Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 11 - Section 11.4 - Conservation of Angular Momentum - Example: 11.2

Answer

$2.5 \times 10^5 $ revolutions per day

Work Step by Step

We know angular momentum is conserved, so we find: $ \frac{2}{5}MR_1^2\omega_1=\frac{2}{5}MR_2^2\omega_2^2 \\ \omega_2 = \frac{R_1^2}{R_2^2}\times \omega_1 = \frac{(2\times 10^7)^2}{(6 \times 10^3)^2} \times \frac{1}{45} = 2.5 \times 10^5 $
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.