Inelastic collision of two squares

[billtodd]

Homework Statement

Two squares with mass $m$ and velocity $v_0$ move to one another, they do an Inelastic collision as seen in the pic. After the collision both squares get attached to one another with a length of $\alpha \ell$.
Where $\alpha$ is between 0 and 1.
They ask to find the speed of the centre of mass of the system of squares and angular velocity around the centre of mass after the collision. They ask what will happen when $\alpha \to 1$?

Relevant Equations

Angular momentum conservation (perhaps?).

I know that the speed of the centre of mass is $v_{cm}=(mv_0+mv_0)/(2m)=v_0$.

But I don't know how to proceed from here with the angular speed around the centre of mass of the system.

Any help will be appreciated.

 

[kuruman]

I know that the speed of the centre of mass is $v_{cm}=(mv_0+mv_0)/(2m)=v_0$.

How do you know that? It looks like you are forgetting that velocity is a vector.

 

[erobz]

Homework Statement: Two squares with mass $m$ and velocity $v_0$ move to one another, they do an Inelastic collision as seen in the pic. After the collision both squares get attached to one another with a length of $\alpha \ell$.
Where $\alpha$ is between 0 and 1.
They ask to find the speed of the centre of mass of the system of squares and angular velocity around the centre of mass after the collision. They ask what will happen when $\alpha \to 1$?
Relevant Equations: Angular momentum conservation (perhaps?).

I know that the speed of the centre of mass is $v_{cm}=(mv_0+mv_0)/(2m)=v_0$.

But I don't know how to proceed from here with the angular speed around the centre of mass of the system.

Any help will be appreciated.

To start, I think you need to check you center of mass speed after this collision? It’s not $v_o$

 

[billtodd]

To start, I think you need to check you center of mass speed after this collision? It’s not $v_o$

Ah, yeah you are correct. is $v_0$ the speed of cm before the collision then?
After the collision we have $mv_0+mv_0= mv_1+mv_2$ $v_1=v_2$, so $v_1=2v_0$ and $v_1=v_{cm}$ after the collision, am I right or have I mucked around with my knowledge which is a lot rusty.
Thanks!

 

[billtodd]

No, I mucked around.

I didn't touch Newtonian Mechanics since 2007.
Reading too much advanced stuff in QM QFT and other stuff messed around with my basics.

 

[billtodd]

How do you know that? It looks like you are forgetting that velocity is a vector.

So it suppose to be zero then, i.e: $v_{cm}=(mv_0-mv_0)/(2m)$.
But then also the angular velocity is zero, because there's no motion after the collision, am I right or yet wrong again...

 

[kuruman]

The collision conserves linear and angular momentum. The total linear momentum before the collision is zero as you have found out. There is no linear motion of the CM after the collision. What about the angular momentum? There could be rotation about the stationary CM after the collision.

Something to try. Place a book flat on a table top. Give it a quick jolt at the middle of the long side. Then give it a quick jolt near the corner. What do you see? Here, each of the squares receives a jolt (impulse more correctly) near its corner when the collision occurs.

 

[erobz]

So it suppose to be zero then, i.e: $v_{cm}=(mv_0-mv_0)/(2m)$.
But then also the angular velocity is zero, because there's no motion after the collision, am I right or yet wrong again...

There will be angular velocity( angular momentum), but the center of mass won’t be translating.

 

[jbriggs444]

So it suppose to be zero then, i.e: $v_{cm}=(mv_0-mv_0)/(2m)$.
But then also the angular velocity is zero, because there's no motion after the collision, am I right or yet wrong again...

Do you know the formula for the angular momentum of a moving (but non-rotating) object about a chosen reference axis?

 

[billtodd]

The collision conserves linear and angular momentum. The total linear momentum before the collision is zero as you have found out. There is no linear motion of the CM after the collision. What about the angular momentum? There could be rotation about the stationary CM after the collision.

Something to try. Place a book flat on a table top. Give it a quick jolt at the middle of the long side. Then give it a quick jolt near the corner. What do you see? Here, each of the squares receives a jolt (impulse more correctly) near its corner when the collision occurs.

OK there's no translation.
The definition of angular momentum is $L=\vec{r}\times p$. Angular momentum is conserved then the derivative of $L$ wrt time is zero. I need perhaps some drawing if you have.

 

[jbriggs444]

OK there's no translation.
The definition of angular momentum is $L=\vec{r}\times p$.

Perfect. So you do know the formula. Now, you should understand as well that both $\vec{r}$ and $\vec{p}$ are vectors. The displacement ($\vec{r}$) of the one block from a reference axis at the center of mass is the opposite of the displacement of the other. The linear momenta ($\vec{p}$) of the two blocks are also opposite.

So the angular momenta of the two blocks. Are they equal? Or opposite? Or something else?

 

[kuruman]

OK there's no translation.
The definition of angular momentum is $L=\vec{r}\times p$. Angular momentum is conserved then the derivative of $L$ wrt time is zero. I need perhaps some drawing if you have.

It so happens that I have a drawing (see figure on the right). The angular momentum of the plane about the base of the tree has magnitude
$L=rp\sin\theta=mvr\sin\theta=mvh.$ Note that if the plane flies at constant altitude $h$ with constant speed $v$, the angular momentum does not change with respect to time, i.e. is conserved.

Here the total angular momentum of the squares is conserved before the collision for the reason I just explained. During the collision the squares exert equal and opposite torques on each other which means that the net torque on the two square system is zero. This conserves the total angular momentum of the two-square system.

 

[billtodd]

Perfect. So you do know the formula. Now, you should understand as well that both $\vec{r}$ and $\vec{p}$ are vectors. The displacement ($\vec{r}$) of the one block from a reference axis at the center of mass is the opposite of the displacement of the other. The linear momenta ($\vec{p}$) of the two blocks are also opposite.

So the angular momenta of the two blocks. Are they equal? Or opposite? Or something else?

The angular momentum of each block before collision is zero since there's no rotation before collision only translation movement.
So it should be something like this: $L=rp\sin \theta$, I think that $r=(1-\alpha)\ell$ and $\sin \theta=\theta$, so basically: $dL/dt=-rp\cos \theta \omega+dr/dt p \sin\theta=0$.

I really need to return to learning Mechanics.
Harmonic Oscillator I know by heart (a figure of speech it should by mind).

 

[Steve4Physics]

Hi @billtodd. A couple of general points...

The system (2 identical squares with equal magnitude, opposite direction, initial velocities) is symmetric. And there are no external forces acting. It follows that the system’s centre of mass must be initially stationary and remain stationary. It’s important (IMO) to understand this. No maths is needed when the underlying physics is understood.

An object does not need to be spinning to have angular momentum. For example, @kuruman ’s diagram in Post #12 shows a (non-spinning) plane with angular momentum L= mvh with respect to the base of the tree.

 

[kuruman]

Distance $r=(1-\alpha)\ell$ is the perpendicular distance between the centers of the squares. That does not change as the squares move before they collide. In order to write an expression for the angular momentum of the system of two squares, first you need to define a point of reference. This is the point where you put the tail of vector $r$ In the figure I posted the reference point is the base of the tree.

Here you have two masses with angular momentum so you need two vectors $r$ with their tails at a common reference point O that you must first choose. You can choose that point anywhere you want. The tip of each arrow, however. should be at the center of each square because you are calculating the angular momentum of the center of mass of each square about point O. Sometimes this is called the "orbital angular momentum".

Just to practice. In the figure I posted, imagine a second airplane flying at altitude $2h$ in the opposite direction with the same speed $v$. What is the angular momentum of the two-airplane system about the base of the tree?