Essential University Physics: Volume 1 (3rd Edition)

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

Chapter 13 - Exercises and Problems: 31

Answer

1.64 seconds

Work Step by Step

Substituting the value of $I=\frac{4}{3}mL^2$ into the equation, we find that the following equation gives the period of the meter stick: $T = 2\pi\sqrt{\frac{\frac{4}{3}mL^2}{mgL}}$ Recall, the length is equal to half of the length of the meter stick, which is half of a meter. Thus, we find: $T = 2\pi\sqrt{\frac{\frac{4}{3}mL^2}{mgL}}$ $T = 2\pi\sqrt{\frac{\frac{4}{3}L}{g}}$ $T = 2\pi\sqrt{\frac{\frac{4}{3}(.5)}{9.81}}=\fbox{1.64 seconds}$
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