position L/2 i, it detaches from the rod leaving with the same instantaneous speed that it had at that moment. Calculate all the new velocities.

Relevant Equations

Angular momentum = I x w + r x mv

First, I can say that the velocity of the mass leaving the system is equal to $w_o$ (k) x $L/2$ (i) which tells me that its velocity will be w_o L/2 (j)

Now, since the net external force is equal to 0, (linear) momentum is conserved, so:
At first the velocity of the center of mass was 0 and after it leaves the system, it will still be 0, so I can state:

0 = mv_1 + 2mv_2

(being v_1 = velocity of the mass leaving the rod and v_2 the velocity of the rod and the other mass together).
So:

v_2 = -mv_1/2m

v_2 $= - v_1/2 = -w_o L/4 (j)----------Now, I need to find the angular velocities:
since the sum of torques is 0, angular momentum is conserved (assume it is true).
So,

L_o = L_f
I can calculate L_o using my Center-of-momentum frame in the center of mass of the system mass-rod-mass.

L_o = I w = (1/12 mL^2 + 2 m(L/2)^2)w_o = 7/12 mL^2

Now, here I had a few doubts.
First of all, I found that the rod-mass system (with only 1 mass) will have a new center of mass in i-axis equals to $x_{cm1} = -L/4$ i (this comes from, xcm = m (-L/2)/2m, with (0,0) in the center of the rod)
But then, to find the final Angular Momentum I had doubts:

Do I have to change the moment of inertia of the rod-mass with respect to an axis that passes through the center of mass of the rod-mass system? Something like:
I_f = (1/12 mL^2 + m (L/4)^2) + m(3/4 L)^2 = 17/24 m L^2
(Or Do I have to keep the same moment of inertia?).

Then, I can write:

L_f = 17/24 m L^2 w_f + (L/2) (i) x m v_1 (j) + (-L/4) (i) x 2m v_2 (-j)

L_f = 17/24 m L^2 w_f + (L/2) m v_1 (k) + (L/4) 2m v_2 (k)

But, since the modules of v_1 and v_2 are: $v_1 = w_oL/2$, and $v_2 = w_oL/4$

L_f = (17/24) m L^2 w_f + (L/2) m (w_oL/2) (k)+ (L/4) 2m (w_oL/4) (k)

L_f= 17/24 m L^2 w_f + (Lm w_oL)/4 (k)+ (Lm w_oL)/8 (k)And then, setting it equal to the initial angular momentum, I can determine wf. My main doubt arises from what I mentioned above, I don't know if changing the moment of inertia of the rod-mass is correct (and I also have doubts, could I have placed the center of moments at another point to do something easier?). I´d apreciatte some help, thanks in advance. (Note: I have not solved it rigorously and it is likely that some things are not well justified, but I shared it to solve specific doubts)

 

[jbriggs444]

Going slowly through your post to make sure that we are on the same page...

Homework Statement: A rod with mass M (without a fixed axis) is attached to two masses m (they are point masses) located at its ends. The system rotates about the center of mass with velocity $w_o$ k .

Ahhh, this took me a bit to comprehend. You are using $k$ as a unit vector, $\hat{k}$ to provide a direction for the angular momentum pseudo-vector $\vec{\omega_0}$ whose magnitude is $\omega_0$.

At the instant in which a mass is in the position L/2 i, it detaches from the rod leaving with the same instantaneous speed that it had at that moment. Calculate all the new velocities.

OK, fine. We have a dumbell shaped object which loses the mass on one end.

First, I can say that the velocity of the mass leaving the system is equal to $w_o$ (k) x $L/2$ (i) which tells me that its velocity will be w_o L/2 (j)

Sure. This makes sense. If you want to exercise your $\LaTeX$ skills, you could write ##\omega_0 \hat{k} \times \frac L 2 \hat{i}## which renders as: $\omega_0 \hat{k} \times \frac L 2 \hat{i}$

Now, since the net external force is equal to 0, (linear) momentum is conserved, so:

The separation is presumably instantaneous. So the requirement for momentum conservation is not even that strict. We just require that the impulse is negligible. Which only requires that the force not be infinite.

But yes, linear momentum is conserved across the separation event.

At first the velocity of the center of mass was 0 and after it leaves the system, it will still be 0, so I can state:

0 = mv_1 + 2mv_2

(being v_1 = velocity of the mass leaving the rod and v_2 the velocity of the rod and the other mass together).

Wait a minute. The rod does not have mass m.

 

[Jeavenasti]

Hi, thanks for taking the time to read it. Yes, in the problem the rod has mass too. I posted this in another page and I could write with latex (iḿ not sure why it didn´t work here, maybe itś another formt). Thanks in advance. My doubt in general is when I have to calculate the final angular momentum

 

[TSny]

Yes, in the problem the rod has mass too.

But the problem statement has $M$ for the mass of the rod and $m$ for each point mass. It looks like you assumed $M = m$ in your work.

 

[Jeavenasti]

But the problem statement has $M$ for the mass of the rod and $m$ for each point mass. It looks like you assumed $M = m$ in your work.

Yes sorry, it was a typo. The mass of the rod is m too.

 

[TSny]

Thanks in advance. My doubt in general is when I have to calculate the final angular momentum

The problem statement does not mention finding the final angular momentum. Is there more to the problem than what you typed in your original post? When posting a problem statement, we kindly ask that you post the entire problem exactly as given to you.

Is it the final angular momentum of the entire system (the rod and both point masses) that you want to find?

For the entire system are there any external torques acting on the system at any time? Does the answer to this question help in finding the final angular momentum of the entire system?

 

[Jeavenasti]

The problem statement does not mention finding the final angular momentum. Is there more to the problem than what you typed in your original post? When posting a problem statement, we kindly ask that you post the entire problem exactly as given to you.

Is it the final angular momentum of the entire system (the rod and both point masses) that you want to find?

For the entire system are there any external torques acting on the system at any time? Does the answer to this question help in finding the final angular momentum of the entire system?

No, The problem doesn't ask me for the final angular momentum, but I need it in order to find the new angular velocity of the rod-mass. The problem is that I'm not sure if I wrote the Final Angular Momentum properly (this: ...L_f = 17/24 m L^2 w_f + (L/2) (i) x m v_1 (j) + (-L/4) (i) x 2m v_2 (-j)... " Or if I made a mistake writing that line because I got confused.
For the entire system I saw that there werent torques acting so the angular momentum doesnt change. (Thanks for your help)

 

[jbriggs444]

No, The problem doesn't ask me for the final angular momentum, but I need it in order to find the new angular velocity of the rod-mass. The problem is that I'm not sure if I wrote the Final Angular Momentum properly (this: ...L_f = 17/24 m L^2 w_f + (L/2) (i) x m v_1 (j) + (-L/4) (i) x 2m v_2 (-j)... " Or if I made a mistake writing that line because I got confused.
For the entire system I saw that there werent torques acting so the angular momentum doesnt change. (Thanks for your help)

It is difficult reading a formula and trying to reverse engineer the thinking that went into it.

Here we have a formula for the final angular momentum, $L_f$. The final angular momentum of what, exactly? Computed about what reference axis? Justified how?

The unit vectors scattered through the formula are not helping matters much. So let us clear that clutter.

$L_f = \frac{17}{24} m L^2 \omega_f + \frac{L}{2} m v_1 - \frac{L}{4} 2m v_2$

Guessing... I think we are using the original center of mass of the complete assembly as the reference axis.

I think the first term there is the angular momentum from the rotation of the rod plus one mass assembly about that assembly's center of mass.

The second term is the angular momentum from the linear momentum of the solo mass about the original COM axis.

The third term appears to be the angular momentum from the linear momentum of the rod plus one mass assembly about the original COM axis.

If that interpretation is correct, that equation makes sense.I am not 100% confident in the formula for the first term. It seems plausible, but I will need to work it out myself... $\frac{17}{24} m L^2 \omega_f$... So the moment of inertia of the assembly about its center of mass would then be $\frac{17}{24} mL^2$.

No. I do not agree with that moment of inertia for the rod plus one mass about that assembly's center of mass.

 

[Jeavenasti]

It is difficult reading a formula and trying to reverse engineer the thinking that went into it.

Here we have a formula for the final angular momentum, $L_f$. The final angular momentum of what, exactly? Computed about what reference axis? Justified how?

The unit vectors scattered through the formula are not helping matters much. So let us clear that clutter.

$L_f = \frac{17}{24} m L^2 \omega_f + \frac{L}{2} m v_1 - \frac{L}{4} 2m v_2$

Guessing... I think we are using the original center of mass of the complete assembly as the reference axis.

I think the first term there is the angular momentum from the rotation of the rod plus one mass assembly about that assembly's center of mass.

The second term is the angular momentum from the linear momentum of the solo mass about the original COM axis.

The third term appears to be the angular momentum from the linear momentum of the rod plus one mass assembly about the original COM axis.

If that interpretation is correct, that equation makes sense.I am not 100% confident in the formula for the first term. It seems plausible, but I will need to work it out myself... $\frac{17}{24} m L^2 \omega_f$... So the moment of inertia of the assembly about its center of mass would then be $\frac{17}{24} mL^2$.

No. I do not agree with that moment of inertia for the rod plus one mass about that assembly's center of mass.

Thank you, you were very helpful! I'm going to check the moment of Inertia again. Thanks!

 

[kuruman]

I cannot tell what you have been doing so I will outline what you should be doing. You have the initial angular momentum about the midpoint of the rod, point O
$L_i=\dfrac{7}{12}mL^2\omega_0~~$ and I agree.

When the mass is detached, the rod and remaining mass rotate about the new center of mass. This means that if you want to conserve angular momentum, you better do it about the midpoint of the rod which is now moving. The angular momentum after the detachment comes in three pieces that you have to calculate.
1. Angular momentum of the detached mass.
2. Angular momentum of the center of mass of the rod + remaining mass relative to the initial point O.
3. Angular momentum about the center of mass of the rod + remaining mass.

The velocities that you are asked to calculate in terms of $\omega_0$ are
1. The velocity of the detached mass.
2. The velocity of the center of mass of the rod + remaining mass
3. The angular velocity of the rod + remaining mass lollipop as it spins about its center of mass.

Please use LaTeX.

 

[haruspex]

Calculate all the new velocities.

Is that the exact wording? I see no reason why any part of the system will instantly change velocity. What will happen is that different parts will become the mass centres of rigid components.