Physics (10th Edition)

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

Chapter 6 - Work and Energy - Problems - Page 171: 86

Answer

$m_{max}=40.8kg$

Work Step by Step

As the rope swings, there is a centripetal force pointing towards the center of rotation. Contributing to this centripetal force, as we can see in the figure a) below, are the tension $T$ and $mg\cos\theta$, opposing each other. $$T-mg\cos\theta=\frac{mv^2}{r}$$ $$T=mg\cos\theta+\frac{mv^2}{r}$$ As the rope swings from its initial position, $\theta$ changes. We have $T_{max}$ when $\cos\theta=1$, or $\theta=0$. $$T_{max}=mg+\frac{mv^2}{r}$$ $$m\Big(g+\frac{v^2}{r}\Big)=800N (1)$$ We do not know $v$ and $r$, so we use the principle of energy conservation to find these. Here, the principle of energy conservation applies as gravitational force, which affects the rope's motion, is conservative. $$E_f-E_0=0$$ $$\frac{1}{2}m(v_f^2-v_0^2)+mg(h_f-h_0)=0$$ $$\frac{1}{2}(v^2-v_0^2)+g(h_f-h_0)=0$$ Figure b) below shows that the rope's motion makes an equilateral triangle, which, as a result, means $h_f-h_0=-r/2$. We also have $v_0=0$ and $g=9.8m/s^2$: $$\frac{1}{2}v^2+(9.8\times (-r/2))=0$$ $$v^2=9.8r$$ Plug this back to (1): $$m\Big(9.8+\frac{9.8r}{r}\Big)=800$$ $$m_{max}=40.8kg$$
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