Angular Momentum of Ice Skater: 881.93kg m2/s

AI Thread Summary
The discussion revolves around calculating the initial angular momentum of an ice skater with a mass of 55.5 kg, who is making one revolution every 5.44 seconds while holding onto a 3.75 m rope. The calculated moment of inertia is approximately 780.47 kg m², and the angular velocity was initially found to be 1.13 rad/s, later corrected to about 1.15 rad/s. The skater's angular momentum was determined to be 881.93 kg m²/s, which is not among the provided answer options. The discrepancy was attributed to rounding errors during calculations. The conversation highlights the importance of precision in mathematical computations.
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Homework Statement


An ice skater with a mass of 55.5kg is holding on to the end of a 3.75 m long rope tha is attached to a pole protruding out of a frictionless ice covered pond. Intially the skater is making one revolution every 5.44seconds.
What is the initial angular momentum of the skater?

A) 780kg m2/s
B) 901kg m2/s
c) 11,200 kg m2/s
d) 18600 kg m2/s

Homework Equations





The Attempt at a Solution


I have found the moment of interia, which i found to be 780.47kg m2
I found the angular velocity to be 1.13rad/s
The answer i get is not one of the options. I get 881.93kg m2/s

 
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It's just a rounding error.
 
What step is my error in? i have been redoing to equations and keep getting the same
 
Angular velocity is more like 1.15 rads/sec. How did you get 1.13? 1 rev=2*pi, and 2*pi/5.44=1.155 rad/sec
 
i did x/1=1/5.44 to get .18revolutions a second and then multiplied that by 2pie to convert to radians a second
i found my error while doing this it was a rounding error thanks for the help
 
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