Question

calculate the moment of inertia of a circular solid disk of radius R from which another disk from the top most part of radius R/3 has been cut out

Answer 1

The axis about which MI is to be found passes through center???

Answer 2

yes through the center

Answer 3

Do u know the answer???

Answer 4

yes its 4MR^2 where M is the mass of the solid disk

Answer 5

The answer should be (4MR^2)/9

Answer 6

ohh sorry its mass is 9M

Answer 7

please answer I am sorry for the mistake

Answer 8

yup then my answer is correct

See, MI = (MR^2)/2 (normal disc)
\[Let\ \sigma\ \]be the Mass per unit area

So, \[\sigma = M/\pi r ^{2}\]
So the mass of disk given becomes 8/9(M).

So acc. to formula \[[8/9(M) \times \pi r ^{2}]/2 = 4/9Mr ^{2}\]

Answer 9

Your mass M=9M

Answer 10

yeah

Answer 11

from did (8/9)M come from

Answer 12

ur area is pi(r^2) - pi{(r/3)^2} = 8/9(pi)(r^2)

Answer 13

whats the formulae u used for last step

Answer 14

MI = (MR^2)/2 (for a disc)

Answer 15

why did you multiply pi

Answer 16

ok thanks