In a long jump, an athlete leaves the ground with an initial angular momentum th

Question

In a long 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 0.940 s, one arm sweeps through 0.110 rev and the other arm sweeps through 0.220 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 is the magnitude of the total angular momentum of the arms around the common rotation axis through the shoulders?

Explanation / Answer

angular speed

W1=2pin1/t=2pi*0.11/0.94 =0.7353 rad/s

W2=2pin2/t =2pi*0.22/0.94=1.47 rad/s

Moment of inertia

I=I1=I2=(1/3)ML2=(1/3)*4*0.62=0.48 kg-m2

Total Angular momentum

L=I(W1+W2)=0.48*(1.47+0.7353)

L=1.06 kg-m2/s

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