College Physics (4th Edition)

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

Chapter 5 - Problems - Page 190: 85


The angular speed is $1.04~rad/s$

Work Step by Step

We can find the rotation radius for each car: $r = 6.00~m+(4.25~m)~sin~45^{\circ}$ $r = 9.0~m$ Let $m$ be the mass of one car. Let $T$ be the tension in the cable attached to one car. The vertical component of tension is equal in magnitude to the car's weight: $T~cos~45^{\circ} = mg$ $T = \frac{mg}{cos~45^{\circ}}$ The horizontal component of the tension provides the centripetal force to keep the car moving in a circle. We can find the angular speed $\omega$: $T~sin~45^{\circ} = m~\omega^2~r$ $(\frac{mg}{cos~45^{\circ}})~sin~45^{\circ} = m~\omega^2~r$ $g~tan~45^{\circ} = \omega^2~r$ $\omega^2 = \frac{g~tan~45^{\circ}}{r}$ $\omega = \sqrt{\frac{g~tan~45^{\circ}}{r}}$ $\omega = \sqrt{\frac{(9.80~m/s^2)~tan~45^{\circ}}{9.0~m}}$ $\omega = 1.04~rad/s$ The angular speed is $1.04~rad/s$
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