Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 15 - Exercises and Problems - Page 282: 75


$2.1\times10^{12} \ Nm$

Work Step by Step

We know the equation for torque: $\tau = rFsin\theta$ We found the force to be $6.64\times10^{10}N$ in problem 42. Since the integral of $r^2$ is $\frac{r^3}{3}$, we know that the radius should be divided by three, for $\frac{r^3}{3r^2}=\frac{r}{3}$. The angle is 90 degrees, so we obtain: $\tau = (\frac{95}{3})(6.64\times10^{10})sin90^{\circ}=2.1\times10^{12} \ Nm$
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