A uniform solid cube of mass M has edge length a. The moment of inertia of the cube about its edge diagonal will be
(a) (3)1/2/ 2 Ma2 (b) 1 / 2 Ma2 (c) 5 / 12 Ma2 (d) 7 / 12 Ma2
The ans. is given as c, Please reply soon.
Hey Sooraj, do u mean the diagonal of on of its faces? If so, We can use parallel axes theorem. MI = Ma2/6 + M(a/2)2 = 5/12 Ma2 .
I hope u understood.
Why M(a/2)2 is to be added, its not clear.
It's probably too late to answer, but, anyway, here goes...
Moment of Inertia of a cube about its centre of mass is Ma^2/6. But we want MI about one of its diagonals, which is situated at a dist. of a/2 from the CM. So, according to Parallel Axes theorem, I(abt. a given axis) = I(abt. CM) + Md^2. In this case, d = a/2. Got it?
But how is I(abt. CM) derived?
I(about CM) has been derived earlier .You can see it in older discussions.