Question
Ina along jump, an athlete leaves the ground with an initial angular momentum that tends to rotate his body forward, threatening to ruin his landing. To counter this tendency, he rotates his outstretched arms to "take up" the angular momentum (see the figure). In 0.537 s, one arm sweeps thro 0.807 rev and the other arm sweeps through 1.614 rev. Treat each arm as a thin rod of mass 4.0 kg and length 0.60 m, rotating around one end. In the athlete's reference frame, what the magnitude of the total angular momentum of the arms around the common rotation axis through the shoulders?
Explanation / Answer
Find the moment of inertia of each arm about the pivot point = m*L²/3. Then find the angular velocity of each arm.
If n is the no of revs in t seconds, the angular velocity is 2n/t. The total angular momentum of each arm is its angular velocity times its moment of inertia. The total angular momentum is the sum from both arms.