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Around which points to make T=I(alpha)

  • Thread starter Karol
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  • #1
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Homework Statement


I was told that I can use the rigid body angular momentum formula T=I[itex]\alpha[/itex]
Aroud 3 points: a fixed one (I assume the temporary center of rotation), or the center of mass or a point moving parallel to the c.m.
I have tested it with the case shown in the picture: a cylinder rotating on a rough surface, a steady force acting on it's center.
When calculating the angular acceleration [itex]\alpha[/itex] round the points A and C I get the same result, but when calculating around B-a different one, meaning the above is correct.
Why is that? the formula L=I[itex]\omega[/itex] is derived this way:
[tex]L=\sum_{n=1}^N L_n=\sum_{n=1}^N m_nv_n\cdot r_n=\sum_{n=1}^N m_nr^2_n \omega=I\omega[/tex]
So, I understand, in order to get the velocity of the particle correct, I have to calculate only round the stationary point of rotation. The formula T=I[itex]\alpha[/itex] is only the derivative of the former.
Can anyone explain a bit, and give a name of a book that explains this in detail, since some ordinary books talk only in general about this.
 

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  • #2
tiny-tim
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Hi Karol! :smile:
When calculating the angular acceleration [itex]\alpha[/itex] round the points A and C I get the same result, but when calculating around B-a different one, meaning the above is correct.
No, the velocity of B is parallel to that of the centre of mass (C), so the result should be the same.

Have you included the friction force at A (not in the diagram) … if there's no friction, then why would the angular speed change? :wink:

(and sorry, I don't know any book that goes into this :redface:)
 
  • #3
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There's friction and I included it in my calculation for point B. Around point A:

[tex]T_A=I_A\alpha[/itex]
[tex]FR=(kmR^2+mR^2)\alpha\Rightarrow\alpha=\frac{F}{mR(k+1)}[/itex]

I get the same [itex]\alpha[/itex] when I calculate round point C.
What is the friction force f ?
[tex]\sum F=ma\Rightarrow F-f=ma=mR\alpha=mR\frac{F}{mR(k+1)}[/tex]
[tex]\Rightarrow f=F\frac{k}{k+1}[/tex]

Around point B:

[tex]T_B=I_B\alpha[/itex]

[tex]F\frac{R}{2}+f\frac{R}{2}=(kmR^2+m\frac{R^2}{4})\alpha[/tex]
[tex]\Rightarrow \alpha=\frac{2F(2k+1)}{mR(k+1)(4k+1)}[/tex]

And it's not the same as the first.
How do you, Tiny-Tim, know this subject with no book to offer?
 

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  • #4
tiny-tim
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Hi Karol! :smile:

Your formula τB = IBα is wrong :redface:

The angular momentum about any point P is Iω about the centre of mass (C) plus the (ordinary) moment about B of the total momentum (mvc.o.m.) as if it were at C.

ie LP = ICω + m PC x vc.o.m.

If P is on the axis of rotation, then vc.o.m. = PC x ω, and so LP = (IC + md2)ω = IPω.

In any other case, LP is not IPω

(and τP is not IPα) :wink:

In other words:

The parallel axis theorem (IP = IC + md2) is only useful at the centre of rotation* (or, if you have a body made up of parts, you can apply it at the centre of mass of the whole body for each part for which you know the moment of inertia )

* to be precise: on the axis of rotation
 
  • #5
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I assume LP = ICω + m PC x vc.o.m. comes from the transformation of vectors between moving coordinate systems:
[itex]\Vec{A_2}[/itex] is vector A2 relative to the ground
[itex]\Vec{A_1}[/itex] is vector A1 relative to the ground
[itex]\Vec{A_{21}}[/itex] is vector A2 relative to vector A1
[tex]\Vec{A_2}=\Vec{A_1}+\Vec{A_{21}}[/tex]
But I think you switched the places of the vectors in:
vc.o.m. = PC x ω
It should be:
vc.o.m. = ω x PC
Since then the direction of vc.o.m. comes out correct.
 
  • #6
tiny-tim
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Yeah, I can never remember which way round that is ! :biggrin:
 
  • #7
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How is this equation derived?
LP = ICω + m PC x vc.o.m.
 
  • #8
tiny-tim
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How is this equation derived?
LP = ICω + m PC x vc.o.m.
LP = ∫∫∫ (r - rP) x ρ v dxdydz

= ∫∫∫ (r - rP) x ρ (ω x (r - rc.o.m.) + vc.o.m.) dxdydz

= ∫∫∫ (r - rP) x ρ (ω x (r - rc.o.m.) dxdydz + (∫∫∫ ρ (r - rP) dxdydz) x vc.o.m.

= ∫∫∫ (r - rc.o.m.) x ρ (ω x (r - rc.o.m.) dxdydz + (mvc.o.m. - mrP) x vc.o.m.

(because ∫∫∫ ρ (r - rc.o.m.) dxdydz = 0)

= Ic.o.m.ω + m PC x vc.o.m. :smile:

More indirectly, but shorter (with vP being the velocity of the part of the body at position P) …

LP = ∫∫∫ (r - rP) x ρ v dxdydz

= ∫∫∫ (r - rP) x ρ (ω x (r - rP) + vP) dxdydz

= ∫∫∫ (r - rP) x ρ (ω x (r - rP)) dxdydz + (∫∫∫ ρ (r - rP) dxdydz) x vP

= IPω + m PC x vP

and then applying the parallel axis theorem …

= Ic.o.m.ω + m PC x vc.o.m. :wink:

(and if P is the centre of rotation, so that vP = vc.o.r. = 0, then:

= Ic.o.r.ω)​
 
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  • #9
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First, Tiny-Tim, I thank you very much. You feed my curiosity...

But I don't understand the transition between:
∫∫∫ (r - rP) x ρ (ω x (r - rc.o.m.) dxdydz + (∫∫∫ ρ (r - rP) dxdydz) x vc.o.m.
And:
∫∫∫ (r - rc.o.m.) x ρ (ω x (r - rc.o.m.) dxdydz + (mvc.o.m. - mrP) x vc.o.m.

In the first integral, I guess you did:
r - rP=(rc.o.m.-rP)+(r - rc.o.m.)=PC+(r - rc.o.m.)

Taking PC and returning to the integral:

∫∫∫PC x ρ (ω x (r - rc.o.m.) dxdydz=
∫∫∫PC x ρ (ω x r - ω x rc.o.m.) dxdydz=
∫∫∫ρ PC x (ω x r) - PC x (ω x rc.o.m.) dxdydz

None of these yields ∫∫∫ ρ (r - rc.o.m.) dxdydz.

In the second integral:

∫∫∫ρrPdxdydz=mrP

But:

∫∫∫ρrdxdydz Is not equal to mvc.o.m.
 
  • #10
tiny-tim
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Hi Karol! :smile:

In the second integral, you're right … I mistyped mvc.o.m. instead of mrc.o.m.

the next line is correct, though, since (mrc.o.m. - mrP) x vc.o.m. = m PC x vc.o.m. :wink:

In the first integral, rc.o.m. rP and ω are all independent of x y and z, so they come outside the ∫∫∫, leaving only ∫∫∫ (r - rc.o.m.) dxdydx = 0 :smile:
 
  • #11
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∫∫∫ (r - rP) x ρ (ω x (r - rc.o.m.) dxdydz=
ρ ω x ∫∫∫ (r - rP) x (r - rc.o.m.) dxdydz=
ρ ω x ∫∫∫ r x (r - rc.o.m.)dxdydz - ∫∫∫rP x (r - rc.o.m.)dxdydz=
ρ ω x ∫∫∫ r x (r - rc.o.m.)dxdydz+0=
ρ ω x (∫∫∫ r x r - ∫∫∫ r x rc.o.m.)=
ω x ∫∫∫ ρ r x rc.o.m.

But this does not give ωIc.o.m., or does it?
 
  • #12
tiny-tim
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∫∫∫ (r - rP) x ρ (ω x (r - rc.o.m.) dxdydz=
ρ ω x ∫∫∫ (r - rP) x (r - rc.o.m.) dxdydz=
You can't do that!! :redface:

A x (B x C) is nothing like B x (A x C) :frown:
 
  • #13
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∫∫∫ (r - rP) x ρ ω x (r - rc.o.m.) dxdydz=
∫∫∫ r x ρ ω x (r - rc.o.m.) - ∫∫∫ rP x ρ ω x (r - rc.o.m.)=
∫∫∫ r x ρ ω x (r - rc.o.m.) - 0=
∫∫∫ r x ρ ω x r - ∫∫∫ r x ρ ω x rc.o.m.=
∫∫∫ r x ρ ω x r - (∫∫∫ r x ρ ω) x rc.o.m.
 
  • #14
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Please, Tiny-Tim, complete the explanation.
In my previous post I wrote the equations, but didn't get the required result: Ic[itex]\omega[/itex]
 
  • #15
tiny-tim
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It's at the start of my post #8 …

you need to have an (r - rc.o.m.) on the extreme left for it to work (to get Ic.o.m.):wink:

for some reason you decided not to do so! :confused:
 
  • #16
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But that's exactly my question!
At the extreme left on the first formula:
∫∫∫ (r - rP) x ρ ω x (r - rc.o.m.) dxdydz + ...
There is: (r - rP)
But then the extreme left member changes to: (r - rc.o.m.) in the next equation:
∫∫∫ (r - rc.o.m.) x ρ ω x (r - rc.o.m.) dxdydz + ...
And the rest of these two formulas is the same. how to make the transition of (r - rP) to (r - rc.o.m.)?
 
  • #17
tiny-tim
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how to make the transition of (r - rP) to (r - rc.o.m.)?
Because …
(because ∫∫∫ ρ (r - rc.o.m.) dxdydz = 0)
… (and rP is constant).
 
  • #18
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I now understand I shouldn't open the parentheses. I understand that From your second example with vP.
But in your first analysis I don't understand how you got the same member twice, as needed and as described in your second example.
That is, from:
∫∫∫ (r - rP) x ρ ω x (r - rc.o.m.) dxdydz + ...
There is: (r - rP)
But then the extreme left member changes to: (r - rc.o.m.) in:
∫∫∫ (r - rc.o.m.) x ρ ω x (r - rc.o.m.) dxdydz + ...
how to make the transition of (r - rP) to (r - rc.o.m.)?
And you write that
(because ∫∫∫ ρ (r - rc.o.m.) dxdydz = 0)
Where do you get this without opening parentheses?
 
  • #19
tiny-tim
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it's the definition of the centre of mass
 
  • #20
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I know the definition of C.M.. I ask how, in what way, you arrived to the partial expression:
∫∫∫ ρ (r - rc.o.m.) dxdydz
From the formula:
∫∫∫ (r - rP) x ρ ω x (r - rc.o.m.) dxdydz

But my main question is this:
You have changed the first member, (r - rP), in the formula:
∫∫∫ (r - rP) x ρ ω x (r - rc.o.m.) dxdydz
To (r - rc.o.m.) In the formula:
∫∫∫ (r - rc.o.m.) x ρ ω x (r - rc.o.m.) dxdydz
By what right? what's the process you have made for that.
From there, I guess, I'll go on my own.
Thanks.
 
  • #21
tiny-tim
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You have changed the first member, (r - rP), in the formula:
∫∫∫ (r - rP) x ρ ω x (r - rc.o.m.) dxdydz
To (r - rc.o.m.) In the formula:
∫∫∫ (r - rc.o.m.) x ρ ω x (r - rc.o.m.) dxdydz
Because I subtracted ∫∫∫ (rP - rc.o.m.) x ρ ω x (r - rc.o.m.) dxdydz,

which is a constant "times" ∫∫∫ ρ (r - rc.o.m.) dxdydz, which = 0.
 
  • #22
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Thank you very much, Tiny-Tim. It's a relief.

I think you should add ∫∫∫ (rP - rc.o.m.) x ρ ω x (r - rc.o.m.) dxdydz, instead of subtracting, since then you get ∫∫∫ (r - rc.o.m.) x ρ ω x (r + rc.o.m.) dxdydz.
 
  • #23
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I try to calculate [itex]\alpha[/itex] around point P, as drawn.
I don't get the right answer, since I get a different [itex]\alpha[/itex] at points O and C, which is (round point O):
[tex]M_O=I_O\alpha\Rightarrow FR=(KmR^2+mR^2)\alpha
\Rightarrow \alpha=\frac{F}{mR(K+1)}[/tex]

I take the derivative of LP = ICω + m PC x vc.o.m. and get MP = ICα + m PC x ac.o.m..

I find the friction force, with the help of [itex]\alpha[/itex]:
[tex]\Sigma F=ma \Rightarrow f=F\frac{K}{K+1}[/tex]

MP = ICα + m PC x ac.o.m.

[tex]\Rightarrow\left(F\frac{R}{2}\cdot f \frac{R}{2}\right)\hat{z}=mR^2\left(K+\frac{1}{4}\right)\vec{\alpha}+\left(m\frac{R}{2}\cdot\frac{F}{m(K+1)}\right)\hat{z}[/tex]
[tex]\alpha=\frac{4FK}{mR(K+1)(4K+1)}[/tex]

Which is not the same as the previous I found.
 

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  • #24
tiny-tim
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Hi Karol! :smile:

The dot should be a +, and the "+ 1/4" shouldn't be there.

Then your equation becomes

(2K+1)R/2(K+1) = (Rα/a)2KR/2(K+1) + R/2(K+1) :smile:
 
  • #25
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The "+ 1/4" should be there since point P is on the half of the radius.
And there is no F in your equation.
 

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