College Physics (7th Edition)

Published by Pearson
ISBN 10: 0-32160-183-1
ISBN 13: 978-0-32160-183-4

Chapter 8 - Rotational Motion and Equilibrium - Learning Path Questions and Exercises - Exercises - Page 307: 41

Answer

$I=I_{CM}+M(\frac{1}{2})^{2}=\frac{1}{12}M(1)^{2}+\frac{1}{4}M=\frac{1}{3}M$ $T=\frac{Mg}{2}$ Thus, angular acceleration =$\frac{T}{I}=\frac{3g}{2}$ a). Tangential acceleration of the 100 cm position = $ 1\times \frac{3g}{2}=1.5g$ Yes, the result is surprising as it is more than g. b). $a_{t}=g$ $g=r \times \frac{3g}{2}$ Thus, $r=\frac{200}{3} cm=66.67cm$. Hence, tangential acceleration is g at position 66.67 cm.

Work Step by Step

$I=I_{CM}+M(\frac{1}{2})^{2}=\frac{1}{12}M(1)^{2}+\frac{1}{4}M=\frac{1}{3}M$ $T=\frac{Mg}{2}$ Thus, angular acceleration =$\frac{T}{I}=\frac{3g}{2}$ a). Tangential acceleration of the 100 cm position = $ 1\times \frac{3g}{2}=1.5g$ Yes, the result is surprising as it is more than g. b). $a_{t}=g$ $g=r \times \frac{3g}{2}$ Thus, $r=\frac{200}{3} cm=66.67cm$. Hence, tangential acceleration is g at position 66.67 cm.
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.