Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 8 - Dynamics II: Motion in a Plane - Exercises and Problems - Page 214: 61

Answer

$v=\sqrt{gy}$

Work Step by Step

First of all, we need to draw the force diagram of the ball as seen below. Now we can apply Newton's second law on the ball in the $x$-direction which is the radial direction here. $$\sum F_x=F_n\cos\theta=ma_r=m\dfrac{v^2}{R}$$ $$F_n\cos\theta= m\dfrac{v^2}{R}\tag 2$$ where $R$, from the two right triangles where one of them is part of the bigger one, is given by $$\dfrac{R}{a}=\dfrac{y}{h}$$ Thus, $$R=\dfrac{ya}{h}$$ Plugging into (2); $$F_n\cos\theta= m\dfrac{v^2h}{ya}\tag 3$$ Now we need to find the normal force which is given by $$\sum F_y=F_n\sin\theta-mg=ma_y=m(0)=0$$ Thus, $$F_n\sin\theta=mg$$ $$F_n=\dfrac{mg}{\sin\theta}$$ Plugging into (3); $$\dfrac{\color{red}{\bf\not} mg}{\sin\theta}\cos\theta= \color{red}{\bf\not} m\dfrac{v^2h}{ya} $$ $$ g\tan\theta= \dfrac{v^2h}{ya} $$ where $\tan\theta=\dfrac{h}{a}$; $$ g\dfrac{\color{red}{\bf\not} h}{\color{red}{\bf\not} a}= \dfrac{v^2\color{red}{\bf\not} h}{y\color{red}{\bf\not} a} $$ Therefore, $$\boxed{v=\sqrt{gy}}$$
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