A m=0.24kg mass is attached to end of a thin uniform rod of mass M =1.22kg and length L=4.38m which is positioned vertically. The rod is pivoted at point O thatis L/4 away from its center. The system is released and allowed to fall. Determine the speed of center of mass of the rod when the rod passes the horizontalposition. Take g=9.81m/s 2 ,Irod_12ML and express your answer using two decimal places.12CML/2LL/4pivotm

m=0.24 kg M=1.22 kg L=4.38 m Since, we knowEi=3L4mg+L4mgEf=12mv2+12Iω2+-3L4mg+-L4Mgω=v/L'and…

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