Angular velocity of a loop after being struck by bullet

[brochesspro]

Homework Statement

A circular hoop of mass m and radius R rests flat on a horizontal frictionless surface. A bullet, also of mass m and moving with velocity v, strikes the hoop and gets embedded in it. The thickness of the hoop is much smaller than R. The angular velocity with which the system rotates after the bullet strikes the hoop is:

Relevant Equations

Conservation of Angular Momentum
Parallel axis theorem to find Moment of Inertia

First off, I do know how to solve this problem. We use the principle of conservation of angular momentum about the centre of mass of the system which comprises of the loop and the bullet to obtain option B. My doubt is, why do we just not use the principle about the centre of the loop? Where is the problem in that?

 

[haruspex]

Homework Statement: A circular hoop of mass m and radius R rests flat on a horizontal frictionless surface. A bullet, also of mass m and moving with velocity v, strikes the hoop and gets embedded in it. The thickness of the hoop is much smaller than R. The angular velocity with which the system rotates after the bullet strikes the hoop is:
Relevant Equations: Conservation of Angular Momentum
Parallel axis theorem to find Moment of Inertia

View attachment 337586
First off, I do know how to solve this problem. We use the principle of conservation of angular momentum about the centre of mass of the system which comprises of the loop and the bullet to obtain option B. My doubt is, why do we just not use the principle about the centre of the loop? Where is the problem in that?

In general, you can choose the centre of mass of the system or any fixed point in space.

 

[Orodruin]

In general, you can choose the centre of mass of the system or any fixed point in space.

The center of the hoop of course being neither of those.

 

[haruspex]

The center of the hoop of course being neither of those.

It was not clear to me how @brochesspro was defining that; could have meant the point where that is initially.

 

[brochesspro]

It was not clear to me how @brochesspro was defining that; could have meant the point where that is initially.

What do you mean? The y- co-ordinate of the CoM won't change, cuz there is no net force on the system, so all is fine, right? Since we only need the vertical distance from each individual object's centre of mass.

 

[Orodruin]

It was not clear to me how @brochesspro was defining that; could have meant the point where that is initially.

”Centre of the loop” to me sounds like the centre of the loop, not the CoM of the full system. Unless OP specifies with actual equations we can’t know for sure of course.

Edit: The OP also seems to make a distinction between the two …

 

[brochesspro]

”Centre of the loop” to me sounds like the centre of the loop, not the CoM of the full system.

It does.

Unless OP specifies with actual equations we can’t know for sure of course.

Equations for what?

 

[Orodruin]

Equations for what?

The equations you are actually writing down to model the evolution of the system. What we have so far is a verbal description

 

[brochesspro]

The equations you are actually writing down to model the evolution of the system. What we have so far is a verbal description

Ok, so here it is.

We conserve angular momentum about the centre of mass of the loop, since all forces are internal, there are no external torques and hence this principle can be applied. The point where the bullet strikes the loop lies at a distance R from our reference point.

Initial angular momentum is given as $MvR$. ie. before the bullet strikes the loop.

The moment of inertia of the system about the same point is given as $mR^2 + mR^2 = 2mR^2$.

So, final angular momentum is given as $mR^2 ω$.

Equating the two, we get $ω$ as $ω = \frac v {2R}$.

Which does not match the actual answer of $ω = \frac v {3R}$.

By the way, I have posted my approach in which I have a doubt, and not the correct one, I will post it if it is needed.

 

[jbriggs444]

This centre of mass lies at a distance $R$ from the centre between the centre of the loop and the point where the bullet strikes the loop.

Initial angular momentum is given as $MvR$. ie. before the bullet strikes the loop.

What point are you taking as your reference for the angular momentum calculation?

Is that point moving?

To be clear, you can do the analysis about any reference you choose. That reference can be in any state of motion that you choose. But you need to choose. And you need to make that choice clear so that we can follow along and understand your work.

It appears that you have chosen the initial position of the center of mass of the loop as your reference and chosen a standard of rest in which the loop begins stationary.

 

[erobz]

Ok, so here it is.

We conserve angular momentum about the centre of mass of the loop, since all forces are internal, there are no external torques and hence this principle can be applied. This centre of mass lies at a distance $R$ from the centre between the centre of the loop and the point where the bullet strikes the loop.

Initial angular momentum is given as $MvR$. ie. before the bullet strikes the loop.

The moment of inertia of the system about the same point is given as $mR^2 + mR^2 = 2mR^2$.

So, final angular momentum is given as $mR^2 ω$.

Equating the two, we get $ω$ as $ω = \frac v {2R}$.

Which does not match the actual answer of $ω = \frac v {3R}$.

By the way, I have posted my approach in which I have a doubt, and not the correct one, I will post it if it is needed.

It's most concise to work around the combined systems center of mass as the reference for conservation of angular momentum here. The center of mass of the thin ring (alone - without the embedded bullet) is not the axis of rotation of the system after impact.

 

[brochesspro]

The center of mass of the thin ring (alone) is not the axis of rotation of the system after impact.

How do we show this? And, can we use the same reference point and still get the correct answer? I have probably missed a few terms.

 

[erobz]

How do we show this?

The center of mass of the system is where the instant of impact? The masses of the bullet $m$ and the ring $m$.

And, can we use the same reference point and still get the correct answer? I have probably missed a few terms.

Yes, either way as long as you are careful about the accounting.

 

[Orodruin]

We conserve angular momentum about the centre of mass of the loop, since all forces are internal, there are no external torques and hence this principle can be applied. This centre of mass lies at a distance R from the centre between the centre of the loop and the point where the bullet strikes the loop.

Your language here is extremely ambiguous.
The center of mass of the loop is some distance away from the center of mass of the loop? That distance is the same as the radius of the loop? I find it basically impossible to parse what you want to say.

Initial angular momentum is given as MvR. ie. before the bullet strikes the loop.

Now you are using R as the orthogonal distance from the center of the loop to where the bullet hits. This is compatible with using the center of the loop but not the center of mass of the system as the reference.

If you are not using the com as the reference, then you are missing terms due to the linear momentum of the system.

 

[brochesspro]

The center of mass of the loop is some distance away from the center of mass of the loop? That distance is the same as the radius of the loop?

Sorry, I was thinking about the other method. let me correct it.

 

[PeroK]

Sorry, I was thinking about the other method. let me correct it.

I suggest trying to visualise the rotation of the loop and bullet after the collision. You can do this in the rest frame of the CoM, which must be moving inertially after the collision. Draw a few diagrams if necessary.

That also allows you to solve the problem by considering the rotation the bullet and the loop separately. That gives you a way to double-check your answer.

 

[brochesspro]

What point are you taking as your reference for the angular momentum calculation?

I have corrected my sentence in my procedure, please check it.

The center of mass of the system is where the instant of impact?

I did not get this.

If you are not using the com as the reference, then you are missing terms due to the linear momentum of the system.

How do we obtain the expression for that?

 

[erobz]

I did not get this.

I'm asking you to use the system center of mass as the reference.

 

[brochesspro]

I'm asking you to use the system center of mass as the reference.

View attachment 337597

I did that and got the correct answer, but that is not my question. My question is how to get the correct answer by taking the reference as the centre of the loop.

 

[PeroK]

I did that and got the correct answer, but that is not my question. My question is how to get the correct answer by taking the reference as the centre of the loop.

The centre of the loop is accelerating after the collision. That makes things tricky.

 

[erobz]

I did that and got the correct answer, but that is not my question. My question is how to get the correct answer by taking the reference as the centre of the loop.

I defer to the experts on that one. I always took it a face value that it can be computed about any point, but never tried to investigate how difficult that may be. I'm interested to see it too.

 

[Orodruin]

Consider a rigid body with masses $m_i$ that have time dependent positions $\vec x_i$. For such a body, the velocity field may be described by
$$
\vec v = \vec v_r + \vec \omega \times (\vec x - \vec x_r)
$$
where the $r$ subscript denotes a reference point. By definition, the angular momentum relative to the reference point is
$$
\vec L_r = \sum_i m_i (\vec x_i - \vec x_r) \times \vec v_i
= \sum_i m_i (\vec x_i - \vec x_r) \times(\vec v_r + \vec \omega \times (\vec x_i - \vec x_r)).
$$
We now use that $\sum_i m_i \vec x_i = M \vec x_{cm}$ defines the center of mass (with $M$ being the total mass):
$$
\vec L_r = M (\vec x_{cm} - \vec x_r) \times \vec v_r + I_r (\vec \omega)
$$
where $I_r$ is the moment of inertia tensor relative to $\vec x_r$. The first term is zero in the the following cases:

• $\vec v_r = 0$: The reference point is chosen such that it has zero instantaneous velocity.
• $\vec x_r = \vec x_{cm}$: The reference point is the center of mass.
• $(\vec x_{cm} - \vec x_r)\times \vec v_r = 0$: The offset from the CoM to the reference point is parallel to $\vec v_r$.

None of the above is applicable to the center of the hoop in your case so you cannot ignore the first term for the angular momentum after the collision.

 

[Orodruin]

Coming to the application to your problem:

You have two unknowns: The velocity $\vec v_r$ as well as the angular velocity $\vec\omega$. To solve for those you will need both the conservation of angular momentum as well as the conservation of linear momentum. Using the CoM as reference point allows you to solve for $\vec\omega$ without involving the equation for linear momentum (because the first term above does not appear).

If you insist on using the center of the loop as reference, the linear momentum of the loop will be $\vec v_r$ and that of the bullet $\vec v_r + \vec\omega \times(-R\vec e_y)$ where the y-direction is chosen in the up direction in your figure. We can also assume $\vec\omega = \omega\vec e_z$ since angular velocity is restricted to be orthogonal to the horizontal plane. The linear momentum conservation becomes:
$$
mv \vec e_x = 2 m\vec v_r + m\omega R \vec e_x
$$
and the angular momentum conservation
$$
mvR \vec e_z = 2m (-R\vec e_y/2)\times \vec v_r + I_r \omega \vec e_z
$$
(assuming I didn’t do something whacky with the arithmetics - on mobile)

From the linear equation:
$$
\vec v_r = \vec e_x (v-\omega R)/2
$$
The angular equation therefore turns into
$$
mvR = mR(v-\omega R)/2 + 2mR^2 \omega
\quad
\Longrightarrow \quad
mvR/2 = 3 mR^2\omega /2
$$
and therefore
$$
\omega = v/3R
$$

 

[brochesspro]

The centre of the loop is accelerating after the collision. That makes things tricky.

Why do you say so? Did it not just gain velocity because of the impulse acting on it by the bullet? There is no other force during that time.

 

[PeroK]

Why do you say so? Did it not just gain velocity because of the impulse acting on it by the bullet? There is no other force during that time.

If you are saying the the hoop moves inertially, then that is wrong. That's why previously I advised you to:

I suggest trying to visualise the rotation of the loop and bullet after the collision. You can do this in the rest frame of the CoM, which must be moving inertially after the collision. Draw a few diagrams if necessary.

 

[Orodruin]

Why do you say so? Did it not just gain velocity because of the impulse acting on it by the bullet? There is no other force during that time.

The center of mass moves inertially. The center of the loop will move in a circular motion around the center of mass because there is angular velocity.

 

[haruspex]

What do you mean? The y- co-ordinate of the CoM won't change, cuz there is no net force on the system, so all is fine, right? Since we only need the vertical distance from each individual object's centre of mass.

I wasn't referring to the CoM.
You wrote that you want to use the centre of the hoop as the axis to take angular momentum about. You can do that if you mean the fixed point in space where the centre of the hoop is initially. It will give the wrong answer if you mean it in a dynamic sense, i.e. relative to where the centre of the hoop is and how it is moving at each instant.

”Centre of the loop” to me sounds like the centre of the loop, not the CoM of the full system.

See above.

 

[Orodruin]

You can do that if you mean the fixed point in space where the centre of the hoop is initially.

You can, but not in the way OP did it. (See #9)
You need an additional linear term apart from the moment of inertia term and you need to involve the conservation of linear momentum as well. (See #22 and #23)

 

[haruspex]

You can, but not in the way OP did it. (See #9)

Agreed

You need an additional linear term apart from the moment of inertia term

I'd put it differently. The angular momentum of the bullet about the initial hoop centre position is $mv'r$ where $v'$ is the new speed of the bullet in the ground frame. That equals $mr^2\omega+mvr''$, where $v''$ is the linear speed of the hoop.
Post #9 omitted that second term.

and you need to involve the conservation of linear momentum as well. (See #22 and #23)

Sure, but I still see it as less effort than finding the CoM, as is often the case.

 

[Elj]

Ok, so here it is.

We conserve angular momentum about the centre of mass of the loop, since all forces are internal, there are no external torques and hence this principle can be applied. The point where the bullet strikes the loop lies at a distance R from our reference point.

Initial angular momentum is given as $MvR$. ie. before the bullet strikes the loop.

The moment of inertia of the system about the same point is given as $mR^2 + mR^2 = 2mR^2$.

So, final angular momentum is given as $mR^2 ω$.

Equating the two, we get $ω$ as $ω = \frac v {2R}$.

Which does not match the actual answer of $ω = \frac v {3R}$.

By the way, I have posted my approach in which I have a doubt, and not the correct one, I will post it if it is needed.

Why does the bullet initially have an R value? assuming that mvr=(12/4mr^2)ω

 

[haruspex]

Why does the bullet initially have an R value?

Do you mean, why does it have initial angular momentum mvR?

 

[Orodruin]

Yes why does the bullet initially have angular momentum, if it is traveling linearly

Anything moving linearly on a path offset from the reference point has angular momentum.

assuming it is a point mass with some radius r,

This is nonsense. A point mass is a point and has no radius.

why does it cancel out with the radius of the loop

It … doesn’t?

 

[Elj]

Anything moving linearly on a path offset from the reference point has angular momentum.This is nonsense. A point mass is a point and has no radius.It … doesn’t?

alright then can you explain how to solve it, Im genuinely confused

 

[Orodruin]

alright then can you explain how to solve it, Im genuinely confused

I already did. See #22 and #23.

 

[PeroK]

alright then can you explain how to solve it, Im genuinely confused

Using conservation of angular momentum about the centre of mass, which is $\frac R 2$ below the centre of the hoop:
$$L = mv\frac R 2 = m(\frac R 2)^2 \omega + I \omega = (\frac{mR^2}{4} + I)\omega$$Where $I$ is the moment of inertia of the hoop about the centre of mass of the system. Using the parallel axis theorem we have:
$$I = mR^2 + m(\frac R 2)^2 = \frac{5mR^2}{4}$$This gives:
$$\frac{6mR^2}{4}\omega = mv\frac R 2$$$$\omega = \frac{v}{3R}$$

 

[Orodruin]

Using conservation of angular momentum about the centre of mass, which is $\frac R 2$ below the centre of the hoop:
$$L = mv\frac R 2 = m(\frac R 2)^2 \omega + I \omega = (\frac{mR^2}{4} + I)\omega$$Where $I$ is the moment of inertia of the hoop about the centre of mass of the system. Using the parallel axis theorem we have:
$$I = mR^2 + m(\frac R 2)^2 = \frac{5mR^2}{4}$$This gives:
$$\frac{6mR^2}{4}\omega = mv\frac R 2$$$$\omega = \frac{v}{3R}$$

I should add that the only reason I did not do it like this is that the OP already did that. My treatment in #22 and #23 was based on conservation of linear momentum and the angular momentum relative to a point that is not the center of mass of the system - the point being to show that it is also possible, but more involved.

 

[haruspex]

I should add that the only reason I did not do it like this is that the OP already did that. My treatment in #22 and #23 was based on conservation of linear momentum and the angular momentum relative to a point that is not the center of mass of the system - the point being to show that it is also possible, but more involved.

Right, but @Elj is not the OP, and was blocked at an earlier point: the understanding that linear momentum also acts as angular momentum about any axis not in the line of motion (which you addressed in post #32).

@Elj, if you are still struggling with that, it's like forces and torques. A linear force also acts as a torque about any axis not in its line of action.
Having explained that, the next step should be for you to show some attempt.

 

[Orodruin]

Right, but @Elj is not the OP, and was blocked at an earlier point: the understanding that linear momentum also acts as angular momentum about any axis not in the line of motion (which you addressed in post #32).

In #36 I was just clarifying why my solution is more involved that the one presented by @PeroK - noting that I would certainly not suggest doing it that way in the normal case - but did so for the benefit of the OP.

 

[brochesspro]

Consider a rigid body with masses $m_i$ that have time dependent positions $\vec x_i$. For such a body, the velocity field may be described by
$$
\vec v = \vec v_r + \vec \omega \times (\vec x - \vec x_r)
$$
where the $r$ subscript denotes a reference point. By definition, the angular momentum relative to the reference point is
$$
\vec L_r = \sum_i m_i (\vec x_i - \vec x_r) \times \vec v_i
= \sum_i m_i (\vec x_i - \vec x_r) \times(\vec v_r + \vec \omega \times (\vec x_i - \vec x_r)).
$$
We now use that $\sum_i m_i \vec x_i = M \vec x_{cm}$ defines the center of mass (with $M$ being the total mass):
$$
\vec L_r = M (\vec x_{cm} - \vec x_r) \times \vec v_r + I_r (\vec \omega)
$$
where $I_r$ is the moment of inertia tensor relative to $\vec x_r$. The first term is zero in the the following cases:

• $\vec v_r = 0$: The reference point is chosen such that it has zero instantaneous velocity.
• $\vec x_r = \vec x_{cm}$: The reference point is the center of mass.
• $(\vec x_{cm} - \vec x_r)\times \vec v_r = 0$: The offset from the CoM to the reference point is parallel to $\vec v_r$.

None of the above is applicable to the center of the hoop in your case so you cannot ignore the first term for the angular momentum after the collision.

I didn't get the 1st line. $\vec v = \vec v_r + \vec \omega \times (\vec x - \vec x_r)$ How do we know whether or not the object moves radially to the reference point?
Also, I did not get how you obtained the 3rd line from the 2nd line.

If you are saying the the hoop moves inertially, then that is wrong.

I am not familiar with the term "moves inertially". Does it mean that the object moves with uniform velocity?

Why does the bullet initially have an R value? assuming that mvr=(12/4mr^2)ω

Because it is at a distance R from the centre of the loop.
Also, I apologise for my late response to all the helpful messages. I was studying chemistry for my college entrances and it was really hard to switch my mind to physics mode.
Also, I tried another approach to solve this question, but I still get an incorrect, albeit different answer. I will post my procedure in a couple of hours.

 

[Orodruin]

How do we know whether or not the object moves radially to the reference point?

It doesn’t necessarily. The $r$ subscript is for $r$eference, not radial. Note the vector arrow on top of $\vec v_r$, it is a vector, not a vector component.

Also, I did not get how you obtained the 3rd line from the 2nd line.

I used the definition of the center of mass as a sum over all particles. Many of the factors in the expression are constant and can therefore be taken out of the sum.

I also used the definition of the moment of inertia tensor relative to the reference point.

 

[PeroK]

I am not familiar with the term "moves inertially". Does it mean that the object moves with uniform velocity?

Yes. I meant that the centre of mass of the hoop, which is in fact its centre, does not move inertially, but orbits the centre of mass of the hoop/bullet system.

That's why I encouraged you to draw some diagrams. To get an insight into the "wobbly" motion of the hoop.

 

[brochesspro]

It doesn’t necessarily. The r subscript is for reference, not radial. Note the vector arrow on top of v→r, it is a vector, not a vector component.

I understood that, but my doubt was how we would know the radial velocity of the object with reference to the reference point. Or do we not need it as it is not needed for the calculation of the angular momentum vector?

I used the definition of the center of mass as a sum over all particles. Many of the factors in the expression are constant and can therefore be taken out of the sum.

Okay, I shall try it.

That's why I encouraged you to draw some diagrams. To get an insight into the "wobbly" motion of the hoop.

Got it, but how will it help to solve the problem?
Also, here is my new procedure.

 

[Orodruin]

I understood that, but my doubt was how we would know the radial velocity of the object with reference to the reference point. Or do we not need it as it is not needed for the calculation of the angular momentum vector?

You have to silve for $\vec v_r$ from the conservation laws just as you have to solve for $\omega$. In the solution I just expressed it in terms of $\omega$ using the linear momentum conservation in order to eliminate it from the angular momentum conservation. That way the angular momentum conservation was expressable in $\omega$ only and it could be solved for. As the question did not ask about the linear velocity (only angular), I did not care about taking the time to explicitly express $\vec v_r$ in terms of the input.

 

[Orodruin]

Also, here is my new procedure.

Please use the forum LaTeX features rather than posting images. Images are impossible to quote relevant parts of.

I do not understand what you have done to write down the angular momentum conservation. The correct procedure is in #23.

 

[haruspex]

I understood that, but my doubt was how we would know the radial velocity of the object with reference to the reference point. Or do we not need it as it is not needed for the calculation of the angular momentum vector?

Okay, I shall try it.

Got it, but how will it help to solve the problem?
Also, here is my new procedure.
View attachment 337922

I think you have confused yourself by, on the one hand, taking moments about O and, on the other, defining v' as the velocity of the CoM.
For clarity, I will refer to the original position of the centre of the hoop as the fixed point O and the moving centre as O'.
Since, for angular momentum purposes, you are treating the subsequent motion of the bullet as the sum of rotation about O' and the linear motion of O', and you have already accounted for the rotational contribution of the bullet about O' in $2mr\omega^2$, you just need the contribution from the bullet in respect of the linear motion of O'.
What is the velocity of O'?

 

[brochesspro]

you just need the contribution from the bullet in respect of the linear motion of O'.

I do not get this part.

What is the velocity of O'?

How do we find that out?

Please use the forum LaTeX features rather than posting images. Images are impossible to quote relevant parts of.

Sorry, I did it due to time constraints, I shall take care from next time.

I also used the definition of the moment of inertia tensor relative to the reference point.

Could you please confirm if I am correct here, after I split the terms $\vec v_r$ and the term $\vec \omega \times (\vec x_i - \vec x_r)$ with the vector product, I get a vector triple product in the 2nd term. But since the vectors $\vec \omega$ and $\vec x_i - \vec x_r$ are always perpendicular(since we are dealing with 2D motion), their dot product is zero, and then we are left with this guy, $I_r (\vec \omega)$. I still do not get the centre of mass part though.

 

[Orodruin]

Could you please confirm if I am correct here, after I split the terms $\vec v_r$ and the term $\vec \omega \times (\vec x_i - \vec x_r)$ with the vector product, I get a vector triple product in the 2nd term. But since the vectors $\vec \omega$ and $\vec x_i - \vec x_r$ are always perpendicular(since we are dealing with 2D motion), their dot product is zero, and then we are left with this guy, $I_r (\vec \omega)$. I still do not get the centre of mass part though.

There is no dot product. Just cross products. The first term is very relevant.

 

[haruspex]

How do we find that out?

Calculate it from v', r and ω. Call it v".
For the purpose of finding the angular momentum about O, you can think of the motion of the hoop+bullet system immediately after impact as the sum of the linear motion of O' (velocity v") and a rotation about O'. For each of hoop and bullet, each of these motions makes a contribution to the angular momentum, potentially making four terms in all.
For the hoop, the linear motion of the mass centre is in a line through O, so that term is zero. The rotational term is $mr^2\omega$.
The rotational term for the bullet is also $mr^2\omega$, as you posted.
That leaves the linear term for the bullet: a mass m moving at velocity v" along a path that is distance r from O at the nearest: $mv''r$

 

[Orodruin]

As for the com issue. Consider
$$
\sum_i m_i (\vec x_i - \vec x_r) \times \vec v_r
=
M \left[ \underbrace{\frac 1M \sum_i m_i \vec x_i}_{\equiv \vec x_{cm}} - \vec x_r \frac 1M \underbrace{\sum_i m_i}_{\equiv M} \right] \times \vec v_r
=
(M \vec x_{cm} - M \vec x_r)\times \vec v_r
$$

 

[brochesspro]

As for the com issue. Consider
$$
\sum_i m_i (\vec x_i - \vec x_r) \times \vec v_r
=
M \left[ \underbrace{\frac 1M \sum_i m_i \vec x_i}_{\equiv \vec x_{cm} - \vec x_r \frac 1M \underbrace{\sum_i m_i}_{\equiv M} \right] \times \vec v_r
=
(M \vec x_{cm} - M \vec x_r)\times \vec v_r
$$

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