Problem 9.37-MC Part A Ice skaters often end their performances with spin turns,

Question

Problem 9.37-MC Part A Ice skaters often end their performances with spin turns, where they spin very fast about their center of mass with their arms folded in and legs together Upon ending, their arms extend outward proclaiming their finish. Not quite as noticeably one leg goes out as well Suppose that the initial moment of inertia of a skater for arms and legs in is 0.70 kg- m2 while with the final moment of inertia with arms out and one leg extended is 3.2 kg·m2 . If she starts out spinning at 5.8 rev/s, what is her angular speed (in rev/s) when her arms and one leg open outward? Express your answer using three significant figures. Hints w 127 rev/s EEE11m MyAnswer Give Up Correct Part B- How much thermal energy was created while she was spinning? Express your answer using three significant figures. Hints O

Explanation / Answer

Part A -

Consider the following -

Ii = Initial moment of inertia

wi = initial angular velocity

If = Final moment of inertia

wf = final angular velocity

Now apply conservation of angular momentum -

Ii * wi = If * wf

=> 0.70 * 5.8 = 3.2 * wf

=> wf = (0.70 * 5.8) / 3.2 = 1.27 rev/s

Part B-

Convert the angular velocity in rad/s -

wi = 5.8 rev/s = (5.8*2*3.141) = 36.44 rad/s

wf = 1.27 rev/s= (1.27*2*3.141) = 7.98 rad/s

So, initial kinetic energy, KEi = (1/2)*Ii*wi^2 = 0.5*0.70*36.44^2 = 465 J

Final kinetic energy, KEf = (1/2)*If*wf^2 = 0.5*3.2*7.98^2 = 102 J

Therefore, amount of thermal energy created = KEi - KEf = 465 - 102 = 363 J

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