Engineering Mechanics: Statics & Dynamics (14th Edition)

Published by Pearson
ISBN 10: 0133915425
ISBN 13: 978-0-13391-542-6

Chapter 14 - Kinetics of a Particle: Work and Energy - Section 14.3 - Principle of Work and Energy for a System of Particles - Problems - Page 195: 5


$h=39.3m$, $\rho=26.2m$

Work Step by Step

We can determine the required height and the minimum radius of curvature as follows: $\frac{1}{2}mv_1^2+mgh=\frac{1}{2}mv_2^2$ We plug in the known values to obtain: $\frac{1}{2}(0)^2+9.81h=\frac{1}{2}(\frac{100\times 1000}{3600})^2$ This simplifies to: $\implies h=39.3m$ Now $\Sigma F_n=ma_n$ $\implies 39.24m-mg=m\frac{v^2}{\rho}$ We plug in the known values to obtain: $39.24-9.81=\frac{(\frac{100\times 1000}{3600})^2}{\rho}$ This simplifies to: $\rho=26.2m$
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