Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 10 - Rotation - Questions - Page 286: 3b

Answer

Work done on discs will be same. $W_{a}=W_{b}=W_{c}=W_{d}=\frac{21}{2}\times I$

Work Step by Step

Work done on the disc will be: $W=\Delta K=K_{f}-K_{i}=\dfrac {1}{2}I\left( w^{2}_{f}-w^{2}_{i}\right) $ $I$ is the moment of inertia of the disc, and $w_{f}$ and $w_{i}$ are the final and initial angular speeds of the disc, respectively. Plugging in the known values, we obtain: $a)w_{f}=5\dfrac {rad}{s};w_{i}=-2\dfrac {rad}{s}$ we get $W_{a}=\dfrac {21}{2}I$ $b)w_{f}=5\dfrac {rad}{s};w_{i}=2\dfrac {rad}{s}$ we get $W_{b}=\dfrac {21}{2}I$ $c)w_{f}=-5\dfrac {rad}{s};w_{i}=-2\dfrac {rad}{s}$ we get $W_{c}=\dfrac {21}{2}I$ $d)w_{f}=5\dfrac {rad}{s};w_{i}=2\dfrac {rad}{s}$ we get $W_{d}=\dfrac {21}{2}I$ So, we see that in all 4 conditions, the work done on the disc will be same $W_{a}=W_{b}=W_{c}=W_{d}$
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.