# Angular momentum for a bullet and a rod attached to a ceiling

• kirito
kirito
Homework Statement
a bullet hits a rod attached to a ceiling with a velocity v and then gets out with velocity v/2 ,
how much momentum went from rod to ceiling, and how much initial energy got lost ,
Relevant Equations
Li=Lf

I went and checked is energy conserved the answer is no since the bullet velocity is not the same as it was initially ,

I am not sure how to check Linear momentum conservation and angular momentum conservation , at first I assumed that the system linear momentum would be conserved during collision to say the assuming that what the bullet lost is gained by the rod I think I know now why that's not the case it is possibly because the rod is attached can't go up or down nor right or left so it will gain angular momentum but whatever linear momentum is there it would go to the ceiling?
if it was not attached though falling from the sky and the bullet hit it would the momentum be conserved during collision?

as for angular momentum being conserved I can understand that in the initial states the torque is zero on the rod and on the bullet ,so the angular momentum is conserved before or after , but then when I look at the rod after it changed location so that angle is not zero with the axis of rotation there is an torque on it , so I get confused regarding angular momentum Conservation
should I consider them as 2 different situations such that if I throw a bullet after it moved the angular momentum is not conserved but now for the first situation it is conserved

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kirito said:
I went and checked is energy conserved the answer is no since the bullet velocity is not the same as it was initially ,
The rod gains kinetic energy, so that is not the reason. You cannot assume energy is conserved. Perhaps it is and perhaps it isn't. You have to look for quantities that you know are conserved.
kirito said:
I am not sure how to check momentum conservation I know some is lost because they stated it goes to the ceiling ,
The rod is connected to the ceiling, which may apply a force with a horizontal and vertical components to the system. So, again you cannot assume that linear momentum is conserved.
kirito said:
the tricky part for me is identifying angular momentum as conserved or not , I know that there is an initial angular momentum in respect to the axis of rotation as for the rod there is no angular momentum in respect to that axis as the bullet goes out the rod will have changed location so that angle is not zero and there is an angular momentum for it , I don't know how to identify that its conserved
I think you are confused about angular momentum. The system has angular momentum about the pivot point, where the rod is attached to the ceiling. Note that the bullet's linear momentum also represents angular momentum about any point not on its initial line of motion. The bullet has initial angular momentum about the pivot point.
kirito said:
what I searched for was the forces I found Normal and gravitational force
The gravitational force on the bullet is ignored, because the collision happens very quickly and gravity would only have a small (negligible) affect on the collision.
kirito said:
and they are all perpendicular to the page so they are in the direction of the pivot and don't apply torque
but after it moves it starts applying torque are they 2 different situations such that if I throw a bullet after it moved the angular momentum is not conserved
The thing you are missing is that that angular momentum about the pivot point is conserved. This is because the force at the pivot point applies to torque about that point.

That's the key to these questions.

MatinSAR, TSny and kirito
kirito said:
I went and checked is energy conserved the answer is no since the bullet velocity is not the same as it was initially ,
So, can you calculate how much energy was lost? Is the rod moving after the collision?
kirito said:
I am not sure how to check momentum conservation I know some is lost because they stated it goes to the ceiling ,
If some is lost then it's not conserved. Can you calculate how much? Remember, momentum is a vector quantity.

kirito said:
the tricky part for me is identifying angular momentum as conserved or not , I know that there is an initial angular momentum in respect to the axis of rotation as for the rod there is no angular momentum in respect to that axis as the bullet goes out the rod will have changed location so that angle is not zero and there is an angular momentum for it , I don't know how to identify that its conserved
What is the condition that must be satisfied if angular momentum is conserved?

kirito said:
what I searched for was the forces I found Normal and gravitational force and they are all perpendicular to the page
No! All the forces are coplanar. They all lie in the plane of the page.

kirito said:
so they are in the direction of the pivot and don't apply torque
Correct.

MatinSAR and kirito
PeroK said:
The rod gains kinetic energy, so that is not the reason. You cannot assume energy is conserved. Perhaps it is and perhaps it isn't. You have to look for quantities that you know are conserved.

The rod is connected to the ceiling, which may apply a force with a horizontal and vertical components to the system. So, again you cannot assume that linear momentum is conserved.

I think you are confused about angular momentum. The system has angular momentum about the pivot point, where the rod is attached to the ceiling. Note that the bullet's linear momentum also represents angular momentum about any point not on its initial line of motion. The bullet has initial angular momentum about the pivot point.

The gravitational force on the bullet is ignored, because the collision happens very quickly and gravity would only have a small (negligible) affect on the collision.

The thing you are missing is that that angular momentum about the pivot point is conserved. This is because the force at the pivot point applies to torque about that point.

That's the key to these questions.

just to check what I am missing , in the case of energy I can't be sure if it is conserved or not I know that the bullet lost some kinetic energy and that the rod gained that energy but I don't know what is going on between the ceiling and the rod

The rod is connected to the ceiling, which may apply a force with a horizontal and vertical components to the system. So, again you cannot assume that linear momentum is conserved.
can you go a bit into the ceiling and rod relationship what type of force may it apply ? is it the force when the bullet contacts the rod at the other end and that's being applied by the pivot in the other direction since in a case where it is not attached a force is causing it to roatate around its Center of Mass ?

the gravitational force on the bullet is ignored, because the collision happens very quickly and gravity would only have a small (negligible) affect on the collision, what forces does one take into account that effects conservation during collision from what I learnt every collision that's quick has its linear momentum conserved because gravity is negligible and the forces the bodies apply on each other are equal but opposite but in this case there is another external force that comes into play on the whole system like the pivot force on rod

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kirito said:
The rod is connected to the ceiling, which may apply a force with a horizontal and vertical components to the system. So, again you cannot assume that linear momentum is conserved.
can you go a bit into the ceiling and rod relationship what type of force may it apply ? is it the force when the bullet contacts the rod at the other end and that's being applied on the other end ?
Imagine that the rod was only very loosely connected to the ceiling when it gets hit by a bullet! What would happen? The rod would get ripped out of position and probably go flying off to the right somewhere.

If that doesn't happen, then you know that there must have been a significant force at the pivot point to hold the rod in place. It's not necessary to calculate that force. But, you know that the linear momentum of the system is not conserved.
kirito said:
the gravitational force on the bullet is ignored, because the collision happens very quickly and gravity would only have a small (negligible) affect on the collision.
This problem analyses a collision that takes place over a short time, where gravity has no time to act significantly.

kirito
PeroK said:
Imagine that the rod was only very loosely connected to the ceiling when it gets hit by a bullet! What would happen? The rod would get ripped out of position and probably go flying off to the right somewhere.

If that doesn't happen, then you know that there must have been a significant force at the pivot point to hold the rod in place. It's not necessary to calculate that force. But, you know that the linear momentum of the system is not conserved.

This problem analyses a collision that takes place over a short time, where gravity has no time to act significantly.
thank you for the clear examples I just have one more point of confusion when I look at the rod after the collision and see if there is torque on it I see there is the gravitational force applying torque which kind of made me get a bit hesitant between saying rod and bullet system angular momentum is conserved

Mister T said:
What is the condition that must be satisfied if angular momentum is conserved?
the torque should be zero or the forces are in the direction of r ?

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kirito said:
thank you for the clear examples I just have one more point of confusion when I look at the rod after the collision and see if there is torque on it
After the collision, gravity will act through the centre of mass and the rod will swing back and forward as you would expect. But, that's beyond the time of the initial analysis.

MatinSAR and kirito
PeroK said:
After the collision, the only force on the rod is at the pivoy point. So, there is no torque about the pivot point.

Note that torque and angular momentum are always relative to some point or axis.
oh , as for gravity I assumed that gravity it is acting on rod and at the end of it, but I am wrong for assuming that because gravity is being applied on the Center of Mass and it can't cause an angular momentum since any force on Center of Mass can case a linear movement ,but thinking again it can cause it to rotate around the pivot so it seems I am not totally sure of why the only force is that of the pivot

kirito said:
it can't cause an angular momentum
It can't cause a change in angular momentum about the pivot point…
kirito said:
since any force on Center of Mass can case a linear movement
… but that is not the reason. The reason is the pivot point lies on the line of action of the gravitational force when the bullet strikes. So the gravitational force exerts no torque about the pivot during the collision.

MatinSAR
haruspex said:
It can't cause a change in angular momentum about the pivot point…

… but that is not the reason. The reason is the pivot point lies on the line of action of the gravitational force when the bullet strikes. So the gravitational force exerts no torque about the pivot during the collision.
so I don't look at its affect after the bullet strikes?

kirito said:
oh , as for gravity I assumed that gravity it is acting on rod and at the end of it, but I am wrong for assuming that because gravity is being applied on the Center of Mass and it can't cause an angular momentum since any force on Center of Mass can case a linear movement ,but thinking again it can cause it to rotate around the pivot so it seems I am not totally sure of why the only force is that of the pivot
The impact of the bullet will give the rod an initial angular velocity. After that gravity will take over. The centre of mass remains almost exactly below the pivot point for the duration of the collision.

Here's another example. If you hit a golf ball, it goes in a parabolic trajectory (approximately). But, the impact of the golf club on the ball is almost instantaneous. You don't have to take gravity into account to analyse the club striking the ball.

MatinSAR
kirito said:
so I don't look at its affect after the bullet strikes?
If you did, you would be analysing the free rotation of the rod about the pivot point after it's been given an initial impulse by the bullet.

PeroK said:
If you did, you would be analysing the free rotation of the rod about the pivot point after it's been given an initial impulse by the bullet.
I see that's what was confusing me so I am now looking at the force that will cause it to do some harmonic motion until it returns to rest

kirito said:
I know that the bullet lost some kinetic energy and that the rod gained that energy
That would be true only if the collision were perfectly elastic. In other words, the bullet loses some energy and the rod gains some energy, but the two amounts of energy are equal only if the collision is perfectly elastic (which it isn't).
kirito said:
the torque should be zero
The net torque.
kirito said:
or the forces are in the direction of r ?
That just means the torque due to those forces is zero (because the lever arm is zero).

MatinSAR
Mister T said:
That would be true only if the collision were perfectly elastic.

The net torque.

That just means the torque due to those forces is zero (because the lever arm is zero).
true I did make that assumption ,
is there any conditions other than the net torque being zero ? since I phrased the same condition differently

kirito said:
true I did make that assumption ,
is there any conditions other than the net torque being zero ? since I phrased the same condition differently
Angular momentum of a system is conserved when the net torque on the system is zero. You should be able to find a statement like this in the textbook section on conservation of angular momentum.

Mister T said:
Angular momentum of a system is conserved when the net torque on the system is zero. You should be able to find a statement like this in the textbook section on conservation of angular momentum.
I prefer "Angular momentum of a system about a given point is conserved when the net torque on the system about that point is zero."

weirdoguy and MatinSAR

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