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

[markosheehan]

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

 

[I like Serena]

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 $mr^2$.