Confusion in choosing an origin point for angular momentum

[Rikudo]

Homework Statement

A mass m1 hits a stick (it has mass m2) at its end with velocity v0. After the collision, the stick both slides horizontally with velocity v2' and rotates. Find the angular velocity!

Relevant Equations

Linear momentum and angular momentum

I am currently reading David Morin book and found this statement :

$\,\,\,\,\,\,\,\,$ "It is important to remember that you are free to choose your origin from the legal possibilities of fixed points or the CM"

Is it really alright to choose the center of a rigid body as an origin even though this body is fixed to a point? because when solving the problem above, I found that both methods yield different result.

Here is my attempt

First, we need to calculate its Linear momentum. (This will be useful when we set the fixed point as an origin)
$$m_1\,V_0\,=\,m_2\,V'_2$$

$1.$Now, If I use the center of the stick as the origin, the angular momentum will be like this :
$$m_1\,V_0\frac L 2 \,= \frac {m_2\,L^2\,\omega}{12}$$
$$\frac {6m_1V_0} {m_2\,L} = \omega \,\,\,\,\,\,\,\,\,\,. . . (1)$$

$2.$However, If I am to use the fixed point as the origin :
$$m_1\,V_0\, L \,= \frac {m_2\,L^2\,\omega}{3} + m_2V'_2\frac L 2$$
Substituting the Linear momentum equation into this will yields:
$$m_1\,V_0\, L = \frac {m_2\,L^2\,\omega}{3} + m_1V_0\frac L 2$$
$$m_1\,V_0\frac L 2 = \frac {m_2\,L^2\,\omega}{3} $$
$$\frac {3m_1V_0} {2m_2\,L} = \omega \,\,\,\,\,\,\,\,\,\,. . . (2)$$

As you can see, both methods yield different results. Though, it is really unlikely for the book to make a mistake. So, are there any concepts that I did wrong?

 

[PeroK]

$1.$Now, If I use the center of the stick as the origin, the angular momentum will be like this :
$$m_1\,V_0\frac L 2 \,= \frac {m_2\,L^2\,\omega}{12}$$
$$\frac {6m_1V_0} {m_2\,L} = \omega \,\,\,\,\,\,\,\,\,\,. . . (1)$$

What happens to the mass of the bullet?

The centre of the rod moves in a curved trajectory after the collision - that physical point on the rod is accelerating after the collision. So, you should really take the horizontal axis that runs through the point where the centre of the rod is initially. The motion of the rod relative to this axis is not a simple rotation, so your simple formula does not look right.

 

[Rikudo]

Ah...sorry, I forgot to said that the bullet stops moving after the collision.

Why the center of the rod is accelerating? Isn't the impulse only happen in the process of collision? (Which means that after the collision, there is no force acting on the rod, and therefore,it has no acceleration)

 

[PeroK]

Ah...sorry, I forgot to said that the bullet stops moving after the collision.

Why the center of the rod is accelerating? Isn't the impulse only happen in the process of collision? (Which means that after the collision, there is no force acting on the rod, and therefore,it has no acceleration)

If you draw a diagram of the motion of the rod after the collision, then clearly the centre of the rod moves in an upwards arc (which implies acceleration).

The only point on the rod that is moving inertially after the collisio is the point at the pivot, which is moving at constant velocity. Every other point on the rod is accelerating.

Edit: the pivot is being pulled backwards and forwards by the swinging rod, so it's not moving inertially either.

 

[Rikudo]

If you draw a diagram of the motion of the rod after the collision, then clearly the centre of the rod moves in an upwards arc (which implies acceleration).

Ah! You mean centripetal acceleration?

 

[PeroK]

Ah! You mean centripetal acceleration?

I meant simply acceleration as the rate of change of velocity. If you adopt an inertial reference frame in which the pivot is at rest, then indeed the rod executes circular motion about the pivot. That actually gives you another way to view the problem.

 

[Lnewqban]

... Is it really alright to choose the center of a rigid body as an origin even though this body is fixed to a point? because when solving the problem above, I found that both methods yield different result.
...
As you can see, both methods yield different results. Though, it is really unlikely for the book to make a mistake. So, are there any concepts that I did wrong?

I would use the actual pivot point to do the calculation.
If choosing the CM of the stick, I would need to compensate for the reverse movement of the slider towards the right, which would reduce the velocity of A towards the left.

In that case, the amount of momentum that the bullet transferred to linear movement would be reduced, while more of it would be transferred to the rotation of the stick (having less moment of inertia when rotating around its CM than around the pivot, since $I=(m_1+m_2)L^2$).

Imagine a light stick having a huge moment of inertia hit by same bullet; it would have a resistance to rotation greater than its resistance to slide forward.
Now imagine a heavy stick having a very small moment of inertia; it would quickly rotate about the pivot without sliding much.

The linear momentum of the bullet (m1) before impact is distributed in specific amounts of linear and angular momentum of the system (m1+m2), according to the specific inertia of the mechanism.

 

[bob012345]

I meant simply acceleration as the rate of change of velocity. If you adopt an inertial reference frame in which the pivot is at rest, then indeed the rod executes circular motion about the pivot. That actually gives you another way to view the problem.

This acceleration is only the direction since velocity is a vector but the angular speed of the rod does not change except if gravity is included and I think the problem statement is after the initial angular velocity immediately after the collision so we can ignore gravity.

I concur with @Lnewqban, the problem is very simple if looked at that way.

 

[haruspex]

Your mistake in method 2 is that you are "double dipping". When you write 1/3 instead of 1/12 you are taking rotation about the endpoint, so that includes the contribution to angular momentum of the linear momentum of the mass centre. And simply dropping the linear contribution doesn't fix it because you don't know the linear speed of the hinged end.
To use the 1/3 form you need to decompose the motion of the stick into a rotation about its endpoint plus a linear motion of that endpoint. If that linear motion is speed u to the left then the actual motion of the mass centre is $u+\omega L/2$ to the left. Equating that to $V_2'$ gives u in terms of the other variables.

 

[haruspex]

If you adopt an inertial reference frame in which the pivot is at rest, then indeed the rod executes circular motion about the pivot.

Not sure what you are suggesting here.
We are only concerned with the arbitrarily brief duration of the impact and the velocities immediately after. The accelerations, including that of the slider, are large and unknowable, and are not relevant. Even after that, I see no guarantee that the slider represents an inertial frame.

 

[bob012345]

Not sure what you are suggesting here.
We are only concerned with the arbitrarily brief duration of the impact and the velocities immediately after. The accelerations, including that of the slider, are large and unknowable, and are not relevant. Even after that, I see no guarantee that the slider represents an inertial frame.

If you look at the problem from the pivot at $A$ things are simple. At the instant where $t>0$ there is an angular velocity $w$ which we can compute even though nothing has actually moved yet. We don't care what happens after that.

 

[haruspex]

If you look at the problem from the pivot at $A$ things are simple. At the instant where $t>0$ there is an angular velocity $w$ which we can compute even though nothing has actually moved yet. We don't care what happens after that.

@PeroK described the pivot as an inertial frame. It is not. The pivot is mounted on a slider which will undergo a large acceleration during the impact.
You can use the original location of the pivot as the reference point, as @Rikudo did in method 2, but you need to allow for the velocity of the slider. See post #9.

 

[bob012345]

@PeroK described the pivot as an inertial frame. It is not. The pivot is mounted on a slider which will undergo a large acceleration during the impact.
You can use the original location of the pivot as the reference point, as @Rikudo did in method 2, but you need to allow for the velocity of the slider. See post #9.

The mass $m1$ has been called a bullet. Is it? Does it imbed into the stick or not? I don't think the problem statement is clear at all and that has to be clear before a solution is reached.

 

[Rikudo]

Does it imbed into the stick or not?

No, it is not

 

[bob012345]

No, it is not

Do you have final answer yet? If so what is it?

 

[Rikudo]

Do you have final answer yet? If so what is it?

No,sorry. I'm still trying to understand the concepts that you guys told me

 

[Rikudo]

I would need to compensate for the reverse movement of the slider towards the right, which would reduce the velocity of A towards the left.

what do you mean by "reverse movement"?

 

[Rikudo]

Your mistake in method 2 is that you are "double dipping". When you write 1/3 instead of 1/12 you are taking rotation about the endpoint, so that includes the contribution to angular momentum of the linear momentum of the mass centre. And simply dropping the linear contribution doesn't fix it because you don't know the linear speed of the hinged end.
To use the 1/3 form you need to decompose the motion of the stick into a rotation about its endpoint plus a linear motion of that endpoint. If that linear motion is speed u to the left then the actual motion of the mass centre is $u+\omega L/2$ to the left. Equating that to $V_2'$ gives u in terms of the other variables.

Alright. So, firstly, let us assume that V2' is the translational velocity of the slider (the stick end point). So:
$$m_1\,V_0\, L \,= \frac {m_2\,L^2\,\omega}{3} + m_2(V2' + \omega L)\frac L 2$$
Is this true?

 

[bob012345]

No,sorry. I'm still trying to understand the concepts that you guys told me

It would help to have the exact problem statement as you were given it.

 

[haruspex]

No, it is not

If it does not imbed into the stick then in order to solve the problem correctly we either need to know what its subsequent motion is or have to assume its mass is sufficiently small compared to that of the stick that we can ignore it thereafter.

 

[Rikudo]

If it does not imbed into the stick then in order to solve the problem correctly we either need to know what its subsequent motion is or have to assume its mass is sufficiently small compared to that of the stick that we can ignore it thereafter.

Let's assume that it is sufficiently small so that we can ignore it after the collision.

However, do this information has anything to do with our calculation?
At least for me, it does not matter, since the the bullet velocity after the collision is 0

 

[bob012345]

Homework Statement:: A mass m1 hits a stick (it has mass m2) at its end with velocity v0. After the collision, the stick both slides horizontally with velocity v2' and rotates. Find the angular velocity!
Relevant Equations:: Linear momentum and angular momentum

This is confusing. We are given v0. Are we given v2' or is it an unknown? This statement seems like v2' is given and we just need to compute $w$.

 

[Rikudo]

This is confusing. We are given v0. Are we given v2' or is it an unknown? The statement seems like v2' is given and we just need to compute $w$.

$V'_2$ is unknown

 

[bob012345]

$V'_2$ is a given value. so, we only need to find the angular velocity

Then that should be straight forward. You know the total angular momentum and that will be conserved. If you pick the right point all the complications should go away.

 

[Rikudo]

Hm. that is true... but I want to explore other possibilities in solving the problem

 

[bob012345]

Hm. that is true... but I want to explore other possibilities in solving the problem

Get the correct answer the most straightforward way first then explore other ways. How do you interpret the statement that the stick has a linear velocity v2? Is that the velocity of the center of mass, of the slider or of the end that is hanging?

 

[haruspex]

Alright. So, firstly, let us assume that V2' is the translational velocity of the slider (the stick end point). So:
$$m_1\,V_0\, L \,= \frac {m_2\,L^2\,\omega}{3} + m_2(V2' + \omega L)\frac L 2$$
Is this true?

No, you misunderstand me.
If we decompose the motion of the stick into a rotation about an endpoint plus a linear velocity then that will be the linear velocity of that endpoint.
I'm going to call the impulse from the bullet p, the slider's velocity u and the stick centre's velocity v.
$mv=p$
$Lp=\frac 13mL^2\omega+\frac L2mu$
(So $p=\frac 13mL\omega+\frac 12mu$)
$v=u+\frac L2\omega$
Whence $u=-\frac 13L\omega$. Note the sign. This is the "reverse movement " of the slider, i.e. it moves to the right.
And $\omega=\frac{6p}{mL}$, as in method 1.

(The m's above should all be $m_2$)

 

[haruspex]

$V'_2$ is a given value. so, we only need to find the angular velocity

That puts a different complexion on it.
If we are to take that as a given then the fate of the bullet is implied.
So instead, just take it as an impulse, p, and otherwise ignore the bullet completely. Use $p=m_2v_2'$.

 

[bob012345]

That puts a different complexion on it.
If we are to take that as a given then the fate of the bullet is implied.
So instead, just take it as an impulse, p, and otherwise ignore the bullet completely. Use $p=m_2v_2'$.

We need to know what v2 is the velocity of such as the CM, the slider or the end.

 

[Rikudo]

$Lp=\frac 13mL^2\omega+\frac L2mu$

Why not $Lp=\frac 13mL^2\omega+\frac L2mv$ ?(since angular momentum is the summation of $I \omega$ ,and linear momentum of the center of mass times the distance, in this case $mv \frac L 2$)

 

[Delta2]

I am not good in problems involving angular momentum (because it wasn't included in our physics cirruculum when I was giving exams for the universities, 30 years ago, and I entered in the mathematics department), but reading this thread I want to make a question to all:
The quantity $$\int_0^L \rho\omega x dx$$ is counted as additional linear momentum (in addition to $m_2V_2'$) when we make the linear momentum equation? ($\rho=\frac{m_2}{L}$).

 

[Rikudo]

We need to know what v2 is the velocity of such as the CM, the slider or the end.

you mean $V'_2$? It is the slider's velocity

 

[haruspex]

you mean $V'_2$? It is the slider's velocity

That is not clear from post #1. Are you sure?

 

[haruspex]

Why not $Lp=\frac 13mL^2\omega+\frac L2mv$ ?(since angular momentum is the summation of $I \omega$ ,and linear momentum of the center of mass times the distance, in this case $mv \frac L 2$)

You have to decide how to decompose the motion of the stick into linear and rotational components. You can pick any point and consider the total motion as the sum of the linear motion of that plus the rotation of the stick about it.
To use the 1/3 form of the MoI you are necessarily taking rotation as being about an endpoint, so the linear component must be the motion of that endpoint.

 

[Rikudo]

My bad. I didn't write the question clearly
Alright, let me explain the question with a little bit more details.

there is a stick with mass $m_2$hanging on a slider. Initially, the rod is not moving at all. After that, a mass $m_1$ hits the rod with velocity $V_0$. As a result, the m1 stopped moving, while the rod's slider translating with velocity $V_2'$.
Note : the stick is translating and rotating after the collision

Does this help you understanding the question more?