Question

A 30-cm-diameter, 90-cm-high vertical cylindrical container is partially filled with 60-cm-high water. Now the cylinder is rotated at a constant angular speed of 180 rpm. Determine how much the liquid level at the center of the cylinder will drop as a result of this rotational motion.

Answer 1

Answer:

$\triangle h_c =0.204m$

Explanation:

Diameter $d=30cm$

Height $h=90cm$

Fill height $h_f=60cm$

Angular speed $N=180rpm$

Generally the equation for Angular velocity is mathematically given by

$\omega=\frac{2 \pi*N}{60}$

$\omega=\frac{2 \pi*180}{60}$

$\omega=18.85rads/s$

Generally the equation for Liquid surface is mathematically given by

$\mu_s=h*\frac{\omega^2*0.15^2}{4*9.81}$

$\mu_s=0.396m$

Therefore the liquid drop at center due to rotation is

$\triangle h_c =h-\mu_s$

$\triangle h_c =0.60-0.396$

$\triangle h_c =0.204m$