Find speed of CoM after collision between ball and "square structure"

[songoku]

Homework Statement

Please see below

Relevant Equations

$\bar x=\frac{1}{M}\int_{a}^{b}x\lambda(x)dx$
$L=I\omega$
$L=r\times p$
Conservation of linear momentum
Conservation of angular momentum

(a)
$$\bar x=\frac{\int_{0}^{L} \frac{\alpha_o}{L}x^2dx}{\int_{0}^{L}\frac{\alpha_o}{L}xdx}$$
$$=\frac{2}{3}L$$

(b)
$$x=\frac{x_1+x_2}{2}=\frac{1}{6}L$$

$$y=\frac{y_1+y_2}{2}=\frac{1}{3}L$$

(c) I am not sure about this part. Do I need to divide the conservation of momentum into two directions?

In x-direction:
$$M.v_o\sin\theta=3M.v_{f,CM}x$$
$$v_{f,CM}x=\frac{2\sqrt{5}}{15}v_o$$

In y-direction:
$$M.v_o\cos\theta=3M.v_{f,CM}y$$
$$v_{f,CM}y=\frac{\sqrt{5}}{15}v_o$$

So:
$$v_{f,CM}=\sqrt{(v_{f,CM}x)^2+(v_{f,CM}y)^2}$$
$$=\frac{1}{3}v_o$$

Is that correct?

Thanks

 

[kuruman]

(c) I am not sure about this part. Do I need to divide the conservation of momentum into two directions?

Conservation of what kind of momentum in two directions?
The linear momentum of the CM of the system consisting of the two rod structure plus the particle is initially only in the x-direction. Momentum conservation means that it is the same after the collision, i.e. still only in the x-direction.

There is another kind of momentum that is conserved through the collision. What kind might that be? Hint: It's in a direction perpendicular to the xy-plane.

 

[haruspex]

Momentum conservation means that it is the same after the collision, i.e. still only in the x-direction.

Except that there will be a vertical impulse from the table.

 

[songoku]

Except that there will be a vertical impulse from the table.

Why is there vertical impulse from the table?

Conservation of what kind of momentum in two directions?
The linear momentum of the CM of the system consisting of the two rod structure plus the particle is initially only in the x-direction. Momentum conservation means that it is the same after the collision, i.e. still only in the x-direction.

Conservation of linear momentum, because I thought there is no net external force acting on the system (particle and L-rod) so the linear momentum should be conserved but based on hint from @haruspex there would be vertical impulse so linear momentum won't be conserved.

There is another kind of momentum that is conserved through the collision. What kind might that be? Hint: It's in a direction perpendicular to the xy-plane.

I suppose conservation of angular momentum can be applied to point (0, 0).

I think maybe during the collision there will be force acting on L-rod to the right, causing it to rotate counterclockwise.

$$L_{initial}=L_{final} ~\text{at origin}$$
$$0=r\times p+I\omega$$

Am I even on the right direction?

Thanks

 

[kuruman]

Except that there will be a vertical impulse from the table.

I read the problem to mean that the plane of the L-shaped structure is parallel to the (frictionless) table. I interpreted the figure to be a top view of the structure resting on the table. The statement "The square structure then rests along the x-axis on a frictionless, horizontal table" says nothing about the y-direction being along the vertical.

I think maybe during the collision there will be force acting on L-rod to the right, causing it to rotate counterclockwise.

You mean that there is a torque exerted by the particle on the L-shaped structure to cause the counterclockwise rotation. Where do you think the axis of this rotation will be?

 

[songoku]

You mean that there is a torque exerted by the particle on the L-shaped structure to cause the counterclockwise rotation. Where do you think the axis of this rotation will be?

I think the particle will rotate about the center of mass of the whole system.

I already have the CM of the L-rod, which is $\left(\frac{1}{6}L, \frac{1}{3}L\right)$. Adding the particle of mass $M$ will change the CM of the system to $\left(\frac{1}{18}L, \frac{1}{9}L\right)$

The final CM is supposed to be the axis of rotation

 

[haruspex]

I read the problem to mean that the plane of the L-shaped structure is parallel to the (frictionless) table. I interpreted the figure to be a top view of the structure resting on the table.

You are probably right. I read "rests along the x axis" as meaning it only rests on that. Otherwise, why not say it rests along the x and y axes?
Also, if it is lying flat on the table, doesn’t question c become trivial (and @songoku's answer correct, but done very much the hard way)?

 

[songoku]

Also, if it is lying flat on the table, doesn’t question c become trivial (and @songoku's answer correct, but done very much the hard way)?

I just realized I got the same answer by doing simple calculation:
$$M.V_o=3M.v_{f,CM}$$
$$v_{f,CM}=\frac{1}{3}v_o$$

My first thought was I need to consider the collision where the CM of particle and CM of L-rod are in one line, that's why I divide the working in two directions but apparently it is not needed. So conservation of momentum can be applied even though the CM of two objects are not in a straight line?

And if let say the case is the plane of L-rod is perpendicular to the table, why is there vertical impulse from the table?

I imagine the L-rod will still rotate counterclockwise due to the force exerted by the particle during collision. Is the rotation causing the vertical impulse?

 

[haruspex]

Is the rotation causing the vertical impulse?

Yes. If this were in the absence of gravity and no table then the rotation would cause the junction of the two rods to move into the negative y half of the plane. To avoid being penetrated, the table exerts a vertically upward impulse.

 

[songoku]

Yes. If this were in the absence of gravity and no table then the rotation would cause the junction of the two rods to move into the negative y half of the plane. To avoid being penetrated, the table exerts a vertically upward impulse.

If there is no particle hitting the L-rod, the table also exerts a vertical force on the rod, which is normal force, and the magnitude is equal to the weight of the rod.

After being hit, is the vertical force from the table equal to weight of particle + weight of L-rod?

 

[haruspex]

After being hit, is the vertical force from the table equal to weight of particle + weight of L-rod?

There is an impulse, just as the particle exerts an impulse on the rods. Don’t think of it as a force since that will be of unknown magnitude and duration. In the idealisation it is an unbounded force acting for an infinitesimal time. Consequently, gravity is irrelevant during the impulse.

 

[songoku]

There is an impulse, just as the particle exerts an impulse on the rods. Don’t think of it as a force since that will be of unknown magnitude and duration. In the idealisation it is an unbounded force acting for an infinitesimal time. Consequently, gravity is irrelevant during the impulse.

Is my method in post #4 correct?

 

[kuruman]

You are probably right. I read "rests along the x axis" as meaning it only rests on that. Otherwise, why not say it rests along the x and y axes?

Good question.

Also, if it is lying flat on the table, doesn’t question c become trivial . . .

It does for the expert but not for the novice. I think there is educational value here. In linear moment conservation problems the autopilot approach is to write $$\mathbf P_{\text{before}}=\mathbf P_{\text{after}}$$, find expressions for each side of the equation in terms of the velocities of the moving parts and solve for what is asked. The origin $$\Delta (\mathbf V_{\text{CM}}) =0$$ of this equation is forgotten as we can see from @songoku's convoluted, but correct, answer for the velocity of the CM.

To @songoku:
Linear momentum conservation means that the the linear velocity of the CM of the three-mass isolated system does not change: if you know it at one point in time, you know it at all points in time. Here, before the collision, you have mass $M$ moving with velocity $\mathbf v_0$ and two masses $M$ each at rest. The momentum of the CM before the collision is $$\mathbf V_{\text{CM}}=\frac{M \mathbf v_0+M*0+M*0}{M+M+M}=\frac{1}{3}\mathbf v_0.$$ Since this velocity doesn't change, it has the same value after the collision. It's as simple as that.

 

[kuruman]

Is my method in post #4 correct?

The method to reach what goal? Please specify your goal and what the symbols in the equations stand for.

 

[Lnewqban]

My first thought was I need to consider the collision where the CM of particle and CM of L-rod are in one line,

But the movement direction of the colliding mass is not in line with the CM of the pre-collision L-shaped rod.

Because of the above, the non-elastic collision should induce simultaneous rotation and translation of the newly formed three-mass system respect to the flat horizontal frictionless surface.

 

[kuruman]

But the movement direction of the colliding mass is not in line with the CM of the pre-collision L-shaped rod.

It doesn't have to be. Since there is no external torque on the L-shaped structure + particle system, the angular momentum about any point is conserved. In this problem, one would choose the CM of the system as the reference for angular momentum because this simplifies the conservation equation to $$Mv_0~d=I\omega$$ where $~d=~$distance of the CM to the x-axis; $I=~$moment of inertia of the composite object after the collision about its CM. In this form, there is only orbital angular momentum before the collision and only spin angular momentum after the collision. OP's choice was to start with zero angular momentum effectively moving the orbital part from the left hand side to the right hand side.

 

[songoku]

The method to reach what goal? Please specify your goal and what the symbols in the equations stand for.

I want to find the speed of CM of the system $(v_{f,CM})$ for the case considered by @haruspex scenario.

The coordinate of CM of L-rod is $\left(\frac{1}{6}L,\frac{1}{3}L\right)$ and after adding the particle, the CM becomes $\left(\frac{1}{18}L,\frac{1}{9}L\right)$

I am taking $\left(\frac{1}{18}L,\frac{1}{9}L\right)$ as the location of axis of rotation.

Consider the conservation of angular momentum:
$$\text{total initial angular momentum}=\text{total final angular momentum}$$
$$M.v_0.\frac{1}{9}L=(I_{particle}+I_{L-rod})\omega...(1)$$

Next, my idea is:
a) Find the moment of inertia of one rod using $I=\int_{0}^{L}r^2 dm$, taking the axis of rotation at origin, then using parallel axis theorem, I move the axis of rotation to $\left(\frac{1}{18}L,\frac{1}{9}L\right)$

b) Solve equation (1) for $\omega$

c) Use conservation of kinetic energy to find $v_{f, CM}$
$$\frac{1}{2}M(v_0)^2=\frac{1}{2}.3M.(v_{f,CM})^2+\frac{1}{2}.(I_{particle}+I_{L-rod})\omega^2$$

Is my approach correct?

 

[erobz]

c) Use conservation of kinetic energy to find $v_{f, CM}$
$$\frac{1}{2}M(v_0)^2=\frac{1}{2}.3M.(v_{f,CM})^2+\frac{1}{2}.(I_{particle}+I_{L-rod})\omega^2$$

Is my approach correct?

Isn't the ball sticking to the "L" structure? Meaning it experiences a certain amount of plastic deformation (perfectly inelastic collision)?

BTW does anyone see the optical illusion with the rods? To me it appears as though the lines are not everywhere parallel because of that gradient.

 

[haruspex]

I am taking $\left(\frac{1}{18}L,\frac{1}{9}L\right)$ as the location of axis of rotation.

That would not produce the correct motion for the point where the rods meet. We know that must move horizontally.
I would take the motion as the sum of the linear motion of that point and a rotation about it.

Consider the conservation of angular momentum:

With angular motion you need to specify the axis you are using. For the angular momentum to be conserved there must be no external impulse about that axis.
And to avoid pitfalls, it is best to choose an axis fixed in space, not fixed in the moving body.
Since we expect there to be an impulse from the table, that means you should choose the point (0,0).

c) Use conservation of kinetic energy

As @erobz notes, KE is not conserved here.

 

[kuruman]

BTW does anyone see the optical illusion with the rods? To me it appears as though the lines are not everywhere parallel because of that gradient.

The look parallel to me.

 

[erobz]

The look parallel to me.

View attachment 354172

I meant with the image of a single rod. The rod appears like it gets wider at the white portion, and is kind of curved in the transition between black and gray. I know the lines are parallel after putting a ruler up to the screen, but I think my brain/eyes system are not sure of it.

Its not related to the problem, just seeing if others see the apparent optical illusion.

 

[haruspex]

I meant with the image of a single rod. The rod appears like it gets wider at the white portion, and is kind of curved in the transition between black and gray.

Yes, I see that. I believe it is an example of "Helmholtz irradiation".

 

[kuruman]

Yes, I see that. I believe it is an example of "Helmholtz irradiation".

Oh that. My wife calls it the "you-look-slimmer-in-dark-clothes" illusion.

 

[Lnewqban]

In this form, there is only orbital angular momentum before the collision and only spin angular momentum after the collision.

Sorry @kuruman, I don't understand the concepts of orbital angular momentum and spin angular momentum.
Could you please explain those a little more?

 

[songoku]

That would not produce the correct motion for the point where the rods meet. We know that must move horizontally.
I would take the motion as the sum of the linear motion of that point and a rotation about it.

With angular motion you need to specify the axis you are using. For the angular momentum to be conserved there must be no external impulse about that axis.
And to avoid pitfalls, it is best to choose an axis fixed in space, not fixed in the moving body.
Since we expect there to be an impulse from the table, that means you should choose the point (0,0).

As @erobz notes, KE is not conserved here.

Taking (0,0) as axis of rotation:
$$0=-3M.v_{f,CM}.\frac{1}{9}L+(I_{L-rod})\omega$$

Then how to proceed? I have 2 unknowns

 

[haruspex]

I don't understand the concepts of orbital angular momentum and spin angular momentum.

The rotational motion of a body wrt an axis can be thought of as made up of a rotation of its mass centre about the axis (orbital angular momentum $=m\vec r\times\vec v$) plus a rotation of the body about its mass centre (spin angular momentum $=I\vec \omega$).

Taking (0,0) as axis of rotation:
$$0=-3M.v_{f,CM}.\frac{1}{9}L+(I_{L-rod})\omega$$

The mass centre will have an X component and a Y component of velocity.
I would represent the motion just after impact as a velocity of the particle and an angular velocity. Since the particle still has no rotation around (0,0) we can leave that out and just consider the angular momentum of the rods about (0,0). It might be simpler to calculate for each rod separately and sum them.

You also still have conservation of linear momentum in the X direction.

 

[kuruman]

Sorry @kuruman, I don't understand the concepts of orbital angular momentum and spin angular momentum.
Could you please explain those a little more?

Example
(a) Orbital angular momentum or angular momentum of the center of mass.
The Earth has orbital angular momentum about the center of the Sun given by $~\vec L_{\text{orb.}}=M_E \vec V_E \times \vec R_{SE}~$, where
$M_E=~$ mass of theEarth; $\vec V_E=$velocity of the Earth in its orbit; $\vec R_{SE}=$ Earth's position relative to the Sun. Note that one can write $\vec V_E=\vec{\Omega} \times \vec R_{SE}$ where $~\Omega =\dfrac{2\pi}{365.25}~\dfrac{\text{rad}}{\text{day}}~$ is the angular speed of the Earth in its orbit. Substituting, $$\vec L_{\text{orb.}}=M_E \vec V_E \times \vec R_{SE}=M_E (\vec{\Omega} \times \vec R_{SE}) \times \vec R_{SE}=M_ER_{SE}^2\vec {\Omega}.$$ The direction of $\vec \Omega$ is perpendicular to the plane of the orbit also known as the ecliptic. Note that one can also write $\vec L_{\text{orb.}}=I_{\text{orb.}}~\vec {\Omega}$ where $I_{\text{orb.}}=M_ER_{SE}^2$ is the moment of inertia of a point-mass Earth relative to the center of the Sun.

(b) Spin angular momentum or angular momentum about the center of mass.
As it orbits the Sun, the Earth also rotates about an axis that goes through its CM with an angular speed $~\omega =\dfrac{2\pi}{1}~\dfrac{\text{rad}}{\text{day}}.$ The spin angular momentum is $$\vec L_{\text{spin}}=I_{CM} \vec {\omega}.$$ Here $I_{\text{CM}}$ is the moment of inertia of the Earth about its axis of rotation. If we approximate the Earth as a smooth sphere, $I_{\text{CM}}=\frac{2}{5}M_E R_E^2.$ The direction $\vec {\omega}$ is along the Earth's axis of rotation which is tipped by 23.4° away from the perpendicular to the ecliptic.

Thus, the total angular momentum of the Earth is the vector sum of two terms, $$\vec L_{\text{total}}=M_ER_{SE}^2~\vec {\Omega}+\frac{2}{5}M_E R_E^2~\vec{\omega}.$$

 

[Lnewqban]

Thank you much, @haruspex and @kuruman

 

[songoku]

The mass centre will have an X component and a Y component of velocity.
I would represent the motion just after impact as a velocity of the particle and an angular velocity. Since the particle still has no rotation around (0,0) we can leave that out and just consider the angular momentum of the rods about (0,0). It might be simpler to calculate for each rod separately and sum them.

You also still have conservation of linear momentum in the X direction.

I am sorry for really late reply.

Conservation of linear momentum in the x-direction can be used to find the x-component of $v_{f,CM}$ which would be $\frac{1}{3}v_0$

I try to separate the L-rod into horizontal and vertical part. For the horizontal part of the L-rod and considering conservation of angular momentum at origin, the total angular momentum before collision between the particle and horizontal rod is zero. I am not sure about the total angular momentum after collision.

There would be spin angular momentum $I_{rod}~\omega$ but what about the orbital angular momentum?

Thanks

 

[haruspex]

There would be spin angular momentum $I_{rod}~\omega$ but what about the orbital angular momentum?

You know the horizontal velocity of the mass centre. What, instantaneously, is the horizontal velocity of the particle? Of the horizontal rod? What about the vertical rod?

 

[songoku]

You know the horizontal velocity of the mass centre. What, instantaneously, is the horizontal velocity of the particle? Of the horizontal rod? What about the vertical rod?

The instantaneous horizontal velocity of the particle, horizontal rod and vertical rod will also be $\frac{1}{3}v_o$. But if I take origin as the axis, wouldn't the orbital angular momentum after collision is zero since there is no perpendicular distance between the motion and origin?

Thanks

 

[haruspex]

The instantaneous horizontal velocity of the particle, horizontal rod

Yes

and vertical rod

No. The rotation will affect it. I should have made clear that I mean the horizontal velocity of the mass centre of each.

 

[songoku]

No. The rotation will affect it. I should have made clear that I mean the horizontal velocity of the mass centre of each.

Since the question does not give information about the vertical impulse by the table, how to incorporate the rotation to find the horizontal velocity of vertical rod after collision?

Is it correct to say the total angular momentum at origin after collision will be contributed by three things: spin angular momentum of vertical rod, spin angular momentum of horizontal rod and orbital angular momentum of center of mass of vertical rod, and sum all of them?

Thanks

 

[haruspex]

Here’s what looks to me to be the simplest approach…

Forget about the velocity you found for the mass centre of the system. It is not helpful.
Let the velocity after impact of the ball (and rod joint) be $u$, and the angular velocity of the rods be $\omega$.
For the linear momentum after impact, you have the horizontal velocities of the ball and horizontal rod as $u$. What is the horizontal velocity of the mass centre of the vertical rod? What equation does conservation give you?

For the angular momentum about the (static) point of impact, you have:

• 0 for the ball
• angular momentum of the horizontal rod about the point of impact due to $\omega$. There is no angular momentum due to its linear motion. What is its moment of inertia about that axis?
• angular momentum of the vertical rod about the point of impact

For angular momentum of the vertical rod about the point of impact consider its motion as the sum of the linear motion of its lower end ($u$) and its rotation ($\omega$).
What is its moment of inertia about that axis? What is its angular momentum due to

• its linear motion?
• its rotation?

 

[Lnewqban]

 

[haruspex]

View attachment 354921

View attachment 354928

Nice pictures, but it is not clear what point you are making. The second picture is somewhat irrelevant since we are only concerned with the motion the instant after impact. The frame has not moved yet.

 

[Lnewqban]

@haruspex
I believe that I was not trying to make any point when I posted those diagrams.
I made them for myself at first, while trying to understand this problem.
Perhaps those could help @songoku clarify the ideas he has shared in his posts 29, 31 and 33.

The exaggerated second diagram and the animated Judo movement show how the common center of mass must move up some for the pivoting point (x-y origin) to try to slide horizontally under it.

Assuming that the frame is rigid and that the horizontal and vertical members remain perpendicular to each other, it seems to me that the collision should simultaneously induce that vertical movement, a horizontal movement, and a rotation in the combined CoM.

 

[haruspex]

@haruspex
I believe that I was not trying to make any point when I posted those diagrams.
I made them for myself at first, while trying to understand this problem.
Perhaps those could help @songoku clarify the ideas he has shared in his posts 29, 31 and 33.

The exaggerated second diagram and the animated Judo movement show how the common center of mass must move up some for the pivoting point (x-y origin) to try to slide horizontally under it.

Assuming that the frame is rigid and that the horizontal and vertical members remain perpendicular to each other, it seems to me that the collision should simultaneously induce that vertical movement, a horizontal movement, and a rotation in the combined CoM.

Ok, but I worry that students might think you are implying an error in their working. Perhaps when just posting helpful diagrams you could mention the purpose?
Thanks.

 

[songoku]

Here’s what looks to me to be the simplest approach…

Forget about the velocity you found for the mass centre of the system. It is not helpful.
Let the velocity after impact of the ball (and rod joint) be $u$, and the angular velocity of the rods be $\omega$.
For the linear momentum after impact, you have the horizontal velocities of the ball and horizontal rod as $u$. What is the horizontal velocity of the mass centre of the vertical rod? What equation does conservation give you?

Since there is vertical impulse, I can't use conservation of linear momentum so the approach will be using conservation of angular momentum.

Total angular momentum before collision = total angular momentum after collision
$$0=-\frac{2}{3}L . v_{CM, vertical rod}+I_{L-rod}.\omega$$

For the angular momentum about the (static) point of impact, you have:

• 0 for the ball
• angular momentum of the horizontal rod about the point of impact due to $\omega$. There is no angular momentum due to its linear motion. What is its moment of inertia about that axis?

$$I_o=\int_0^L r^2 dm$$
$$=\int_0^L x^2 \frac{\alpha_o}{L}x ~ dx$$
$$=\frac{\alpha_o.L^3}{4}$$

Because the axis of rotation of horizontal rod is at darker region, by using parallel axis theorem:
$$I=I_o+ML^2$$
$$=\frac{\alpha_o.L^3}{4}+\int_0^L \frac{\alpha_o}{L}x~dx . L^2$$
$$=\frac{3}{4}\alpha_o.L^3$$

For angular momentum of the vertical rod about the point of impact consider its motion as the sum of the linear motion of its lower end ($u$) and its rotation ($\omega$).
What is its moment of inertia about that axis? What is its angular momentum due to

• its linear motion?
• its rotation?

Moment of inertia = $I_o = \frac{\alpha_o.L^3}{4}$

Angular momentum due to velocity $u$ is zero because there is no perpendicular distance to the origin.

Angular momentum due to its rotation is $I_o.\omega = \frac{\alpha_o.L^3}{4} . \omega$

 

[haruspex]

Since there is vertical impulse, I can't use conservation of linear momentum

You only can’t use conservation of the vertical component of the linear momentum. The horizontal component is conserved.

by using parallel axis theorem:

Your integral used distance from the lighter end of the rod, not distance from the mass centre of the rod. The $I_o$ you computed is the MoI about that end of the rod, so it gave the right value but for the wrong rod. No need for the parallel axis theorem.

For the horizontal rod, use the same integral but with the mass density as $\alpha_0(1-\frac xL)$.

Once you have both of those, as a check, you can use the parallel axis theorem in reverse to find the MoI about the mass centre in two ways:
$I_{heavy end}-M(\frac L3)^2=I_c =I_{light end}-M(\frac {2L}3)^2$

Angular momentum due to velocity $u$ is zero because there is no perpendicular distance to the origin.

No.
We are treating the motion of the vertical rod as it rotating about its lower end while moving as a whole linearly to the right at speed $u$. In this view, the whole rod has that linear velocity, so for finding the contribution to its angular momentum we need to take its mass as being at the rod's mass centre.

 

[songoku]

You only can’t use conservation of the vertical component of the linear momentum. The horizontal component is conserved.

Wouldn't the horizontal velocity of center of mass of vertical rod also be $u$?

For the horizontal rod, use the same integral but with the mass density as $\alpha_0(1-\frac xL)$.

Sorry I don't understand why the mass density is this for horizontal rod

 

[haruspex]

Wouldn't the horizontal velocity of center of mass of vertical rod also be $u$?

No. The lower end of the rod is moving to the right with velocity $u$, but the rod is rotating anticlockwise, so the mass centre’s horizontal velocity must be less, or even negative.
However, for the purpose of calculating its angular momentum, I am suggesting to consider the motion of that rod as a linear motion speed $u$ for the rod as a whole, plus an anticlockwise rotation rate $\omega$ about the lower end.
What is the angular momentum contribution from each?

Sorry I don't understand why the mass density is this for horizontal rod

Because the density is $\alpha_0$ at $x=0$ and $0$ at $x=L$.
Don’t confuse the $x$ in the diagram for question a) with the $x$ in the diagrams for the other parts.

 

[songoku]

No. The lower end of the rod is moving to the right with velocity $u$, but the rod is rotating anticlockwise, so the mass centre’s horizontal velocity must be less, or even negative.
However, for the purpose of calculating its angular momentum, I am suggesting to consider the motion of that rod as a linear motion speed $u$ for the rod as a whole, plus an anticlockwise rotation rate $\omega$ about the lower end.
What is the angular momentum contribution from each?

Angular momentum contribution from:
(1) linear motion = $\frac{2}{3} MuL$

(2) rotational motion = $\frac{1}{4} \alpha_o L^3 \omega$