Question

A cylinder of radius R full of liquid of density rho is rotated about it's axis at w rad s. The increase in pressure at the centre of cylinder will be

Answer 1

as you know, moment of inertia of a hollow cylinder of radius R about its axis of rotation is given by $I=MR^2$

density of liquid ,$\rho$ is filled in cylinder . now hollow cylinder is changed into solid cylinder of mass ,$m=\pi R^2l\rho$

now, moment of inertia of cylinder fully filled with liquid about its axis of rotation is given by
$I'=MR^2+\frac{1}{2}\pi R^2l\rho R^2$

therefore increased torque = change in momentum of inertia × angular acceleration
= $\frac{1}{2}\pi R^4l\rho\times\alpha$

we know, $R\alpha=\omega^2R$
and $\tau=F.R$

so, $F.R=\frac{1}{2}\pi R^3l\rho\omega^2R$

$F=\frac{1}{2}\pi R^3l\rho\omega^2$

$\frac{F}{A}=\frac{\pi R^3l\rho\omega^2}{2\pi R^2}$

$P=\frac{\rho Rl\omega^2}{2}$

hence, pressure per unit length at centre is $\frac{\rho\omega^2R}{2}$

Answer 2

pw^2R^2/2

since height in the drum due to rotation h= (wR)^2/2g and P = density*g*h-------@

put the value of h in eq. @

so that P= density*g*(wR)^2/2g

P = density* w^2*R^2/2g