Question

On average, both arms and hands together account for 13% of a person's mass, while the head is 7.0% and the trunk and legs account for 80%. We can model a spinning skater with her arms outstretched as a vertical cylinder (head, trunk, and legs) with two solid uniform rods (arms and hands) extended horizontally. Suppose a 70.0 kg skater is 1.80 m tall, has arms that are each 64.0 cm long (including the hands), and a trunk that can be modeled as being 32.0 cm in diameter. If the skater is initially spinning at 68.0 rpm with her arms outstretched, what will her angular velocity wy be (in rpm) after she pulls in her arms and they are at her sides parallel to her trunk? Assume that friction between the skater and the ice is negligble. 96.48

Answer 1

Solution is 100 74- Xto 100 mass 7 arms and hard-13% -70713 - 19.1kg Jm moss of head= 74 - 7*70 = 14 g kg = m2). mass of trunk and legs-gov. - 80x 70 = 156 kg = m3) height 7 &keter=diom cholg station 1,5 m radius= 1s = 10.75 m) = Initial rate of rotation = 0,=68.0pm . and mass = M3= 56 kg at initial stage tatal angular moment = W, XrX (M3tma) at final stage ) lis 2 . and mobo m2=4og ke sum = W₂ xrx (mit M2 +M3) using conservation of ongular momentuna 5 W,X7X (M2+ m2) = WzxDx 1 mxm x + m) > W2= 688 60.9 70 W2 = 59.16 opm