Question

Q.1 A hydraulic circular rod having a 100mm diameter and 2m long is pulled by a force of 200 N. It moves with a constant velocity within a circular hollow tube of 101mm internal diameter. The clearance of 0.5mm between the rod and the tube is filled with oil of specific gravity 0.8 and kinematic viscosity 400 mm2/s as shown in the figure below. Determine the velocity of the rod in units of m/s.

Answer 1

Answer on Question #84695- Physics - Mechanics | Relativity

A hydraulic circular rod having a 100mm diameter and 2m long is pulled by a force of 200 N. It moves with a constant velocity within a circular hollow tube of 101mm internal diameter. The clearance of 0.5mm between the rod and the tube is filled with oil of specific gravity 0.8 and kinematic viscosity 400 mm²/s as shown in the figure below. Determine the velocity of the rod in units of m/s.

Solution.

According to the Newton's second law of motion:

$$F - F _ {\text {viscosity}} = 0,$$

where

$$F _ {\text {viscosity}} = \mu S \frac {d \vartheta}{d h}$$ $$\mu = \nu \rho$$

$\nu$ is kinematic viscosity, $\mu$ is dynamic viscosity, $\rho$ is density of oil.

Substituting (2) and (3) into (1), we receive:

$$F d h = \nu \rho S d \vartheta$$

Taking into account, that velocity of oil layers is changing from 0 near the tube to $\vartheta$ near the rod, let's integrate (4):

$$\int_ {0} ^ {h} F d h = \int_ {0} ^ {\vartheta} v \rho S d \vartheta$$

After integration we receive:

$$F h = \nu \rho S \vartheta$$

Thus:

$$\vartheta = \frac {F h}{\nu \rho S}$$

Specific gravity:

$$S G = \frac {\rho}{\rho_ {H _ {2 O}}}$$ $$\rho = S G \rho_ {H _ {2 O}}$$

Cylinder's lateral area:

$$S = 2 \pi R l = \pi D _ {r o d} l$$

Let's substitute (6) and (7) into (5):

$$\vartheta = \frac {F h}{\nu S G \rho_ {H _ {2 O}} \pi D _ {r o d} l}$$

Finally:

$$\vartheta = \frac {200 \cdot 0.5 \cdot 10^{-3}}{400 \cdot 10^{-6} \cdot 0.8 \cdot 1000 \cdot 3.14 \cdot 100 \cdot 10^{-3} \cdot 2} = 0.5 \ (m/s)$$

Answer: $\vartheta = 0.5 \ (m/s)$.

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