University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 5 - Applying Newton's Laws - Problems - Exercises - Page 163: 5.47

Answer

$T = 3.66~s$

Work Step by Step

Note that $v = \frac{2\pi~r}{T}$ At the bottom of the track: $\sum F = \frac{mv^2}{r}$ $F_N-mg = \frac{m(2\pi~r)^2}{T^2r}$ $2.50~mg-mg = \frac{m(4\pi^2~r)}{T^2}$ $1.50~g = \frac{4\pi^2~r}{T^2}$ $T^2 = \frac{4\pi^2~r}{1.50~g}$ $T = 2\pi~\sqrt{\frac{5.00~m}{(1.50)(9.80~m/s^2)}}$ $T = 3.66~s$
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