Q13. In a hollow cylinder, filled with water till a height \(h\), being rotated at angular velocity \(\omega\). Its radius is \(R\). Consider the density of water to be \(\rho\). Find the moment of inertia of the rotating liquid as function of angular velocity \(\omega\).

Idea:

To find \(H\),

\[V_i = V_f\]

\[\pi R^2 h = V_{\text{till }H} - V_{paraboloid}\]

\[\implies H = h + {R^2 \omega^2 \over 4g}\]

Equation of parabola is as follows(easily derivable):

\[y = {x^2 \omega^2 \over 2g}\]

Similar analysis gives,

\[\lambda = h - {R^2 \omega^2 \over 4g}\]

Which gives the equation of surface in cylindrical coordinates as,

\[z = h + {\omega^2 \over 2g}\left( r^2 - {R^2 \over 2} \right)\]

Observe that heights increased and decreased is symmetrical about the original height.

Moment of inertia of the liquid can be written as:

\[I_{\text{liquid}} = I_{\text{till H}} - I_{\text{paraboloid, } \lambda \text{ to } H}\]

Final result:

\[I = \pi \rho \left( {R^4 h \over 2} + {R^6 \omega^2 \over 24g} \right)\]