Architecture
°ÇÃà Architecture


The shape of the water surface in a rotating cylinder

ÀÛ¼ºÀÚ Uploader : a.c.e. ÀÛ¼ºÀÏ Upload Date: 2019-10-12º¯°æÀÏ Update Date: 2023-07-06Á¶È¸¼ö View : 1988

When a cylinder with radius R (m) rotates with angular velocity ¥ø (rad/s), the slope of the water surface at a distance x from the center of rotation is as follows.

Rotational acceleration: a = ¥ø^2*x
Gravity acceleration: g

Slope of the water surface: tan¥è = ¥ø^2*x/g

since, tan¥è = dy/dx = ¥ø^2*x/g

dy = ¥ø^2*x/g * dx

¡ò dy = ¡ò ¥ø^2*x/g * dx

y = (1/2)(¥ø^2/g)*x^2 + C

The case of C = 0 is when the water surface touches the bottom at the center.

Since C = H0, the water surface height y at a location x away from the center is as follows.

y = (1/2)(¥ø^2/g)*x^2 + H0

When x = R, y = HR, and HR = (1/2)(¥ø^2/g)*R^2 + H0.

Since the volume at the time of still water and at the time of rotation is the same,

¥ðR^2*H = ¥ðR^2*HR - V

Here, V is the volume of the part above water surface.

The volume of the rotating body rotated on the y-axis of the water surface is as follows.

x^2 = (2g/¥ø^2)(y-H0)

Integrate in the interval H0 ¡Â y ¡Â HR.

V = ¡ò ¥ðx^2 dy = ¥ð(2g/¥ø^2) ¡ò (y-H0) dy

 = ¥ð(2g/¥ø^2)[(1/2)y^2 - H0*y]

 = ¥ð(2g/¥ø^2)((1/2)(HR^2-H0^2) - H0*(HR-H0))

 = ¥ð(g/¥ø^2)(HR-H0)^2

thus,

¥ðR^2*H = ¥ðR^2*HR - ¥ð(g/¥ø^2)(HR-H0)^2

Substituting HR = (1/2)(¥ø^2/g)*R^2 + H0,

R^2*H = R^2*((1/2)(¥ø^2/g)*R^2 + H0) - (g/¥ø^2)((1/2)(¥ø^2/g)*R^2)^2

H = (1/2)(¥ø^2/g)*R^2 + H0 - (1/4)(¥ø^2/g)*R^2

H0 = H - (1/4)(¥ø^2/g)R^2

finally the water surface can be expressed as follows.

y = (1/2)(¥ø^2/g)*x^2 + H - (1/4)(¥ø^2/g)R^2

 = (1/2)(¥ø^2/g)*(x^2 - (1/2)R^2) + H


*** Âü°í¹®Çå[References] ***

y = (1/2)*(¥ø^2/g)*(x^2 - (1/2)*R^2) + H
º¯¼ö¸í Variable º¯¼ö°ª Value º¯ ¼ö ¼³ ¸í Description of the variable


¡Ø ÀÌ »çÀÌÆ®´Â ±¤°í¼öÀÍÀ¸·Î ¿î¿µµË´Ï´Ù.

¡Ú ·Î±×ÀÎ ÈÄ ¼ö½ÄÀÛ¼º ¹× Áñ°Üã±â¿¡ Ãß°¡ÇÒ ¼ö ÀÖ½À´Ï´Ù.
¡Ú To make new formula or to add this formula in your bookmark, log on please.


ÄÚ¸àÆ®

´ñ±Û ÀÔ·Â