Answer to Question #157124 in Mechanics | Relativity for wilder

Question

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)

Answer 1

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

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

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

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