A circular opening, 3 m diameter, in the vertical side of a water tank is closed by a disc of 3 m diameter which can rotate about a horizontal diameter. Calculate the force on the disc and the torque required to maintain the disc in equilibrium in the vertical position when the head of water above the horizontal diameter is 4 m.

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Rotating Circular Disc Problem

Problem Statement

A circular opening, 3 m diameter, in the vertical side of a water tank is closed by a disc of 3 m diameter which can rotate about a horizontal diameter. Calculate: (i) the force on the disc, and (ii) the torque required to maintain the disc in equilibrium in the vertical position when the head of water above the horizontal diameter is 4 m.

Given Data

  • Diameter of disc, \( D = 3 \, \text{m} \)
  • Pivot: Horizontal diameter (centroidal axis)
  • Head of water above pivot, \( \bar{h} = 4 \, \text{m} \)
  • Fluid is water, \( \rho = 1000 \, \text{kg/m}^3 \)

Solution

(i) Force on the Disc (\(F\))

First, calculate the area of the disc.

$$ A = \frac{\pi}{4} D^2 $$ $$ A = \frac{\pi}{4} (3)^2 $$ $$ A = \frac{9\pi}{4} \approx 7.0686 \, \text{m}^2 $$

Now, calculate the total force on the disc using the depth to the centroid.

$$ F = \rho g A \bar{h} $$ $$ F = 1000 \times 9.81 \times 7.0686 \times 4 $$ $$ F \approx 277372 \, \text{N} $$

(ii) Torque required to maintain equilibrium (\(T\))

The torque is created because the hydrostatic force \(F\) acts at the centre of pressure, which is below the pivot point (the centroid). The lever arm is the distance between these two points.

First, find the depth of the centre of pressure (\(h^*\)).

$$ I_G = \frac{\pi D^4}{64} $$ $$ I_G = \frac{\pi \times (3)^4}{64} $$ $$ I_G = \frac{81\pi}{64} \approx 3.976 \, \text{m}^4 $$
$$ h^* = \frac{I_G}{A \bar{h}} + \bar{h} $$ $$ h^* = \frac{3.976}{7.0686 \times 4} + 4 $$ $$ h^* = \frac{3.976}{28.2744} + 4 $$ $$ h^* \approx 0.1406 + 4 = 4.1406 \, \text{m} $$

The lever arm is the vertical distance between the centre of pressure and the centroid (pivot).

$$ \text{Lever Arm} = h^* - \bar{h} $$ $$ \text{Lever Arm} = 4.1406 - 4.0 $$ $$ \text{Lever Arm} = 0.1406 \, \text{m} $$

Finally, calculate the torque.

$$ T = F \times (\text{Lever Arm}) $$ $$ T = 277372 \times 0.1406 $$ $$ T \approx 38995 \, \text{N-m} $$
Final Results:

(i) Total Force on the disc: \( F \approx 277.37 \, \text{kN} \).

(ii) Torque required: \( T \approx 39.0 \, \text{kN-m} \).

Explanation of Concepts

Total Force: This is the total hydrostatic force exerted by the water on the disc. It is calculated by finding the pressure at the disc's geometric center (centroid) and multiplying it by the disc's total area.

Torque and Centre of Pressure: The disc is pivoted at its centroid, but the total hydrostatic force acts at a slightly lower point called the centre of pressure. This is because the water pressure is greater on the bottom half of the disc than at the top. The offset between the point of force application (centre of pressure) and the pivot point (centroid) creates a lever arm. The torque required to prevent rotation is the moment produced by the hydrostatic force acting over this lever arm (\( T = F \times \text{Lever Arm} \)).

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