on average both arms and hands together account or 13% of a person\'s mass while
ID: 1781076 • Letter: O
Question
on average both arms and hands together account or 13% of a person's mass while the head is 7 0 and the trunk an legs account fo 80% can model a spinning skater with his arms outstretched as a vertical cylinder (headtrunk + legs) with two solid uniform rods (arms + hands) extended horizontally. Suppose a 63-kg skater is 1.9 m tall, has arms that are each 63 cm long (including the hands), and a trunk that can be modeled as being 30 cm in diameter. If the skater is initially spinning at 68 rpm with his arms outstretched what will his angular velocity be (in rpm) when he pulls in his arms until they are at his sides parallel to his trunk? rpm we can model a spinning skater of mass M as cylinders put together-the head + trunk + legs as a vertical cylinder with mass Mtorso 0.87M and dtorso = 30 cm and each arm + hand as a thin rod with mass Marm-0.065M and length L = 63 cm. The skater starts spinning at an angular speed of = 68 rpm with his arms outstretched. The moment of inertia in this case is the sum of the moment of inertia for the torso cylinder and the two moments of inertia for the arms rotated about an axis that is (-m + Rtorso) from their centers; this means we need to use the parallel-axis theorem for the arms. The final moment of inertia is the sum of the moment of inertia for the torso cylinder, which is the same as before, and the two arm rods by the skater's side. Because we are modeling the arms as thin rods, each arm contributes MarmRtorso to the final moment of inertia. Finally, since angular momentum is conserved, we can put all of this information together and find the skater's final angular speedExplanation / Answer
Mtorso = mass of trunk + mass of legs + head = 0.8M + 0.07M = 0.87 M = 0.87*63 = 54.81 kg
mass of each arm = Marm = 6.5% = 0.065 M= 0.065*63 = 4.095 kg
initial moment of inerta
Iinitial = (1/2)*Mtorso*r^2 + 2*( (1/12)*Marm*L^2 + Marm*(r+L/2)^2 )
Iinitial = (1/2)*54.81*0.15^2 + 2*( (1/12)*4.095*0.63^2 + 4.095*(0.15+0.63/2)^2) = 2.66 kg m^2
initial angular momentum Li = Iinitial *wi = 2.66*68
final moment of inertia
Ifinal = (1/2)*Mtorso*r^2 + 2*Marm*r^2
Ifinal = (1/2)*54.81*0.15^2 + 2*4.095*0.15^2= 0.8 kg m^2
final angular momentum Lf = Ifinal *wf
from momentum conservation
Lf = Li
0.8*w2 = 2.66*68
w2 = 226.1 rpm
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