Homework Help: Conservation of angular momentum

1. Mar 21, 2016

JulienB

Hi everybody! I'm preparing myself for upcoming exams, and I struggle a little with conservation of angular momentum. Can anybody help me understand how to solve such problems?

1. The problem statement, all variables and given/known data

(for a better comprehension, see the attached image)

We have a wooden cylinder of mass mZ = 600g and of radius r0 = 5cm, which can rotate around its symmetry axis. Someone shoots on it, and the projectile has the mass mG = 5.0g and initial velocity v = 80m/s. The distance between the linear trajectory of the projectile and the rotation axis of the cylinder is r1 = 3.0cm. The projectile penetrates the cylinder and stays stuck at a distance of r2 = 3.5cm from the rotation axis of the cylinder.

a) What is the frequency of rotation f of the cylinder after the impact? Where and how should you shoot the projectile in order to obtain maximum/minimum frequency?
b) Which part of the kinetic energy is used to deform the wooden cylinder?
c) If the cylinder was not fixed on a rotation axis but on a thread, what would be the differences to previous case when the projectile hits the cylinder?

2. Relevant equations

So I imagine both conservation of linear momentum and of angular momentum are important. We also know that ω = 2πƒ.

3. The attempt at a solution

Okay I give it a go:

We know that the linear momentum is conserved, that the cylinder is not moving before the collision and that the two objects are moving together after the collision:

mG ⋅ v = (mG + mZ) ⋅ v'

Here I already see a problem: v' is supposed to be the tangential velocity of the system cylinder-projectile after the collision, but I believe a projectile located at r2 = 3.5cm does not have the same tangential velocity as a point located at r0 = 5.0cm. Is that correct? Then we would have mG ⋅ v = mG ⋅ vG' + mZ ⋅ vZ', which is also not so great.

I encounter the same problem with the conservation of angular momentum:

mG ⋅ v ⋅ r0 = (mG + mZ) ⋅ v' ⋅ r0
or
mG ⋅ v ⋅ r0 = mG ⋅ vG' ⋅ r2 + mZ ⋅ vZ' ⋅ r0
?

I feel like I'm missing something, since none of those equations lead me anywhere :( Furthermore, I never manage to involve r1 in the equations, which obviously plays a role because of the 2nd part of the question. Can someone give me a clue so that I clarify my misunderstandings?

Thank you very much in advance.

Julien.

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2. Mar 21, 2016

ZapperZ

Staff Emeritus
You appear to have completely ignored the moment of inertia of the cylinder.

Zz.

3. Mar 21, 2016

PeroK

I would assume from the problem statement that the cylinder is fixed and can only rotate about its axis of symmetry, not move linearly.

For part c) I would assume it is hanging by a thread, and is free to swing and rotate. Although I am not totally confident how to interpret part c). I guess you had to translate the question for us.

4. Mar 21, 2016

ZapperZ

Staff Emeritus
What does this have anything to do with not including the moment of inertia? If the cylinder is rotating under any circumstances, its moment of inertia comes into play!

Zz.

5. Mar 21, 2016

PeroK

If the cylinder is fixed, linear momentum is not conserved. If that is the case, the OP is off on the wrong foot.

6. Mar 21, 2016

JulienB

@ZapperZ :
Yes I realised that, but don't have I to also include the moment of inertia of the projectile then? I don't know its shape though.

Or may I just write the linear momentum of the projectile on the left side of the equation and the moment of inertia of the cylinder on the right side, like that:
mG ⋅ v = IZ ⋅ ω?

@PeroK :
Yes I had to translate the problem from German, sorry :) Your assumption is correct, but if you don't mind I will first focus on question a) to make sure I understand how to use the equation first.

7. Mar 21, 2016

JulienB

@PeroK :
Yes I believe the cylinder is fixed. So the linear momentum is not conserved??

8. Mar 21, 2016

PeroK

I would assume not. I would assume for parts a) and b) it is only free to rotate about a fixed axis.

9. Mar 21, 2016

ZapperZ

Staff Emeritus
The moment of inertia of the projectile can be assume to be the same as that for a particle at some fixed radius once it is embedded in the cylinder. So you have to consider TWO different angular momentum after the "collision": from the spinning cylinder and the projectile. Otherwise, you have missed the entire problem.

I still don't understand why, just because you don't know the "shape" of the projectile, that you are not even including the moment of inertia of the cylinder. This is a major part of the physics, and you missed it.

Zz.

10. Mar 21, 2016

JulienB

I was (and am still) a bit confused about the whole thing. If linear momentum is not conserved, then may I write the following:

mG ⋅ v ⋅ r0 = mG ⋅ vG' ⋅ r2 + mZ ⋅ r0 ⋅ vZ' ?

On the left hand side I consider the angular moment at the moment of the collision, and on the left hand side the angular momentum of the projectile and of the cylinder once the projectile is stuck in the cylinder. The moment of inertia of the cylinder being mG ⋅ r02 and since ω = vZ'/r0, I would say that LZ = mZ ⋅ r0 ⋅ vZ'. Am I getting somewhere?

11. Mar 21, 2016

JulienB

I went further with that idea, and reached the following:

mG ⋅ v ⋅ r0 = mG ⋅ r22 ⋅ ω + mZ ⋅ r02 ⋅ ω
⇒ ω = (mG ⋅ v ⋅ r0) / (mG ⋅ r22 + mZ ⋅ r02) = 13.28 rad/s2
⇒ ƒ = ω/2π = 2.1 Hz

Does that make sense? mZ ⋅ r02 is the moment of inertia of the cylinder.

12. Mar 21, 2016

PeroK

I'm not sure I follow what you are doing. You seem to be unsure what is moment of inertia. In this problem, you have two examples of angular momentum:

1) Angular momentum of a point mass about a point/axis, which I think you understand.

2) Angular momentum of a rigid body about an axis of rotation. In this case you need the moment of inertia (MoI) of the rigid body about that axis. This is not the mass of the body, but a measure of the mass distribution of the body in terms of distance from the axis.

You should know or be able to calculate the MoI of a solid cylinder. This is what you are missing. Your last equation looks wrong as well.

First step: we need the MoI of a solid cylinder about its axis of symmetry.

Last edited: Mar 21, 2016
13. Mar 21, 2016

PeroK

That's the MoI of a hollow cylinder, where all the mass is $r_0$ from the centre.

14. Mar 21, 2016

JulienB

@PeroK
Sounds bad, but I won't give up :)

Yes I indeed used the wrong formula since the beginning. The moment of inertia of a full cylinder is I believe:

IZ = ½ πhρr4 = ½ mZ ⋅ r2

Now for the rest, I will try to explain you thoroughly my train of thoughts:

I assume the angular momentum of the point mass "projectile" about the axis of rotation of the cylinder just before the collision should be equal to the angular momentum of the cylinder about its axis of rotation + the angular momentum of the bullet embedded inside the cylinder after the collision:

LG = LG' + LZ'
⇔ mG ⋅ v ⋅ r0 = IG ⋅ ω + IZ ⋅ ω
⇔ mG ⋅ v ⋅ r0 = mG ⋅ r22 ⋅ ω + ½ ⋅ mZ ⋅ r02 ⋅ ω

ω being the angular velocity, it should be the same at r0 for the cylinder and at r2 for the bullet, right? At least that's how I attempted to resolve:

ω = (mG ⋅ v ⋅ r0)/(mG ⋅ r22 + ½ mZ ⋅ r02)

Unfortunately I get a bit of a crazy result. I probably still mess something up.

15. Mar 21, 2016

PeroK

That looks right. What did you get for $\omega$?

Sorry, just saw you used $r_0$ on the left-hand side. Can you see what it should be?

16. Mar 21, 2016

JulienB

@PeroK
Well... Around 1468 rad/s2, which would give a frequency of 233.6 Hz.

17. Mar 21, 2016

JulienB

Nevermind I made a mistake: now ω = 26.45 and ƒ = 4.2 Hz.

18. Mar 21, 2016

PeroK

That's not right, because you are using $r_0$ for the initial angular momentum.

19. Mar 21, 2016

JulienB

Aaaah that should be r1, right? I used r0 because I thought the initial velocity would be affected while penetrating the cylinder.

20. Mar 21, 2016

PeroK

Yes, it must be $r_1$.

21. Mar 21, 2016

JulienB

@PeroK
Thank you so much for all of your efforts to help me understand this problem. I now get a frequency ƒ = 2.5 Hz. Would you mind explaining me why we must take r1 on the left side of the equation? I never really got how we are supposed to choose our reference point before a rotational collision, especially in this case of plastic collision.

Thank you very much, I'm gonna try to figure out the rest of the problem now.

Julien.

22. Mar 21, 2016

PeroK

The angular momentum of a projectile about a point is actually a vector quantity and is the cross product of the linear momentum with the displacment from the point:

$\vec{L} = \vec{r} \times m\vec{v}$

This equates to

$L = dmv$

Where $d$ is the perpendicular distance from the point to the line of the projectile.

23. Mar 21, 2016

JulienB

@PeroK
Thank you, that was an amazingly clear answer :)

24. Mar 21, 2016

JulienB

May I ask about question b)? I am not sure I get the point of the question:

Since this collision is inelastic, the first thing that comes to my mind is to say that the translational kinetic energy of the projectile before the collision (½ ⋅ mG ⋅ v2) is bigger than the sum of the rotational kinetic energies of the projectile and of the cylinder after the collision (½ ⋅ mG ⋅ r22 ⋅ ω + ¼ ⋅ mZ ⋅ r02 ⋅ ω).
But the question is asking which part of the initial kinetic energy has been used to deform the cylinder. I have no idea what a "part" of that kinetic energy might be, since to me the initial kinetic energy only comes from its translational motion. Any idea?

Julien.

25. Mar 21, 2016

PeroK

It might mean what proportion of the KE is lost?