f we multiply all design dimensions of an object by a scaling factor f, its volume and mass will be multiplied by f^3.

Part A:
By what factor will its moment of inertia be multiplied?

Question

f we multiply all design dimensions of an object by a scaling factor f, its volume and mass will be multiplied by f^3.

Part A: 
By what factor will its moment of inertia be multiplied?

fellowroot · Answer

Someone jump in if I'm wrong, but doesn't it depend on the type of object you are talking about. A thin hoop for instance has a different moment of inertia than a sphere.

anonymous · Answer

Let's see from the definition of moment of inertia, $I=\int r^2 dm$ (or just use $I=mr^2 $ for this purpose). If we multiply r by f, and m by $f^3$, $I'=fm(f^3r)^2=f^7mr^2$. The new moment of inertia is f^7 times the initial.

anonymous · Answer

Sorry, I'=f^3m(fr)^2=f^5 mr^2
$I'=f^5I$