calculate the work W Against Gravity required to build a cylindrical column of height 4 M and radius 0.6 M out of a lightweight material of density 600 kg meters cubed

To calculate the work \( W \) against gravity required to build a cylindrical column, we can follow these steps:

### Step 1: Determine the Volume of the Cylinder
The volume \( V \) of a cylinder is given by the formula:
\[
V = \pi r^2 h
\]
where:
- \( r = 0.6 \, \text{m} \) is the radius,
- \( h = 4 \, \text{m} \) is the height of the cylinder.

Substituting these values:
\[
V = \pi (0.6)^2 (4) = \pi \times 0.36 \times 4 = 1.44 \pi \, \text{m}^3
\]
Thus, the volume of the cylinder is \( 1.44 \pi \, \text{m}^3 \).

### Step 2: Find the Mass of the Cylinder
The mass \( m \) of the cylinder is found using the formula:
\[
m = \text{density} \times V
\]
Given the density of the material \( \rho = 600 \, \text{kg/m}^3 \), we can substitute the values:
\[
m = 600 \times 1.44 \pi = 864 \pi \, \text{kg}
\]
This is the total mass of the cylindrical column.

### Step 3: Calculate the Work Against Gravity
The work \( W \) done against gravity to build the column involves moving the material to its final height. We use the concept of gravitational potential energy and integration to compute the work.

For an infinitesimal mass element \( dm \) at height \( y \), the work to lift it to height \( y \) is \( g y \, dm \), where \( g \) is the acceleration due to gravity (\( 9.8 \, \text{m/s}^2 \)).

The mass of an infinitesimal disk of thickness \( dy \) at height \( y \) is:
\[
dm = \rho \times \pi r^2 \times dy = 600 \times \pi \times (0.6)^2 \times dy = 216 \pi \, dy
\]
So, the infinitesimal work to move this disk to height \( y \) is:
\[
dW = g \, y \, dm = 9.8 \times y \times 216 \pi \, dy = 2116.8 \pi \, y \, dy
\]

To find the total work, we integrate this from \( y = 0 \) to \( y = 4 \):
\[
W = \int_0^4 2116.8 \pi y \, dy
\]

### Step 4: Perform the Integration
The integral is:
\[
W = 2116.8 \pi \int_0^4 y \, dy = 2116.8 \pi \left[ \frac{y^2}{2} \right]_0^4 = 2116.8 \pi \times \frac{16}{2} = 2116.8 \pi \times 8
\]
\[
W = 16934.4 \pi \, \text{Joules}
\]

### Step 5: Final Answer
The total work required to build the cylindrical column against gravity is:
\[
W \approx 16934.4 \times 3.1416 = 53207.7 \, \text{Joules}
\]

Thus, the work required is approximately \( 53207.7 \, \text{J} \).

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Would you like any further clarifications or details on this calculation?

Here are some related questions for further exploration:
1. How would the work change if the radius were doubled?
2. What is the effect of increasing the height of the column on the work required?
3. How does the density of the material affect the work calculation?
4. What if the column is built on a different planet with different gravitational acceleration?
5. How would we calculate the work if the material had a variable density?

**Tip:** Always double-check units in physical problems to ensure consistency, especially when working with derived quantities like work, energy, and power.

Calculate the work W against gravity required to build a cylindrical column of height 4 meters and radius 0.6 meters out of a lightweight material of density 600 kg/m^3.

Learn how to calculate the work required to build a cylindrical column of height 4 meters and radius 0.6 meters against gravity, using principles of gravitational potential energy, integration, and volume calculations. Suitable for undergraduate-level physics or engineering students.

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