Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 8 - Dynamics II: Motion in a Plane - Exercises and Problems - Page 201: 47


(a) $T = \frac{mgL}{\sqrt{L^2-r^2}}$ (b) $\omega = \sqrt{\frac{T}{m~L}}$ (c) $T = 5.0~N$ $\omega = 30.2~rpm$

Work Step by Step

Let $\theta$ be the angle that the string makes with the vertical. Note that $cos(\theta) = \frac{\sqrt{L^2-r^2}}{L}$ (a) We can find an expression for the tension $T$ in the string. The vertical component of the tension is equal to the ball's weight. $T~cos(\theta) = mg$ $T~(\frac{\sqrt{L^2-r^2}}{L}) = mg$ $T = \frac{mgL}{\sqrt{L^2-r^2}}$ (b) The horizontal component of the tension provides the centripetal force to keep the ball moving in a circle. $T~sin(\theta) = m~\omega^2 ~r$ $T~(\frac{r}{L}) = m~\omega^2 ~r$ $\omega^2 = \frac{T}{m~L}$ $\omega = \sqrt{\frac{T}{m~L}}$ (c) We can find the tension in the string. $T = \frac{mgL}{\sqrt{L^2-r^2}}$ $T = \frac{(0.500~kg)(9.80~m/s^2)(1.0~m)}{\sqrt{(1.0~m)^2-(0.20~m)^2}}$ $T = 5.0~N$ We can find the angular speed. $\omega = \sqrt{\frac{T}{m~L}}$ $\omega = \sqrt{\frac{5.0~N}{(0.500~kg)(1.0~m)}}$ $\omega = 3.16~rad/s$ We can convert the angular speed to units of rpm. $\omega = (3.16~rad/s)(\frac{1~rev}{2\pi ~rad})(\frac{60~s}{1~min})$ $\omega = 30.2~rpm$
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