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I'm confused about this rolling without (or better

There are no forces acting on the disk besides the

Things are ok if I take as pivot point the center of mass.

$$\{\begin{matrix} - \mathbf{f} = m\mathbf{a_{CM}}\\ - \mathbf{r} \times \mathbf{f} =I_{cm} \mathbf{\alpha} \end{matrix}\tag{1}$$

But if I take the point ##O## on the ground, then the kinetic friction has zero torque.

$$\{\begin{matrix} - \mathbf{f} = m\mathbf{a_{CM}}\\ 0 =I_{O} \mathbf{\alpha} \end{matrix}\tag{2}$$

I assumed that the angular velocity (and so $\alpha$) is the same it I take as pivot the center of mass or the point##O##.

If this is the case than parallel axis theorem can be used and $$I_O=I_{cm}+m \mathbf{r}^2$$

But there is a contradiction since I get ##\alpha=0## from ##(2)## and ##\alpha\neq0## from##(1)##.

How can that be? Maybe##\alpha## is not the same in the two cases?

**with**) slipping situation. Suppose to have a disk with initial velocity ##v## and angular velocity ##\omega##. The motion is to the right but the angular velocity is counterclockwise.There are no forces acting on the disk besides the

**kinetic**friction##\mathbf{f}##.Things are ok if I take as pivot point the center of mass.

$$\{\begin{matrix} - \mathbf{f} = m\mathbf{a_{CM}}\\ - \mathbf{r} \times \mathbf{f} =I_{cm} \mathbf{\alpha} \end{matrix}\tag{1}$$

But if I take the point ##O## on the ground, then the kinetic friction has zero torque.

$$\{\begin{matrix} - \mathbf{f} = m\mathbf{a_{CM}}\\ 0 =I_{O} \mathbf{\alpha} \end{matrix}\tag{2}$$

I assumed that the angular velocity (and so $\alpha$) is the same it I take as pivot the center of mass or the point##O##.

If this is the case than parallel axis theorem can be used and $$I_O=I_{cm}+m \mathbf{r}^2$$

But there is a contradiction since I get ##\alpha=0## from ##(2)## and ##\alpha\neq0## from##(1)##.

How can that be? Maybe##\alpha## is not the same in the two cases?

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