Contradiction between equations of angular momentum

[Karol]

Homework Statement

Two equal, parallel and opposite forces at at both sides of a horizontal disk that lies on a smooth table, according to the picture.
The mass is m and the moment of inertia is: kmR²
Angular momentum round the center point A:
$$2FR=kmR^2 \cdot \alpha$$.
Angular momentum round the point B on the outer edge:
$$2FR=mR^2(k+1) \cdot \alpha$$.
It is clear i will get 2 different angular acceleration $\alpha$, how come?

Homework Equations

M=I$\alpha$
Shteiner's theorem of the parallel axis:
I_b=I_c+mb²

The Attempt at a Solution

It is clear that the second equation is wrong, since the first one is right, since it is round a static point.
Maybe i have to compensate, when calculating round point B, for it's acceleration?
How? maybe with D'alamber's sentence?
But then, can i solve only from the point of view of the accelerating system?
I want to solve from the static, inertial system.

 

[tiny-tim]

Hi Karol!

L_P = I_Pω only works if P is the centre of mass or the centre of rotation.

From the PF Library on angular momentum …

about point P, where $\mathbf{v}_P$ is the velocity of the part of the body at position P:

$$\mathbf{L}_{P}\ =\ \tilde{I}_{P}\mathbf{\omega}\ +\ m(\mathbf{r}_{c.o.m.}-\mathbf{r}_P)\times\mathbf{v}_P$$​

 

[Karol]

And what if the center of rotation is itself under acceleration?
For example the new picture here.
Can i use L=I$\alpha$ around point B?
A rope is wound around a falling cylinder.

 

[tiny-tim]

that equation is for a fixed point P,

and v_P is the velocity of the (changing) point on the body that happens to be at P

(so no question of P accelerating )

Can i use L=Iα around point B?

(you mean L = Iω) yes

but a moment later, B will no longer be the centre of rotation, so you'll have problems finding dL/dt …

it'll be much easier to use the centre of mass!

 

[Karol]

I want to make clear when to use equation:
$$\mathbf{L}_P\ =\ I_{c.o.m.}\,\mathbf{\omega}\,+\, \mathbf{r}_{c.o.m.}\times m\mathbf{v}_{c.o.m.}$$
You have a disk rotating round point A, as in the picture.
Round points A, the center and point B:
$$L_A=mR^2(k+1) \cdot \omega$$
$$L_{Center}=kmR^2 \cdot \omega$$
$$L_B=kmR^2 \cdot \omega \,+\, mR\cdot\sqrt{2}R \cdot \omega$$

About B i am confused, since i don't understand the transition between the equations:
$$\mathbf{L}_{P}\ =\ \tilde{I}_{P}\mathbf{\omega}\ +\ m(\mathbf{r}_{c.o.m.}-\mathbf{r}_P)\times\mathbf{v}_P$$
And:
$$=\ \tilde{I}_{c.o.m.}\mathbf{\omega}\ +\ m(\mathbf{r}_{c.o.m.}-\mathbf{r}_P)\times\mathbf{v}_{c.o.m.}$$
In the first we use the moment of inertia and velocity of the moving point P, and in the second-of the center of mass.

 

[tiny-tim]

my computer won't open that .bmp, it says it's corrupt

 

[Karol]

I replaced

 

[tiny-tim]

Hi Karol!

I'm sorry for the delay … I've been working this out!

About B i am confused, since i don't understand the transition between the equations:
$$\mathbf{L}_{P}\ =\ \tilde{I}_{P}\mathbf{\omega}\ +\ m(\mathbf{r}_{c.o.m.}-\mathbf{r}_P)\times\mathbf{v}_P$$
And:
$$=\ \tilde{I}_{c.o.m.}\mathbf{\omega}\ +\ m(\mathbf{r}_{c.o.m.}-\mathbf{r}_P)\times\mathbf{v}_{c.o.m.}$$
In the first we use the moment of inertia and velocity of the moving point P, and in the second-of the center of mass.

You're right , that first equation is far too general …

it only applies if P is the centre of rotation and if the axis of rotation stays parallel to a principal axis of the body.

(So this applies, for example, to a sphere or a cylinder rolling over a step, but not to a cone rolling on a plane, or a wheel rolling on a curved rail.)

Thanks for pointing this out.

I've now corrected the Library entry (and given you the credit!).​