Answer to Question #114151 in Mechanics | Relativity for Feker

Question #114151
A 140kg hoop rolls along a horizontal floor so that the hoops center of mass has a speed of 0.150m/sec, how much work must be done on the hoop to stop it?
1
Expert's answer
2020-05-14T09:19:41-0400

The moment of inertia of a hoop is


"I=mr^2."


The rolling loop has kinetic energy of linear motion and rotation:


"EK=\\frac{1}{2}(mv^2+I\\omega^2)=\\frac{1}{2}(mv^2+mr^2\\omega^2)=mv^2."

According to work-energy theorem, the change from this value of kinetic energy to zero is equal to the work performed to stop the loop:


"W=mv^2=3.15\\text{ J}."

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