Derivation of Moment of Inertia Equation for Solid Discs

[kelli105]

This isn't quite a homework question, but my calculus teacher mentioned to those of us also taking physics that it was possible to prove that I = .5MR^2 using calc. I had some extra time on my hands and decided to give it a try. I've tried doing a summation with a geometric series but then ran into trouble defining a and r [using the equation a/(1-r)]. Does anyone out there know how to do this?

 

[mjsd]

go back to the definition of how moment of inertia is defined for an arbitrary mass distribution of an object... you may find that it is given by something like (off top of my head only)
$$I=\sum_i M_i R_i^2$$
where $$M_i$$ is a mass element at perpendicular distance $$R_i$$ from the axis of rotation.

there should also be a continuous version in your book...if not derive it yourself by introducing $$\rho$$ (density) such that
$$\Delta M_i = \rho \Delta V$$
and when $$\Delta V$$ goes to zero you sum will become an integral.

hint: before tackling the disc, try a hoop first. then you may use that result to help you solve the disk

 

[P3X-018]

As mjsd mentioned there is also an expression for the contineous version, which should be

$$I_{CM} = \int_M r^2\,\textrm{d}m$$

Where r is the distance from center of the disk to the infinitesimal mass dm. If $\sigma$ is the mass pr. unit area, then assuming that there is a uniform distribution of mass over the disk, $\sigma$ would be constant. Change the above integral so that you integrate with respect to the area instead of mass. You would then have a multiple integral over the area corresponding the to surface area of the disk. Using polar coordinates this would be much easier to solve, so you should change from cartesian to polar coordinates (while remembering the Jacobian). You're integrations limits would change to something much simpler.
Don't worry about the $\sigma$, you can later replace an expression including $\sigma$ with the mass M.