Here is the question:
"A rigid body with a cylindrical cross-section is released from the top of a $30^{\text{o}}$ incline. It rolls $10.0 \, \text{m}$ to the bottom in $2.60 \text{s}$. Find the moment of inertia of the body in terms of its mass $m$ and radius $r$."
What I don't quite understand is why you couldn't just use the normal formula for the moment of inertia for a solid cylinder rotating around its cylindrical axis. Then it would be $0.5 m r^{2}$. But using the formula for an object rolling down the incline without slipping gives $0.66 m r^{2}$. I don't understand conceptually why it's different.
Best Answer
The formula $I=\frac{1}{2}mr^2$ for the moment of inertia of a solid cylinder is only valid, if its density is homogeneous. But it is not valid anymore, if the density varies with radius.
For example: If its mass is concentrated near the outer edge, then the moment of inertia is closer to $mr^2$. And if its mass is concentrated near the axis, then the moment of inertia is closer to $0$.