Recent content by archaeosite

Thank you for the explanation. Your second paragraph especially helped!

Ah, I see what you mean. Now I've solved it. And as a courtesy to future answer-seekers, here's how. Use the kinematic equation delta-x = 0.5(v0-v)t 12.0 m = 0.5*v*4.0 s v = 6.00 m/s then, omega = 20.0 rad/s Plugged into the energy equation above, I = 4.09 kg*m2

But why not? There is no friction on the wheel from its bearing or mounting, and so the only external force acting on the block is gravity...?

Homework Statement You are asked to measure the moment of inertia of a large wheel for rotation about an axis perpendicular to the wheel at its center. You measure the diameter of the wheel to be 0.600 m. Then you mount the wheel on frictionless bearings on a horizontal frictionless axle at...