Question

Flat disk of radius R has circular hole of radius R/2 that is positioned as shown on the picture. Mass of the disk is m. Find moment of inertia I of such disk with respect to the axis going through the geometrical center of the disk O perpendicular to the disk’s plane

Answer 1

the picture of the system is in the attached file.

Answer 2

Calculate the moment of inerta of the disc with radius R and subtract the moment of inertia due to the R/2 disc with its axis on the circumference.

Answer 3

The mass of the missing portion = m*(r/2)^2/(r)^2 = m/4.
So,The moment of inertia = \[mr ^{2}/2 - [(m/4)*(r/2)^{2}/2+(m/4)*(r/2)^{2}]\] = \[mr ^{2}/2 - mr ^{2}/8 = 3mr^2/8\]

Answer 4

Sorry I made a mistake. (m/4)∗(r/4)2
the axis of the missing disc is shifted by r/4.
So the corrected answer is:
\[mr ^{2}/2−[(m/4)∗(r/2)^{2}/2+(m/4)∗(r/4)^{2}] \]
= \[mr ^{2}/2-5mr ^{2}/64=27mr ^{2}/64\]

Anyway verify the answer.

Answer 5

thank you!!

Answer 6

You're welcome.

Answer 7

can you explain how you get the mass of the cutout portion? what is that formula?

Answer 8

Let h be the thickness of the disk.
The density = m/(pi * R^2 * h)
So mass of cutout portion = density * (pi * (R/2)^2 * h)
= m* (pi * (R/2)^2 * h)/(pi * R^2 * h) = m/4.

Answer 9

(pi * (R/2)^2 * h) = volume of cutout portion.

Answer 10

thank you! but what if the disk is considered flat, so no 'h'? could we just leave it out?

Answer 11

but i think the disk is not flat and has some h. at least in the books the moment of inertia the disk has some h.

Answer 12

Yea but a perfectly flat disc is not possible.The height of the disc will get cancelled out anyway.No need to worry about it.