A drum of radius R full of liquid of density | vedclass

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(A) Consider a small element of liquid at a distance $r$ from the center of the drum rotating with angular velocity $\omega$.The centripetal force req

Vedclass · Accepted Answer

$\frac{\omega^2 R^2 d}{2}$

Vedclass · Answer

(A) Consider a small element of liquid at a distance $r$ from the center of the drum rotating with angular velocity $\omega$.The centripetal force required for this element of mass $dm$ is $dF = (dm) \omega^2 r$.This force is provided by the pressure gradient: $dP \cdot A = (dm) \omega^2 r$,where $dm = d \cdot A \cdot dr$.Substituting $dm$: $dP \cdot A = (d \cdot A \cdot dr) \omega^2 r$.Simplifying,we get $dP = d \cdot \omega^2 r \cdot dr$.To find the total pressure increase at the center relative to the edge,we integrate from $r = 0$ to $r = R$:$\Delta P = \int_{0}^{R} d \cdot \omega^2 r \cdot dr = d \cdot \omega^2 \left[ \frac{r^2}{2} \right]_{0}^{R}$.$\Delta P = \frac{d \omega^2 R^2}{2}$.

$A$ drum of radius $R$ full of liquid of density $d$ is rotated at angular velocity $\omega \ rad/s$. The increase in pressure at the centre of the drum will be

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