How Does Bullet Collision Affect Hoop Velocity in Rotational Motion?

[coldblood]

Hi friends,
Please help me in solving this problem, I'll appreciate the help.

The problem is as:


https://fbcdn-sphotos-f-a.akamaihd.net/hphotos-ak-ash3/q71/s720x720/575434_1461727284054377_268945819_n.jpg

Attempt -

https://fbcdn-sphotos-h-a.akamaihd.net/hphotos-ak-ash4/1501673_1461727677387671_1024341158_n.jpg


Thank you all in advance.

 

[haruspex]

As with your problem 4, choose a fixed reference point for angular momentum.

 

[coldblood]

As with your problem 4, choose a fixed reference point for angular momentum.

haruspex: Can you please tell me why it is important to take any external reference point while conserving angular momentum?

 

[haruspex]

haruspex: Can you please tell me why it is important to take any external reference point while conserving angular momentum?

If your reference point can accelerate then that clearly has the potential to give a wrong result. Consider a bar floating in space, struck perpendicularly at one end. If we take the point of the bar that was struck as the reference point, the impulse has no moment about that point, so the bar should acquire no angular momentum about it. But from the perspective of that end of the bar, just after impact, the entire of the rest of the bar is rotating about it (in the opposite direction to the impulse).
If, instead, we use the fixed point in space where that end of the bar had been, we find that the part of the bar near that point acquires a moment one way, but the far end of the bar acquires an opposing moment, the two cancelling out.

 

[coldblood]

If your reference point can accelerate then that clearly has the potential to give a wrong result. Consider a bar floating in space, struck perpendicularly at one end. If we take the point of the bar that was struck as the reference point, the impulse has no moment about that point, so the bar should acquire no angular momentum about it. But from the perspective of that end of the bar, just after impact, the entire of the rest of the bar is rotating about it (in the opposite direction to the impulse).
If, instead, we use the fixed point in space where that end of the bar had been, we find that the part of the bar near that point acquires a moment one way, but the far end of the bar acquires an opposing moment, the two cancelling out.

Ok, first let me clear Problem - 4 and all the things behind that, then I'll again try to solve this problem. For any confusion beyond that, I'll again come to you. Thanks a lot in advance.

 

[coldblood]

Ok, first let me clear Problem - 4 and all the things behind that, then I'll again try to solve this problem. For any confusion beyond that, I'll again come to you. Thanks a lot in advance.

Please don't mind

 

[coldblood]

As with your problem 4, choose a fixed reference point for angular momentum.

https://fbcdn-sphotos-e-a.akamaihd.net/hphotos-ak-prn2/1480736_1462448153982290_1020254380_n.jpg
In this case if I take the reference point A, which is momentary at rest after collision,

Conserving angular momentum, mv(2R) = [2MR²(ring) + MR²(particle)] ω

Which gives the result, ω = 2v/3R

Still not the correct one.

 

[haruspex]

https://fbcdn-sphotos-e-a.akamaihd.net/hphotos-ak-prn2/1480736_1462448153982290_1020254380_n.jpg
In this case if I take the reference point A, which is momentary at rest after collision,

Looks like you mean that point of the hoop/loop. How can you be sure that any point of the hoop is momentarily at rest after the collision? Try taking a point fixed in the plane. I would try the point where the hoop's centre was when at rest.

 

[coldblood]

I would try the point where the hoop's centre was when at rest.

I think that point would be point C, the centre of line AB. And if i'll conserve angular momentum then post 1 again comes in front of me.

 

[Tanya Sharma]

Can you calculate the new Center of Mass of the system ,after the bullet gets embedded in the loop ?

You may consider any fixed point in the plane .Two good choices are 1) center of the loop 2) new position of the center of mass in the plane . Both will yield correct answer .

Another thing you need to remember is that - The angular momentum of the system(bullet+loop) about a fixed point,say C, is the sum of angular momentum of the CM about C + angular momentum of the system around the CM

 

[haruspex]

I think that point would be point C, the centre of line AB. And if i'll conserve angular momentum then post 1 again comes in front of me.

In the OP, you took moments about the centre of the hoop as it moved. Taking instead a fixed point, the angular momentum of the particle will be greater. It is not only rotating about the centre of the hoop, but also moving forwards with the hoop.

 

[coldblood]

Can you calculate the new Center of Mass of the system ,after the bullet gets embedded in the loop ?

Yes it will be at the distance R/2 from the centre of the loop.

 

[coldblood]

In the OP, you took moments about the centre of the hoop as it moved. Taking instead a fixed point, the angular momentum of the particle will be greater. It is not only rotating about the centre of the hoop, but also moving forwards with the hoop.

So, will this be the right approach?

conserving linear momentum,

m(v) = 2m v' => v' = v/2

conserving angular momentum about centre of the loop,

mv(R) = mR²(loop) + {mR²(particle)}. ω + (2m). (v/2).(R/2)(particle +loop com)
it gives the answer, ω = v/4R

 

[coldblood]

Can you calculate the new Center of Mass of the system ,after the bullet gets embedded in the loop ?

You may consider any fixed point in the plane .Two good choices are 1) center of the loop 2) new position of the center of mass in the plane . Both will yield correct answer .

Another thing you need to remember is that - The angular momentum of the system(bullet+loop) about a fixed point,say C, is the sum of angular momentum of the CM about C + angular momentum of the system around the CM

In the OP, you took moments about the centre of the hoop as it moved. Taking instead a fixed point, the angular momentum of the particle will be greater. It is not only rotating about the centre of the hoop, but also moving forwards with the hoop.

So, will this be the right approach?

conserving linear momentum,

m(v) = 2m v' => v' = v/2

conserving angular momentum about centre of the loop,

mv(R) = mR²(loop) + {mR²(particle)}. ω + (2m). (v/2).(R/2)(particle +loop com)
it gives the answer, ω = v/4R

 

[coldblood]

So, will this be the right approach?

conserving linear momentum,

m(v) = 2m v' => v' = v/2

conserving angular momentum about centre of the loop,

mv(R) = mR²(loop) + {mR²(particle)}. ω + (2m). (v/2).(R/2)(particle +loop com)
it gives the answer, ω = v/4R

Now I got where I was doing mistake.

conserving linear momentum,

m(v) = 2m v' => v' = v/2

conserving angular momentum about centre of the loop,

mv(R/2) = mR² + m(R²)/4(loop) + {mR²/4(particle)}. ω + (m). (v/2).(R/2) - (m). (v/2).(R/2)(particle +loop com)
it gives the answer, ω = v/3R

 

[Tanya Sharma]

Now I got where I was doing mistake.

conserving linear momentum,

m(v) = 2m v' => v' = v/2

conserving angular momentum about centre of the loop,

mv(R/2) = mR² + m(R²)/4(loop) + {mR²/4(particle)}. ω + (m). (v/2).(R/2) - (m). (v/2).(R/2)(particle +loop com)
it gives the answer, ω = v/3R

Why mv(R/2) in the LHS ? Shouldn't the angular momentum of the bullet about the center of the loop be mvR ?

 

[coldblood]

Why mv(R/2) in the LHS ? Shouldn't the angular momentum of the bullet about the center of the loop be mvR ?

I conserve A.m. about the center of mass of the system.

 

[haruspex]

I conserve A.m. about the center of mass of the system.

You can also do it about the point where the centre of the hoop started:
initial moment of particle = mvr
final moment of hoop = mr²ω
final linear velocity of system = v/2
final moment of particle = mr²ω + (mv/2)r
The particle has moment mr²ω in consequence of the hoop's rotation, and another (mv/2)r on account of the linear motion of the system.
mvr = (mv/2)r + 2mr²ω

 

[coldblood]

You can also do it about the point where the centre of the hoop started:
initial moment of particle = mvr
final moment of hoop = mr²ω
final linear velocity of system = v/2
final moment of particle = mr²ω + (mv/2)r
The particle has moment mr²ω in consequence of the hoop's rotation, and another (mv/2)r on account of the linear motion of the system.
mvr = (mv/2)r + 2mr²ω

But this will give the answer ω = v/ 4r, and answer is v/ 3r

 

[Tanya Sharma]

You can also do it about the point where the centre of the hoop started:
initial moment of particle = mvr
final moment of hoop = mr²ω
final linear velocity of system = v/2
final moment of particle = mr²ω + (mv/2)r
The particle has moment mr²ω in consequence of the hoop's rotation, and another (mv/2)r on account of the linear motion of the system.
mvr = (mv/2)r + 2mr²ω

You have used MI of the system about the center of the hoop ,whereas it should be MI about the new CM .

The item in red should mvr= 2m(v/2)(r/2) + (3/2)mr²ω

 

[haruspex]

You have used MI of the system about the center of the hoop ,whereas it should be MI about the new CM .

No, I was taking moments about the fixed point where the centre of the hoop had been. That is entirely valid. But I did make a mistake. I took the linear velocity of the hoop to be v/2, but it isn't.
If the linear velocity of the hoop after impact is v' then by conservation of linear momentum:
mv = mv' (hoop) + m(v'+ωr) (bullet)
And moments about the old centre of hoop is as I wrote:
mvr = m(r²ω + v'r) (bullet) + mr²ω (hoop)
Giving ω = v/3r.

 

[coldblood]

No, I was taking moments about the fixed point where the centre of the hoop had been. That is entirely valid. But I did make a mistake. I took the linear velocity of the hoop to be v/2, but it isn't.
If the linear velocity of the hoop after impact is v' then by conservation of linear momentum:
mv = mv' (hoop) + m(v'+ωr) (bullet)
And moments about the old centre of hoop is as I wrote:
mvr = m(r²ω + v'r) (bullet) + mr²ω (hoop)
Giving ω = v/3r.

That means when I was conserving linear momentum, it was giving the answer as v'(hoop + bullet) = v/2, which was wrong but giving the correct answer(coincidence). I took conservation of linear momentum in wrong manner. This is the best approach.
What do you say Tanya?

 

[Tanya Sharma]

That means when I was conserving linear momentum, it was giving the answer as v'(hoop + bullet) = v/2, which was wrong but giving the correct answer(coincidence). I took conservation of linear momentum in wrong manner. This is the best approach.
What do you say Tanya?

v' = v/2 is correct (v' is the velocity of center of mass of hoop+bullet after the collision) .

I think haruspex is using v' for the velocity of the hoop .So although there is a slight difference in the approach ,both are yielding correct results .

 

[haruspex]

v' = v/2 is correct (v' is the velocity of center of mass of hoop+bullet after the collision) .

I think haruspex is using v' for the velocity of the hoop .So although there is a slight difference in the approach ,both are yielding correct results .

That's right, my v' is different from that in the OP. Thanks.