Point of application of angular momentum

[serverxeon]

I know the approach is to use conservation of angular momentum.
I also know that I am entirely free to pick which point I would like to calculate my L about. (Although calculating L about the CM will make it much easier, but I'd like to try otherwise)

So, if I were to calculate my L(after collision) about the left end of the rod, how would I express my L?

Am I able to do just L=Iω, where I is the modified $I=\frac{1}{12} \times 2m\times l^{2}+ml^{2}$
The ω I get is then the answer I am looking for? It feels wrong to me.

 

[haruspex]

Am I able to do just L=Iω, where I is the modified $I=\frac{1}{12} \times 2m\times l^{2}+ml^{2}$

You can take moments about wherever you like, but you do need to be careful.
Suppose the rod rotates anticlockwise at rate ω, and moves linearly with speed s. For clarity, I'll write M for the rod's mass and I for its MI about mass centre.
If you take moments about the mass centre of the rod, you can omit s from that part of the equation (but it still shows up for the impacting mass, since its speed after collision is s+lω/2):
$mVl/2 =Iω+m(l/2)(s+ωl/2)$
If you take moments about the left hand end, you have to consider the moment corresponding to the linear motion of the rod:
$mVl =Iω+Msl/2+ml(s+ωl/2)$
If you take moments about the right hand end, you get:
$0 =Iω-Msl/2$
In the question at hand, M=2m, I = Ml²/12, ω=V/l, s = V/6, Iω=mVl/6, s+ωl/2 = 2V/3. I think you'll find that satisfies all three equations.

 

[tiny-tim]

hi serverxeon!

angular momentum = Iω only works about the centre of rotation or the centre of mass

for any other point, you have to add mr_c.o.m x v_c.o.m

(for the centre of rotation, but nowhere else, that happens to equal the parallel axis thingy, md² times ω )

 

[serverxeon]

You can take moments about wherever you like, but you do need to be careful.
Suppose the rod rotates anticlockwise at rate ω, and moves linearly with speed s. For clarity, I'll write M for the rod's mass and I for its MI about mass centre.
If you take moments about the mass centre of the rod, you can omit s from that part of the equation (but it still shows up for the impacting mass, since its speed after collision is s+lω/2):
$mVl/2 =Iω+m(l/2)(s+ωl/2)$
If you take moments about the left hand end, you have to consider the moment corresponding to the linear motion of the rod:
$mVl =Iω+Msl/2+ml(s+ωl/2)$
If you take moments about the right hand end, you get:
$0 =Iω-Msl/2$
In the question at hand, M=2m, I = Ml²/12, ω=V/l, s = V/6, Iω=mVl/6, s+ωl/2 = 2V/3. I think you'll find that satisfies all three equations.

Thanks for the response.

How did you get the speed of the impacting mass after the collision to be s+lω/2, and the rod s=v/6 ?

 

[haruspex]

How did you get the speed of the impacting mass after the collision to be s+lω/2

The centre of the rod moves forward (i.e. in the same direction as the original V) at speed s. It is also rotating at rate ω. Just after impact, the impacted end of the rod will be moving forwards at speed ωl/2 relative to the rod's centre, for a total speed of s + ωl/2. The impacting mass is doing likewise.

, and the rod s=v/6 ?

By taking one (any one) of the three angular momentum equations I wrote out and combining it with the linear momentum equation you already had.

 

[serverxeon]

What is the equation for conservation of linear momentum?
Following your conventions,

I have
$P_{i}=mV$

$P_{f}=Ms+m(s+\frac{lω}{2})$

When i solve them for s, I get
$s=3V-\frac{3lω}{2}$

Doesn't look too correct to me =/

 

[haruspex]

$P_{i}=mV$
$P_{f}=Ms+m(s+\frac{lω}{2})$

Right. That is satisfied by the set of values I quoted before:
ω=V/l, s = V/6, Iω=mVl/6, s+ωl/2 = 2V/3

When i solve them for s, I get
$s=3V-\frac{3lω}{2}$

That is not satisfied by those values, so you must have gone wrong. If you cannot find your error, please post your detailed working.

 

[serverxeon]

Right. That is satisfied by the set of values I quoted before:
ω=V/l, s = V/6, Iω=mVl/6, s+ωl/2 = 2V/3

That is not satisfied by those values, so you must have gone wrong. If you cannot find your error, please post your detailed working.

Oops, I made a careless slip.

The centre of the rod moves forward (i.e. in the same direction as the original V) at speed s. It is also rotating at rate ω. Just after impact, the impacted end of the rod will be moving forwards at speed ωl/2 relative to the rod's centre, for a total speed of s + ωl/2. The impacting mass is doing likewise.

By taking one (any one) of the three angular momentum equations I wrote out and combining it with the linear momentum equation you already had.

I have a conceptual gap here.

You said that the impacted end of the rod will be moving forward at a speed of ωl/2 relative to the rod centre. My understanding is that this ω is around the new centre of mass, and not at the rod centre. So how is it that we can use it to solve question?
Is the quantity ω universal in a sense that it has no respect to any particular point?

 

[haruspex]

Is the quantity ω universal in a sense that it has no respect to any particular point?

Yes. If a rigid body is rotating, every part of it is rotating at the same rate. If you take any two points on it, A and B, you can express B's instantaneous velocity as A's velocity+ vector product of the rotation vector and the displacement vector AB.

 

[serverxeon]

Alright,
I got it now.
Thanks so much!

 

[serverxeon]

I now know how you solved it your way (by breaking up the angular momentum quantity),

but when I went back to try and reconcile my method (using a shifted moment of inertia), I am still kind of confused.

If I consider the angular momentum about the centre of mass of the combined system ($x=\frac{2}{3} l$ from the left end), then I have these equations:

$L_{i}=\frac{mVl}{3}$
$L_{f}=I'ω=(\frac{1}{12} Ml^{2}+MD^{2}+m[\frac{1}{3} l]^{2})ω=\frac{1}{3} ml^{2} ω$

where D is the shifting distance in parallel axis theorem, and in this case $D = \frac{2l}{3} - \frac{l}{2}$

That solves to give $ω=\frac{v}{l}$, which indicate this approach of using a modified moment of inertia, is correct.

---

However, when I use this approach and take angular momentum about the left end, I get these equations:
$L_{i}=mVl$
$L_{f}=I'ω=(\frac{1}{12} Ml^{2}+MD^{2}+ml^{2})ω=\frac{5}{3} ml^{2} ω$

$D=\frac{l}{2}$

which solves to give $ω=\frac{3v}{5l}$.

My gut feeling is that there is something special about using this method at the centre of mass - that it will only work if used there. But I can't pinpoint what exactly is the correction factor required for it to work if not taken at CM.

Any help here?

--------------------------------------------

Also, there's an extension to this question which I missed out.

It asks, determine the position of the point on the rod which remains stationary immediately after the collision

Using the equation $V_{A}=V_{B}+ω×R_{AB}$
Then I'll get the point of 0 velocity to be x=1/3 l from the left end.

However, I have this concept in my mind that any object which is free to rotate will rotate about its centre of mass, which is x=2/3 l from the left end.

So which is the correct answer?

 

[haruspex]

when I use this approach and take angular momentum about the left end, I get these equations:
$L_{i}=mVl$
$L_{f}=I'ω=(\frac{1}{12} Ml^{2}+MD^{2}+ml^{2})ω=\frac{5}{3} ml^{2} ω$

That would be correct if the point you're taking moments about is the instantaneous centre of rotation. That's what the parallel axis theorem is for - when the axis is known and is not the mass centre. Here, we don't know where the centre of rotation is (though we could work it out), so instead do it the way I posted: moment about mass centre + moment of that mass's linear motion about the point you chose.

It asks, determine the position of the point on the rod which remains stationary immediately after the collision

Ha! I wrote the above without reading this far.

any object which is free to rotate will rotate about its centre of mass

Throw a Frisbee. Imagine looking down on it as it flies. The centre is moving forward, but the disc is also spinning, clockwise say. So a point just to the right of centre will be, for an instant, stationary. Of course, that's just in your reference frame. If you were flying along at the same speed as the Frisbee you'd see it as rotating about its centre.

 

[serverxeon]

right, that clears everything.

thanks!