PROPERTIES OF MATTER CYU 7

(a) Cylinder Fixed, Liquid Flowing

The liquid flows through the annular clearance $h$. Since $h \ll R$, this can be modeled as flow between parallel plates (length $l$, width $2\pi R$).

1. Volume Flow Rate ($Q$):
Forces on the liquid in the gap: Pressure difference pushes forward, Viscosity resists at both walls.
Force Balance: $\Delta p (2\pi R h) = 2 \times [\tau \times (2\pi R l)]$.
Using $\tau = \eta \frac{v}{h}$, we get $\Delta p h = \frac{2\eta v l}{h} \implies v = \frac{\Delta p h^2}{2\eta l}$.
Flow rate $Q = \text{Area} \times v = (2\pi R h) \times v$.

$$ Q \approx \frac{\pi \Delta p R h^3}{\eta l} $$

2. Tensile Force in Thread ($T$):
The force exerted by the liquid on the cylinder consists of:

• Pressure force on the face: $F_p = \Delta p (\pi R^2)$.
• Viscous drag on the curved surface: $F_v = \tau \times (2\pi R l) = (\eta \frac{v}{h}) 2\pi R l$.

Substituting $v$: $F_v = \frac{\eta}{h} (\frac{\Delta p h^2}{2\eta l}) 2\pi R l = \pi \Delta p R h$.
Total Force $T = F_p + F_v = \pi \Delta p R^2 + \pi \Delta p R h$.

(b) Cylinder Falling (Denser than Liquid)

Let the cylinder move down with velocity $u$. Since the bottom is closed, liquid must flow **upwards** through the gap to conserve volume.

1. Continuity:
Volume swept by cylinder = Volume flowing up through gap.
$u (\pi R^2) = v_{liq} (2\pi R h) \implies v_{liq} = u \frac{R}{2h}$.
Since $h \ll R$, $v_{liq} \gg u$. The relative velocity gradient in the gap is dominated by $v_{liq}$.

2. Pressure Generation:
To force liquid up through the narrow gap at high speed $v_{liq}$, a large pressure difference $\Delta p$ is generated across the cylinder length.
From part (a), relation between velocity and pressure gradient: $\Delta p \approx \frac{2\eta l v_{liq}}{h^2} = \frac{2\eta l}{h^2} \left( \frac{uR}{2h} \right) = \frac{\eta l R u}{h^3}$.

3. Force Balance on Cylinder:
Downward: Effective Weight $W_{eff} = (\rho – \rho_0) \pi R^2 l g$.
Upward: Pressure Force $F_p = \Delta p (\pi R^2)$. (Viscous drag is negligible compared to pressure force for $h \ll R$).
Equating $W_{eff} = F_p$:

$$ (\rho – \rho_0) \pi R^2 l g = \left( \frac{\eta l R u}{h^3} \right) \pi R^2 $$ $$ (\rho – \rho_0) g = \frac{\eta u R}{h^3} $$

(c) Cylinder Rising (Less Dense)

The physics is identical to part (b), but the buoyancy exceeds weight. The driving force is $(\rho_0 – \rho)g$. The liquid flows downwards as cylinder moves up.