# Around which points to make T=I(alpha)

## Homework Statement

I was told that I can use the rigid body angular momentum formula T=I$\alpha$
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 $\alpha$ 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$\omega$ is derived this way:
$$L=\sum_{n=1}^N L_n=\sum_{n=1}^N m_nv_n\cdot r_n=\sum_{n=1}^N m_nr^2_n \omega=I\omega$$
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$\alpha$ 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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tiny-tim
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Hi Karol! When calculating the angular acceleration $\alpha$ 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? (and sorry, I don't know any book that goes into this )

There's friction and I included it in my calculation for point B. Around point A:

$$T_A=I_A\alpha[/itex] [tex]FR=(kmR^2+mR^2)\alpha\Rightarrow\alpha=\frac{F}{mR(k+1)}[/itex] I get the same $\alpha$ when I calculate round point C. What is the friction force f ? [tex]\sum F=ma\Rightarrow F-f=ma=mR\alpha=mR\frac{F}{mR(k+1)}$$
$$\Rightarrow f=F\frac{k}{k+1}$$

Around point B:

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

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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tiny-tim
Homework Helper
Hi Karol! Your formula τB = IBα is wrong 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) 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

I assume LP = ICω + m PC x vc.o.m. comes from the transformation of vectors between moving coordinate systems:
$\Vec{A_2}$ is vector A2 relative to the ground
$\Vec{A_1}$ is vector A1 relative to the ground
$\Vec{A_{21}}$ is vector A2 relative to vector A1
$$\Vec{A_2}=\Vec{A_1}+\Vec{A_{21}}$$
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.

tiny-tim
Homework Helper
Yeah, I can never remember which way round that is ! How is this equation derived?
LP = ICω + m PC x vc.o.m.

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

= Ic.o.r.ω)​

Last edited:
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.

tiny-tim
Homework Helper
Hi Karol! 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. 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 ∫∫∫ (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?

tiny-tim
Homework Helper
∫∫∫ (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!! A x (B x C) is nothing like B x (A x C) ∫∫∫ (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.

In my previous post I wrote the equations, but didn't get the required result: Ic$\omega$

tiny-tim
Homework Helper
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.) for some reason you decided not to do so! 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.)?

tiny-tim
Homework Helper
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).

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?

tiny-tim
Homework Helper
it's the definition of the centre of mass

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.

tiny-tim
Homework Helper
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.

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.

I try to calculate $\alpha$ around point P, as drawn.
I don't get the right answer, since I get a different $\alpha$ at points O and C, which is (round point O):
$$M_O=I_O\alpha\Rightarrow FR=(KmR^2+mR^2)\alpha \Rightarrow \alpha=\frac{F}{mR(K+1)}$$

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 $\alpha$:
$$\Sigma F=ma \Rightarrow f=F\frac{K}{K+1}$$

MP = ICα + m PC x ac.o.m.

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

Which is not the same as the previous I found.

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tiny-tim
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Hi Karol! The dot should be a +, and the "+ 1/4" shouldn't be there.

(2K+1)R/2(K+1) = (Rα/a)2KR/2(K+1) + R/2(K+1) 