Question #108071

Use conservation of energy to determine the angular speed of the spool shown in Figure P8.58 after the 3.00-kg bucket has fallen 4.00 m, starting from rest. The light string attached to the bucket is wrapped around the spool and does not slip as it unwinds.

Expert's answer

"\\frac{mv^2}{2}+\\frac{I\\omega^2}{2}=mgh"

"v=\\omega r" and "I=\\frac{1}{2}Mr^2" . So,

"\\frac{m(\\omega r)^2}{2}+\\frac{1\/2\\cdot Mr^2\\omega^2}{2}=mgh"

"\\omega=\\sqrt{\\frac{4mgh}{(2m+M)r^2}}=\\sqrt{\\frac{4\\cdot 3\\cdot 9.81\\cdot 4}{(2\\cdot3+5)0.6^2}}=10.9" "rad\/s"

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