A 3.2 kg solid disk with a radius of 0.45 m has a tangential force of 420.4 N applied to it. What is the moment of inertia applied to the disk

Explanation:

Your question makes no sense.

moment of inertia is a property of the disk and its geometry.

The moment of inertia of a uniform solid disk around an axis through its geometric center and perpendicular to its flat ends is

I = ½mR² = ½(3.2)0.45² = 0.324 kg•m²

the applied torque about the same axis would be

τ = FR = 420.4(0.45) = 189.18 N•m

and the angular acceleration about the same axis would be

α = τ/I = 189.18/0.324 = 583.9 rad/s²

answer: i'm not sure which is which for i get confused on it all the time but i can tell you that the answer lies with the convex and concave devices. i hope this you. have a good day and best of luck to you!

explanation: