Answer to Question #260470 in Mechanics | Relativity for isaa

Question

The cylindrical tank containing water accelerates upward with az = 2 m/s². while it rotates about its vertical axis by n= 100 rpm. Determine the pressure at point A.

(there's a diagram, but I wish I could attach it).

Answer 1

$p_z\cdot \pi R^2=mg-ma_z$

pressure at point A in z-direction:

$p_z=\frac{m(g-a_z)}{\pi R^2}$

pressure at point A in radial direction:

$p_r=\frac{ma_x}{2\pi Rh}$

$a_r=\omega^2R=2\pi n=2\pi\cdot 100/60=10.5$ s^-1

$m=2\pi R^2h\rho=2\pi \cdot 0.3^2\cdot 1.2\cdot 1000=678.6$ kg

then:

$p_z=\frac{678.6(9.8-2)}{\pi \cdot 0.03^2}=1872044$ Pa

$p_r=\frac{678.6\cdot 10.5}{2\pi \cdot 0.3\cdot 1.2}=3150$ Pa

pressure at point A:

$p=\sqrt{p_r+p_z}=\sqrt{3150^2+1872044^2}=1872046.65$ Pa