Question

A glass ball of radius 1 cm is placed in a vertical cylindrical vessel of radius 10 cm filled with water and rotating at an angular velocity Ω about the axis of symmetry of the cylinder. After equilibration, the vessel, the water in the vessel and the ball rotate at the same angular velocity 20 rad/s, the ball being on the bottom at the vertical wall of rotating vessel. Find the force with which the ball acts on the vessel. Glass density is 2,5 g/cm^3, water density is 1 g/cm^3. Free fall acceleration is 10 m/s^2. Give your answer in milli- newtons and round it up to the whole.

Answer 1

$$F=\sqrt{\left(mg\right)^2+\left(\frac{4}{3}\pi r^3(\rho-\rho_w)\omega^2R\right)^2}\\F=\sqrt{\left(\frac{4}{3}\pi r^3\rho g\right)^2+\left(\frac{4}{3}\pi r^3(\rho-\rho_w)\omega^2R\right)^2}\\F=\frac{4}{3}\pi (0.01)^3\sqrt{\left(2500(10)\right)^2+\left((2500-1000)(20)^2(0.1)\right)^2}$$

$$F=0.27\ N$$