## finding the moment of inertia of a uniform square lamina

parallel and perpendicular axis theorem for moment of inertias

So i solved the Moment of inertia for the large square through the perpendicular axis through a,

(1/3)*M*(l^2), where l is 4a/2=2a,
using the perpendicular theorem, Ixx+Iyy=Izz,
we have (4/3)*M*(a^2)+(4/3)*M*(a^2)=(8/3)*M*(a^2),

then using the parallel theorem, I,+md^2=I.,
d is the distance AO, which is sqrt.(8)*a therefore d^2= 8a^2
we get (8/3)*M*(a^2) + M*8*(a^2)=(32/3)M*(a^2)

Now I will subtract the moment of inertia of the 2 small squares from the big square's moment of inertia we got.

The small squares moment of inertia through its perpendicular centre is (m/3)*(a/2)^2=ma^2/12, and m is M/9 therefore it is M*a^2/108,

The axis is at vertex A, so we apply the parallel axis theorem, d^2 = 12.5, so we get 25Ma^2/2 + Ma^2/108=1351Ma^2/108

The two small squares are identical so it is 1351Ma^2/54

subtracting we got a different result

Any help is appreciated thanks alot!
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hi kingkong69!
 Quote by kingkong69 … m is M/9 …
nooo
m is M/16
 Hi tiny-tim! Thanks for pointed my mistake out, is the rest correct?

## finding the moment of inertia of a uniform square lamina

Alright I found it! Thanks a ton again!
 (1/3)*M*(l^2), where l is 4a/2=2a, isn't the l here is 32a^2
 why are you using the formula 1/3ml^2 inspite of 1/6ml^2?Then why have u taken the a as 4a/2.Please explain and if u got the right answer,please explain it here.

 Quote by Gauranga why are you using the formula 1/3ml^2 inspite of 1/6ml^2?Then why have u taken the a as 4a/2.Please explain and if u got the right answer,please explain it here.