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It is obvious that there is no change in the mass of it and its radius. But the shape of the object does change. Does it mean its moment of inertia will also change?

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I think the answer to your question depends on how one chooses to model a cut.

The moment of inertia of a two-dimensional object is

$$I= \int_\text{disk} r^2 \rho(r,\theta) r\,dr\,d\theta,$$

where $\rho$ is the areal (two-dimensional) mass density.

If your model for what "cutting" means leaves all of the relevant quantities (which I suppose are the area of integration and $\rho$) unchanged, then the moment of inertia would not change. If, however, cutting does affect some quantity, then the moment would likely change.

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No change of moment of inertia would happen, if the axis of rotation is down the centre of the disk, and the spiral does not expand outwards as it is rotated.

if $I = \xi^2 M$ where $\xi$ is the moment-operator $\xi X = \int r \; dX$. Because the spiral would keep $r$ unchanged over elements of $M$, the size of the inertia would also be constant.

Were the spiral to expand outwards, then the moment of inertia would change, probably involving some $v^2$ term.

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