Question

Use equationI=∫r2dmto calculate the moment of inertia of a uniform, solid disk with mass M and radius R for an axis perpendicular to the plane of the disk and passing through its center. Express your answer in terms of the variables M and R.

Answer 1

Answer:

$I = \frac{1}{2}MR^2$

Explanation:

Let say the ring is made up of small rings of radius "r" and thickness "dr"

so here we have

$dm = \rho (2\pi r dr)$

so we have

$dm = \frac{M}{\pi R^2}(2\pi r dr)$

$dm = \frac{M(2rdr)}{R^2}$

now from above formula we have

$I = \int r^2 dm$

$I = \int r^2(\frac{M(2rdr)}{R^2})$

so we have

$I = \frac{1}{2}MR^2$