I Explanation of parallel axis theorem

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The parallel axis theorem states that for a rotating system with mass m, the moment of inertia I about an axis parallel to the axis through its center of mass (CM) is given by I = I_CM + mx^2, where x is the distance from the CM. The discussion explores whether treating the object as a point mass at its CM when moving the axis x distance away is valid, suggesting it contributes an additional mx^2 to the moment of inertia. However, it is clarified that this approach is only applicable in specific contexts, such as gravitational fields outside the sphere. The moment of inertia of the sphere around its CM differs from that of a point particle, indicating that the simplification cannot be universally applied. Understanding these distinctions is crucial for accurate calculations in rotational dynamics.
Trollfaz
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For a rotating system with mass m this theorem says that if it rotates about an axis distance x from but parallel to the axis through it's natural mass center (CM), then I moment of inertia is
$$I=I_{CM}+mx^2$$
My thinking is if one move the axis x distance away from the axis through it's CM, and we can treat the object as a point mass at it's CM, then it's as though we are moving that point x distance away from the axis of rotation, contributing another ##mx^2## moment of inertia, is this explanation correct?
 
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So for instance I of sphere mass m is ##\frac{2}{5}mr^2## for radius=r. But in Newtonian mechanics, we can treat the sphere as a point mass in its geometrical center. Then if this axis of rotation is x away from it's CM, then the point mass is also x from the axis of rotation add another ##mx^2## to I. Assuming sphere is uniformly distributed in mass
 
Trollfaz said:
But in Newtonian mechanics, we can treat the sphere as a point mass in its geometrical center.
We most certainly cannot. Only for certain things such as the gravitational field outside the sphere does this hold.

In particular, the sphere has a moment of inertia around its CM - which the point particle does not.
 
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