MHB Moment of Inertia: Disc+Point Mass | Help Needed

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To find the moment of inertia of a disc with a point mass on its circumference, the moment of inertia of the disc is calculated as 1/2mr^2. The point mass contributes an additional moment of inertia of mr^2, where r is the radius of the disc. By applying the parallel axes theorem, the total moment of inertia is the sum of both contributions, resulting in I = 1/2mr^2 + mr^2. This simplifies to I = 11/2 ml^2, confirming the answer found in the textbook. Understanding these calculations is essential for accurately determining the moment of inertia in composite systems.
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Find the moment of inertia of a disc of mass m and radius r with a point mass of m on its circumference. i can't work this out can someone help me. i can work out that the moment of inertia of the disc is 1/2mr^2 but i don't know what to do with the point mass. i think it could have to do with some thing with the parallel axes theorem. the answer at the back of the book is 11/2 ml^2
 
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markosheehan said:
Find the moment of inertia of a disc of mass m and radius r with a point mass of m on its circumference. i can't work this out can someone help me. i can work out that the moment of inertia of the disc is 1/2mr^2 but i don't know what to do with the point mass. i think it could have to do with some thing with the parallel axes theorem. the answer at the back of the book is 11/2 ml^2

The moment of inertia is the sum of all mass contributions times their squared distances to the axis.
In formula:
$$I_z = \sum_i m_i r^2$$
or for a body:
$$I_z = \int r^2\,dm $$

It implies that we can sum the two contributions of moment of inertia together (with respect to the same axis).
And it also implies that a point mass $m$ rotating at distance $r$ from the axis has moment of inertia $mr^2$.
 
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