Essential University Physics: Volume 1 (3rd Edition)

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

Chapter 5 - Exercises and Problems - Page 87: 28


$8.46 \ m/s$

Work Step by Step

We know the y-forces must cancel, for the ball remains in the horizontal plane. Thus, we find: $mg = F_t sin12$ We also know that the x-component of tension is the centripetal force: $F_tcos12 =\frac{mv^2}{r}$ Thus, we use substitution to find: $mg/tan12=\frac{mv^2}{r}$ $v=\sqrt{grtan12}=\sqrt{\frac{9.81\times1.55}{tan12}}=8.46 \ m/s$
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