What is the correct expression for the total moment of inertia in this scenario?

[Nexus99]

Homework Statement

A homogeneous rod of lenght L and mass M translate with a constant velocity $\vec{v_0}$ on a smooth plane. At the time $t = t_0$ the rod is hit simultaneously by 2 bodies of mass $m_1$ and $m_2$ that are moving with velocity $\vec{v} = - \vec{v_0}$ The collision is completely anaelastic. Determine;
1) The position of the center of mass of the system after the collision (consider a reference system with origin in the center of the rod)
2) The final speed of the system after the collision
3) the angular velocity after the collision
4) How much should the magnitude of $\vec{v}$ so that the system composed by the rod and the masses stop after the impact?

Relevant Equations

center of mass, conservation of linear and angular momentum

1) I found:
$x_{CM} = z_{CM} = 0$
$y_{CM} = \frac{L}{2} \frac{m_1 - m_2}{M + m_1 + m_2}$

2) Applying the conservation of linear momentum:
$M v_0 - m_1 v_0 - m_2 v_0 = (M + m_1 + m_2) v_f$
$v_f = \frac{M - m_1 - m_2 }{M + m_1 + m_2} v_0$
The velocity should have been found also calculating the velocity of the center of mass of the system, that isn't changed after the collision because there are no external impulsive forces that act on the system.

3) $\omega = \frac{v_f}{y_{CM}}$

4)
$M v_0 + m_1 v+ m_2 v = 0$
$v = -\frac{M v_0}{m_1 + m_2}$

Is it correct?

 

[TSny]

Your answers to parts (1), (2), and (4) look good, assuming I'm interpreting the problem statements correctly.

Part (2) is apparently asking for the speed of the CM of the system, which I think you found correctly.

In part (4) they want the system to "stop". I guess they want the CM of the system to be at rest after the collision (and therefore also at rest before the collision). If $m_1 \neq m_2$ and if $m_1$ and $m_2$ have the same initial speed, then it's not possible for the system to come to rest such that no parts of the system are moving after the collision . Note that they ask for the magnitude of $\vec v$. Should you include the negative sign in your answer?

I don't agree with your result for Part (3). This part looks like it will take quite a bit of algebra. Can you think of a conservation law that you can use?

 

[kuruman]

I don't believe (3) is correct. Instead of $v_f$ I would use $v_{cm}$ for the velocity of the center of mass. After the collision the CM located at $y_{cm}$ relative to the midpoint of the rod is moving in a straight line with velocity $v_{cm}$.

I just noticed that @TSny posted a reply before mine. Follow its advice.

 

[kuruman]

Part (1) is presumably asking for the location of the CM of the system immediately after the collision rather than at some other time after the collision.

Part (1) could also be interpreted to be asking for the position of the center of mass as a function of time.

 

[Nexus99]

@kuruman Part (1) asked for the position on the CM only after the collision

For part 3 i thought also to this solution:

Angular momentum is conserved:
$m_1 v_0 \frac{L}{2} - m_2 v_0 \frac{L}{2} = I_{TOT} \omega + (M + m_1 + m_2) v_f y_{CM}$
where $I_{TOT}$ is the momentum of inertia of the system, and i think is something like that
$I_{TOT} = \frac{ML^2}{12} + m_1 (\frac{L}{2})^2 + m_2 (\frac{L}{2})^2$

am i right?

 

[TSny]

For part 3 i thought also to this solution:

Angular momentum is conserved, calculating the angular momentum in y_{CM}:
$m_1 v_0 \frac{L}{2} - m_2 v_0 \frac{L}{2} = I_{TOT} \omega + (M + m_1 + m_2) v_f y_{CM}$
where $I_{TOT}$ is the momentum of inertia of the system, and i think is something like that
$I_{TOT} = \frac{ML^2}{12} + M{y_{CM}^2} + m_1 (\frac{L}{2} - y_{CM})^2 + m_2 (\frac{L}{2} - y_{CM})^2$

am i right?

Looks like a good approach.

In the angular momentum equation, does the last term have the correct overall sign? Are you taking clockwise or counterclockwise as positive?

In the equation for $I_{TOT}$, you have the same distance $\left( \frac{L}{2} - y_{CM} \right)$ for both $m_1$ and $m_2$. Is that right?

 

[Nexus99]

I modified something. Anyway I'm taking counterclockwise rotation as positive

 

[TSny]

I modified something. Anyway I'm taking counterclockwise rotation as positive

OK. I still think you have the wrong sign for the last term in the angular momentum equation. Don't positive values for $y_{CM}$ and $v_f$ produce a clockwise angular momentum about the origin?

@kuruman Part (1) asked for the position on the CM only after the collision

For part 3 i thought also to this solution:

Angular momentum is conserved:
$m_1 v_0 \frac{L}{2} - m_2 v_0 \frac{L}{2} = I_{TOT} \omega + (M + m_1 + m_2) v_f y_{CM}$
where $I_{TOT}$ is the momentum of inertia of the system, and i think is something like that
$I_{TOT} = \frac{ML^2}{12} + m_1 (\frac{L}{2})^2 + m_2 (\frac{L}{2})^2$

am i right?

I think you were closer to the correct expression for $I_{TOT}$ before you changed it. If you are taking the origin, O, for angular momentum to be at the center of the rod at the time of the collision, then after the collision

$L_{\rm TOT \, about \, O} = L_{\rm rotation \, about \, CM} + L_{\rm due \, to \, motion \, of \, CM \, relative \, to \, O}$.

The first term on the right represents angular momentum due to spinning about the CM of the system. So, this term is $I_{TOT} \omega$, where $I_{TOT}$ is the moment of inertia of the system relative to the CM of the system.

 

[Nexus99]

So it's easier to consider
$I_{TOT} = \frac{ML^2}{12} + My_{CM}^2 + m_1 (\frac{L}{2} - y_{CM})^2 + m_2 (\frac{-L}{2} + y_{CM})^2$
and:
$m_1 v_0 \frac{L}{2} - m_2 v_0 \frac{L}{2} = I_{TOT} \omega - (M + m_1 + m_2) v_f y_{CM}$

 

[kuruman]

So it's easier to consider
$I_{TOT} = \frac{ML^2}{12} + My_{CM}^2 + m_1 (\frac{L}{2} - y_{CM})^2 + m_2 (\frac{-L}{2} + y_{CM})^2$
and:
$m_1 v_0 \frac{L}{2} - m_2 v_0 \frac{L}{2} = I_{TOT} \omega - (M + m_1 + m_2) v_f y_{CM}$

No. That's inconsistent. The point about which you consider angular momentum must match the point about which you have figured out the moment of inertia. In the first equation that point is the CM; in the second equation it is the midpoint of the rod.

I would write $m_1 v_0 y_{cm} - m_2 v_0 y_{cm} = I_{TOT} \omega$ because after the collision the rotation is about the CM and that is the $\omega$ you are looking for.

On Edit: If $I_{TOT}$ is the moment of inertia about the CM then the expression you have for it is incorrect. One of the masses is at distance $\frac{L}{2}+y_{cm}$ and the other at $\frac{L}{2}-y_{cm}$ from it. You can't have a relative negative sign in both squared terms.

 

[Nexus99]

So the equations should be:
$I_{TOT} = \frac{ML^2}{12} + My_{CM}^2 + m_1( \frac{L}{2} - y_{CM})^2 + m_2( \frac{L}{2} + y_{CM})^2$

and
$m_1 v_0 y_{CM} - m_2 v_0 y_{CM} = I_{TOT} \omega$

Why the term $(M + m_1 + m_2) y_{CM} v_f$ should not be inserted in the equation? Isn't the system both rotating and traslating?

 

[kuruman]

So the equations should be:
$I_{TOT} = \frac{ML^2}{12} + My_{CM}^2 + m_1( \frac{L}{2} - y_{CM})^2 + m_2( \frac{L}{2} + y_{CM})^2$

and
$m_1 v_0 y_{CM} - m_2 v_0 y_{CM} = I_{TOT} \omega$

Why the term $(M + m_1 + m_2) y_{CM} v_f$ should not be inserted in the equation? Isn't the system both rotating and traslating?

It is, but if you choose to calculate angular momentum about the CM, the translational term $\vec r\times \vec p_{cm}$ does not contribute to the angular momentum because $\vec r$ and $\vec p_{cm}$ are collinear.

Your equation for $I_{TOT}$ is correct. Good luck with the algebra.

 

[haruspex]

It is, but if you choose to calculate angular momentum about the CM, the translational term $\vec r\times \vec p_{cm}$ does not contribute to the angular momentum because $\vec r=0$.

Your equation for $I_{TOT}$ is correct. Good luck with the algebra.

Right, but the second equation in post #11 is missing the prior angular momentum of the rod about the system's mass centre.

I find it is generally easier to avoid working with the mass centre of such a system. Just take moments about a fixed point in the line of the original velocity of the rod's mass centre.

 

[kuruman]

Right, but the second equation in post #11 is missing the prior angular momentum of the rod about the system's mass centre.

Right. I forgot that the rod is not at rest before the collision.

 

[Nexus99]

I find it is generally easier to avoid working with the mass centre of such a system. Just take moments about a fixed point in the line of the original velocity of the rod's mass centre.

For example? The original center of mass of the rod? Or the points where the two masses collide with the rod?

The equation in post #11 should be:
$m_1 v_0 y_{CM} - m_2 v_0 y_{CM} + M v_0 y_{CM}= I_{TOT} \omega$

and, after a lot of absurd algebra i got:
$\omega = 6v_0 \frac{(m_1 - m_2)(m_1 + M - m_2)}{L(4m_1(M+3m_2) + M(M+4m_2))}$

 

[haruspex]

For example? The original center of mass of the rod? Or the points where the two masses collide with the rod?

Any fixed point in that line will do.

The equation in post #11 should be:
$m_1 v_0 y_{CM} - m_2 v_0 y_{CM} + M v_0 y_{CM}= I_{TOT} \omega$

Also a problem with the $m_1, m_2$ terms? Where are those in relation to the system's mass centre?

 

[kuruman]

$\dots$
and, after a lot of absurd algebra i got:
$\omega = 6v_0 \frac{(m_1 - m_2)(m_1 + M - m_2)}{L(4m_1(M+3m_2) + M(M+4m_2))}$

A simple test, which a correct expression must pass, is that $\omega$ change its sign but retain its magnitude if you swap the positions of the masses. Algebraically this is accomplished by swapping indices 1 and 2. Your expression does not pass this test.

 

[Nexus99]

A simple test, which a correct expression must pass, is that $\omega$ change its sign but retain its magnitude if you swap the positions of the masses. Algebraically this is accomplished by swapping indices 1 and 2. Your expression does not pass this test.

Because the equation is wrong, like haruspex said, the correct one i think is:

$m_1 v_0 ( \frac{L}{2} - y_{CM}) - m_2 v_0 ( \frac{L}{2} + y_{CM}) + M v_0 ( y_{CM}) = I_{TOT} \omega$

 

[kuruman]

Because the equation is wrong, like haruspex said, the correct one i think is:

$m_1 v_0 ( \frac{L}{2} - y_{CM}) - m_2 v_0 ( \frac{L}{2} + y_{CM}) + M v_0 ( y_{CM}) = I_{TOT} \omega$

Yes, that is the correct momentum conservation equation. I am sorry I misled you in post #10 about the angular momentum before the collision. I should not have written down without a diagram.

 

[TSny]

So it's easier to consider
$I_{TOT} = \frac{ML^2}{12} + My_{CM}^2 + m_1 (\frac{L}{2} - y_{CM})^2 + m_2 (\frac{-L}{2} + y_{CM})^2$
and:
$m_1 v_0 \frac{L}{2} - m_2 v_0 \frac{L}{2} = I_{TOT} \omega - (M + m_1 + m_2) v_f y_{CM}$

This looks correct except for the last term in the equation for $I_{TOT}$. I believe you corrected this later with the correct expression $m_2 (\frac{L}{2} + y_{CM})^2$

Note that you can use conservation of linear momentum to replace $(M+m_1+m_2)v_f$ in the last term of the angular momentum equation by $(M-m_1-m_2)v_0$. If you do this and then rearrange terms, you will get the same equation as you wrote in post #18.