Giancoli's Physics: Principles with Applications, 6th Edition

8

Rotational Motion

Change chapter8-1: 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

$KE = 5 \times 0^{16} \textrm{J}$

$L = 3 \times 10^{20} \textrm{kgm}^2\textrm{/s}$**Note:** At 3:23 I mention correctly that the angular velocity is $3.333 \times 10^{-4}$, whereas at 3:46 I incorrectly wrote $3.333 \times 10^{-11}$. My calculations were nevertheless done using $3.333 \times 10^{-4}$, so the results are correct. Where you hear $3.333 \times 10^{-11}$, please think: $3.333 \times 10^{-4}$.

$L = 3 \times 10^{20} \textrm{kgm}^2\textrm{/s}$

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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 $3.333 \times 10^{-4}$, whereas at 3:46 I incorrectly wrote $3.333 \times 10^{-11}$. My calculations were nevertheless done using $3.333 \times 10^{-4}$, so the results are correct. I'll put a note above the video to use $3.333 \times 10^{-4}$ for the angular velocity.

All the best,

Mr. Dychko

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