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 it…

Question

asked Oct 28, 2020 219k views

1 Answer

Anand Vidvat · Answer 1 · 2020-11-02T10:45:02+0000

Answer:

I = (1)/(2)MR^2

Step-by-step 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

$dm = (M)/(\pi R^2)(2\pi r dr)$

dm = (M(2rdr))/(R^2)

now from above formula we have

$I = \int r^2 dm$

$I = \int r^2((M(2rdr))/(R^2))$

so we have

I = (1)/(2)MR^2