1) Calculate the moment of inertia of a CD, including the effect of the hole. For a CD of radius 6.0 cm, estimate the percentage change in the moment of inertia due to a hole of radius 7 mm. (Neglect the difference in mass due to the hole.)


Seems simple but I can't figure it out

The CD's inertia is 1/2mR^2

m = mass
R = radius

you need to get the difference of the moment of difference from the inertia of the CD w w/o hole

CD's radius itself is 1/2 mR^2
hole, let it be 1/2mr^2


moment of inertia of CD = .5 (M/R^2) - .5 (M/r^2) = .5m (R^2 - r^2)
you'll need to know the mass though I think unless there's something else?


do some math
you get the difference of the CD w/o hole = .5 (R^2 - r^2) divided by the CD itself which is 1/2mR^2

Remove the .5's and mass and you get (R^2-r^2) / R^2

--> 1 - r/R^2 = 1 - (7/0.6)^2 = 0.986388889 = 98.6%

Subtract from 100 which is the actual value of the disk inertia without the hole

100 - 98.6/98.63/99 = 1.4%

yeee, thats what I got the first time I did it, but its wrong.

Heh don't know what's wrong then

Physics might be too advanced for me, I don't know what level ur taking

Moment of inertia is pi/4*r^4, not r^2

Since pi/4 cancel out and you only want %diff,

Difference is (7mm^4)/(60mm^4) = 0.00018526234 = 0.0185%

Total moment of inertia would be (60mm^4-7mm^4) = 12,957,599mm^4

We assume you are taking moment of inertia about its centroid (center of disk).



Just have to keep in mind that moment of inertia is always in units^4.

This post was edited by Dontrunaway on Oct 31 2013 09:31pm

second moment of inertia*

Meh, I only use second moment of inertia.