Architecture
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The shape of the water surface in a rotating cylinder, when the water surface touches the bottom at the center

ÀÛ¼ºÀÚ Uploader : a.c.e. ÀÛ¼ºÀÏ Upload Date: 2019-10-11º¯°æÀÏ Update Date: 2023-07-06Á¶È¸¼ö View : 65

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.

y = (1/2)(¥ø^2/g)*x^2

*** Âü°í¹®Çå[References] ***

y = (1/2)*(¥ø^2/g)*x^2
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