An open vessel partly filled with a liquid rotates about a vertical axis at constant angular velocity. Determine the equation of the free surface of the liquid after it has acquired the same angular velocity as the vessel.

Question

Accepted Answer

Figure 5-5(a) represents a section through the rotating vessel, and any particle A is at a distance x from the axis of rotation. Forces acting on mass A are the weight W vertically downward and P, which is normal to the surface of the liquid since no friction is acting. The acceleration of mass A is xω², directed toward the axis of rotation. The direction of the resultant of forces W and P must be in the direction of this acceleration, as shown in Fig. 5-5(b).

From Newton's second law, [latex]F_{x}=Ma_{x} [/latex], or [latex]P\sin heta =\frac{W}{g}{x}\omega ^2 [/latex] (1)

From [latex]\sum{Y=0} [/latex] , [latex]P\cos heta =W [/latex] (2)

Dividing (1) by (2), [latex] an heta =\frac{x\omega ^2}{g} [/latex] (3)

Now θ is also the angle between the X axis and a tangent drawn to the curve at A in Fig. 5-5(a). The slope of this tangent is tanθ or dy/dx. Substituting in (3) above,

[latex]\frac{dy}{dx}=\frac{x\omega ^2}{g} [/latex] from which, by integration [latex]y=\frac{\omega ^2}{2g}x^2 +C_{1} [/latex]

To evaluate the constant of integration, C₁: When x = 0, y = 0 and C₁ = 0.

Question 5.7: An open vessel partly filled with a liquid rotates about a v...

This Question has been Answered.

Already have an account?

Login

Related Answered Questions