Physics (10th Edition)

Published by Wiley
ISBN 10: 1118486897
ISBN 13: 978-1-11848-689-4

Chapter 5 - Dynamics of Uniform Circular Motion - Problems - Page 141: 61


(a) $T=22.6N$ (b) $T=77.4N$

Work Step by Step

Without upward lift, the guideline slopes downward with angle $\theta$. As shown in the figure below, tension $T$ in the string has 2 components: - Vertical component $T\sin\theta$ balancing the plane's weight $mg$ $$T\sin\theta=mg=0.9\times9.8=8.82N (1)$$ - Horizontal component $T\cos\theta$ providing the centripetal force $$T\cos\theta=F_c=\frac{mv^2}{r}$$ Because the guideline slopes, the radius $r$ of the rotation circle this time does not equal $17m$, however. Instead, $r=17\cos\theta$ $$T\cos\theta=\frac{0.9v^2}{17\cos\theta}$$ $$T\cos^2\theta=0.053 v^2 (2)$$ Using these two equations, we can find $T$. (a) For $v=19m/s$, equation (2): $T\cos^2\theta=19.13N$ Dividing (2) over (1), we have $$\frac{T\cos^2\theta}{T\sin\theta}=\frac{1-\sin^2\theta}{\sin\theta}=\frac{19.13}{8.82}=2.17$$ $$1-\sin^2\theta=2.17\sin\theta$$ $$\sin^2\theta+2.17\sin\theta-1=0$$ $$\sin\theta=0.3905$$ $$T=\frac{8.82}{\sin\theta}=22.6N$$ (b) For $v=38m/s$, equation (2): $T\cos^2\theta=76.53N$ We carry out the same steps as in (a) and get $\sin\theta=0.114$ Therefore, $$T=\frac{8.82}{\sin\theta}=77.4N$$
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