- #1
kingkong69
- 22
- 0
http://img20.imageshack.us/img20/9443/ssssnm.png
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 a lot!
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 a lot!
Last edited by a moderator:
… 