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

`\triangle h_c =0.204m`

Step-by-step 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=(2 \pi*N)/(60)`

`\omega=(2 \pi*180)/(60)`

`\omega=18.85rads/s`

Generally the equation for Liquid surface is mathematically given by

`\mu_s=h*(\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`