Answer to Question #157124 in Mechanics | Relativity for wilder

Question #157124
A tyre of mass 15kg is suspended on a rope of length 7.5m and a girl of mass 75kg sits in the tyre. If the girls speed at the lowest point as she swings is 5m/s and the rope extends by 20cm, find the radius of this rope(young modulus = 2.0 x 10^11 Pa)
1
Expert's answer
2021-02-02T09:32:27-0500

From the definition of the Young's modulus, we have:


E=FTL0AΔL=mgL0πr2ΔL.E=\dfrac{F_TL_0}{A\Delta L}=\dfrac{mgL_0}{\pi r^2\Delta L}.

From this formula we can find the radius of the rope:


r=mgL0πEΔL,r=\sqrt{\dfrac{mgL_0}{\pi E\Delta L}},r=90 kg9.8 ms27.5 mπ2.01011 Pa0.2 m=2.29104 m.r=\sqrt{\dfrac{90\ kg\cdot9.8\ \dfrac{m}{s^2}\cdot7.5\ m}{\pi\cdot2.0\cdot10^{11}\ Pa\cdot0.2\ m}}=2.29\cdot10^{-4}\ m.

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