College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 5 - Problems - Page 186: 18


The ball's tangential speed is $5.74~m/s$

Work Step by Step

The vertical component of the rope's tension is equal in magnitude to the ball's weight. We can find an expression for the tension $T$: $T~cos~\theta = mg$ $T = \frac{mg}{cos~\theta}$ The horizontal component of the tension provides the centripetal force to keep the ball moving around in a circle. We can find the ball's tangential speed: $T~sin~\theta = \frac{mv^2}{r}$ $(\frac{mg}{cos~\theta})~sin~\theta = \frac{mv^2}{r}$ $mg~tan~\theta = \frac{mv^2}{r}$ $g~r~tan~\theta = v^2$ $v= \sqrt{g~r~tan~\theta}$ $v= \sqrt{(9.80~m/s^2)(1.30~m)~sin~70.0^{\circ}~tan~70.0^{\circ}}$ $v = 5.74~m/s$ The ball's tangential speed is $5.74~m/s$.
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