Confusion in choosing an origin point for angular momentum

[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?

 

[haruspex]

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?

We've gone from not enough information to too much.
If I know the masses and initial states, where and how the bullet hits the rod, and what happens to it, I don't need to be told the speed of the slider.
The information that the bullet hits the rod square on at one end comes only from the diagram. Is it possible we should not assume it is right at the end?

 

[Rikudo]

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 $w$.

Ah! Really sorry!
$V'_2$ is unknown.
hehehe...
I already edited my answer on your question

 

[bob012345]

Ah! Really sorry!
$V'_2$ is unknown.
hehehe...
I already edited my answer on your question

Well, at least we now know it is the slider velocity. It seems sometimes there is not enough effort to understand the question before information overload on trying to help get the answer.

 

[Rikudo]

The information that the bullet hits the rod square on at one end comes only from the diagram. Is it possible we should not assume it is right at the end?

It collides at the end of the rod

 

[bob012345]

I assume the slider is considered massless?

 

[Rikudo]

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.

I see...so, I only need to take the translational velocity into account. Also, does translational velocity always located at the CM of a rigid body?

I assume the slider is considered massless?

Exactly!

 

[PeroK]

@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.

And, also, after the collision the motion of the slider is not inertial. I didn't think that through.

 

[Delta2]

Guys I am pretty much confused on this problem, but is this supposed to be solved using conservation of linear momentum, conservation of angular momentum and possibly conservation of kinetic energy?

 

[haruspex]

Guys I am pretty much confused on this problem, but is this supposed to be solved using conservation of linear momentum, conservation of angular momentum and possibly conservation of kinetic energy?

You can rule out the last one.

 

[Delta2]

You can rule out the last one.

Can you tell me why KE is not conserved (before and after collision)?

 

[haruspex]

I see...so, I only need to take the translational velocity into account. Also, does translational velocity always located at the CM of a rigid body?

That is not what I wrote.

 

[Rikudo]

That is not what I wrote.

I meant that : $mv \frac L 2$ is not valid because $v$ is the sum of the tangential velocity and translational velocity.(When calculating linear momentum, I only need to multiply translational velocity and its mass)

Perhaps, I misinterpret your reply. Could you please explain at which part that I misinterpret your reply?

 

[Delta2]

(When calculating linear momentum, I only need to multiply translational velocity and its mass)

I think this hold only if the rotation is around the CM, but here the rotation is around one end point so I think the total linear momentum is $mv+m\omega L/2$ where $v$ the translational velocity and $\omega$ the angular velocity.

 

[haruspex]

I meant that : $mv \frac L 2$ is not valid because $v$ is the sum of the tangential velocity and translational velocity.(When calculating linear momentum, I only need to multiply translational velocity and its mass)

Perhaps, I misinterpret your reply. Could you please explain at which part that I misinterpret your reply?

The translational velocity depends on your reference point.
Consider a rod length L rotating at rate w about one endpoint. If that endpoint is your reference then all the motion is rotation. If you take the instantaneous location of the rod's centre as reference then rotation is the same but the translational velocity is wL/2. If you take the instantaneous location of the other endpoint then the translational velocity is wL.

 

[bob012345]

Can you tell me why KE is not conserved (before and after collision)?

I think because the slider would do work on the rod because the motion is constrained.

 

[bob012345]

It might help to work out the problem where the rod is not connected to a slider first as a reference answer then ask how the slider changes things. I believe there is a key insight to this problem that makes it simple. Very simple for both cases. Hint: pick any point along a line going through the slider in the direction it slides from -∞ to +∞ and ask what is the angular momentum of $m_1$ around that point?

 

[haruspex]

I think because the slider would do work on the rod because the motion is constrained.

No, it is nothing to do with the slider. There is no friction, so the force the slider would exert is normal to its motion.

What energy is lost is in the impact. If it were elastic then the relative masses of bullet and rod would dictate their relative shares of KE afterwards. Not sure of the details in this case without working through the algebra, but for point masses the relative velocity reverses. So in principle you could figure out how much KE is lost in terms of the given variables. And with the right values of those variables it could be zero.

 

[bob012345]

No, it is nothing to do with the slider. There is no friction, so the force the slider would exert is normal to its motion.

What energy is lost is in the impact. If it were elastic then the relative masses of bullet and rod would dictate their relative shares of KE afterwards. Not sure of the details in this case without working through the algebra, but for point masses the relative velocity reverses. So in principle you could figure out how much KE is lost in terms of the given variables. And with the right values of those variables it could be zero.

Ok but this may be an elastic collision since the end velocity of $m_1$ is zero yet it is not imbedded in the rod.

 

[haruspex]

Ok but this may be an elastic collision since the end velocity of $m_1$ is zero yet it is not imbedded in the rod.

Yes, that's what I wrote: " with the right values of those variables [the lost KE] could be zero."
Looks like it would be mass of rod = 4 x mass of bullet.

 

[bob012345]

Yes, that's what I wrote: " with the right values of those variables [the lost KE] could be zero."
Looks like it would be mass of rod = 4 x mass of bullet.

Just for fun, consider the subsequent motion of the rod. Obviously it will oscillate and overall move left. Assuming no other loss mechanisms we should be able to write the equations of motion.