# Moment of inertia of a rolling cylinder

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.

• Does the cylinder have a uniform density? Is it solid? Nov 15, 2023 at 20:44

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 $$0$$.