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

• Nexus99
In summary: O)?The second term on the right represents angular momentum due to the motion of the CM of the system relative to the origin. So, this term is the difference of the moment of inertia of the system relative to the origin and the moment of inertia of the system relative to the CM.
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?

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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?

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}##.

TSny said:
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.

TSny
@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?

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Okpluto said:
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?

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

Okpluto said:
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?

Okpluto said:
@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.

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}##

Okpluto said:
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.

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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?

Okpluto said:
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.

kuruman said:
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.

haruspex said:
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.

haruspex said:
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))} ##

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Delta2
Okpluto said:
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.
Okpluto said:
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?

Okpluto said:
##\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.

kuruman said:
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 ##

Okpluto said:
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.

Okpluto said:
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.

## 1. What is the "rod hit by two masses" experiment?

The "rod hit by two masses" experiment is a physics experiment that involves a long rod suspended horizontally from one end. Two masses are then released from the other end of the rod, causing it to rotate.

## 2. What is the purpose of this experiment?

The purpose of this experiment is to demonstrate the principles of conservation of momentum and conservation of energy in a simple and visual way.

## 3. How does the mass of the two masses affect the outcome of the experiment?

The mass of the two masses will affect the velocity and momentum of the rod after the collision. The larger the masses, the greater the velocity and momentum of the rod will be.

## 4. What factors can affect the accuracy of this experiment?

The accuracy of this experiment can be affected by external factors such as air resistance, friction, and the precision of the equipment used. Human error in measuring and releasing the masses can also affect the results.

## 5. What are some real-world applications of this experiment?

This experiment can be used to explain the mechanics of collisions in sports, such as billiards or bowling. It can also be applied in engineering and design to understand the effects of collisions on structures and objects.

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