Giancoli 7th Edition textbook cover
Giancoli's Physics: Principles with Applications, 6th Edition
8
Rotational Motion
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8-1: Angular Quantities
8-2 and 8-3: Constant Angular Acceleration; Rolling
8-4: Torque
8-5 and 8-6: Rotational Dynamics
8-7: Rotational Kinetic Energy
8-8: Angular Momentum
8-9: Angular Quantities as Vectors

Problem 64
A
KE=5×016JKE = 5 \times 0^{16} \textrm{J}
L=3×1020kgm2/sL = 3 \times 10^{20} \textrm{kgm}^2\textrm{/s} Note: At 3:23 I mention correctly that the angular velocity is 3.333×1043.333 \times 10^{-4}, whereas at 3:46 I incorrectly wrote 3.333×10113.333 \times 10^{-11}. My calculations were nevertheless done using 3.333×1043.333 \times 10^{-4}, so the results are correct. Where you hear 3.333×10113.333 \times 10^{-11}, please think: 3.333×1043.333 \times 10^{-4}.
Giancoli 6th Edition, Chapter 8, Problem 64 solution video poster
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COMMENTS
By thesouthportschool on Sat, 2/18/2017 - 11:37 PM

Is the answer not 453.6926 J or 5*10^2 J, I copied the exact calculations from your final answer of part A (3:49) and did not get 5*10^16 J.

By thesouthportschool on Sat, 2/18/2017 - 11:43 PM

And I also you changed the angular velocity from 3.333*10^-4 to 3.333*10^-11 when doing the calculations to work out the KE

By Mr. Dychko on Wed, 2/22/2017 - 8:37 AM

Hello student from thesoutportschool, thank you for spotting the error. At 3:23 I mention correctly that the angular velocity is 3.333×1043.333 \times 10^{-4}, whereas at 3:46 I incorrectly wrote 3.333×10113.333 \times 10^{-11}. My calculations were nevertheless done using 3.333×1043.333 \times 10^{-4}, so the results are correct. I'll put a note above the video to use 3.333×1043.333 \times 10^{-4} for the angular velocity.

All the best,
Mr. Dychko

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