Math Problem Statement

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

Solution

To calculate the work WW against gravity required to build a cylindrical column, we can follow these steps:

Step 1: Determine the Volume of the Cylinder

The volume VV of a cylinder is given by the formula: V=πr2hV = \pi r^2 h where:

  • r=0.6mr = 0.6 \, \text{m} is the radius,
  • h=4mh = 4 \, \text{m} is the height of the cylinder.

Substituting these values: V=π(0.6)2(4)=π×0.36×4=1.44πm3V = \pi (0.6)^2 (4) = \pi \times 0.36 \times 4 = 1.44 \pi \, \text{m}^3 Thus, the volume of the cylinder is 1.44πm31.44 \pi \, \text{m}^3.

Step 2: Find the Mass of the Cylinder

The mass mm of the cylinder is found using the formula: m=density×Vm = \text{density} \times V Given the density of the material ρ=600kg/m3\rho = 600 \, \text{kg/m}^3, we can substitute the values: m=600×1.44π=864πkgm = 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 WW 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 dmdm at height yy, the work to lift it to height yy is gydmg y \, dm, where gg is the acceleration due to gravity (9.8m/s29.8 \, \text{m/s}^2).

The mass of an infinitesimal disk of thickness dydy at height yy is: dm=ρ×πr2×dy=600×π×(0.6)2×dy=216πdydm = \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 yy is: dW=gydm=9.8×y×216πdy=2116.8πydydW = 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=0y = 0 to y=4y = 4: W=042116.8πydyW = \int_0^4 2116.8 \pi y \, dy

Step 4: Perform the Integration

The integral is: W=2116.8π04ydy=2116.8π[y22]04=2116.8π×162=2116.8π×8W = 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=16934.4πJoulesW = 16934.4 \pi \, \text{Joules}

Step 5: Final Answer

The total work required to build the cylindrical column against gravity is: W16934.4×3.1416=53207.7JoulesW \approx 16934.4 \times 3.1416 = 53207.7 \, \text{Joules}

Thus, the work required is approximately 53207.7J53207.7 \, \text{J}.


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.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Volume of a Cylinder
Mass Calculation
Gravitational Potential Energy
Integration

Formulas

Volume of a Cylinder: V = πr^2h
Mass: m = density × volume
Infinitesimal Work: dW = g y dm
Total Work: W = ∫ g y dm

Theorems

Gravitational Potential Energy
Basic Integration

Suitable Grade Level

Undergraduate level (Engineering or Physics)