Physical Applications: Center of Mass, Moments of Inertia and Work
Integrals turn physical sums into exact quantities: the center of mass is x̄ = ∫ x ρ(x) dx / ∫ ρ(x) dx, the moment of inertia is I = ∫ r2 dm, and the work done by a variable force is W = ∫ F(x) dx. In brief: for a rod of length L with constant density, the center of mass is at L/2 and the moment of inertia about its end is (1/3)mL2. Stretching a spring with F = kx from 0 to d needs the work (1/2)kd2. Pumping water out of a tank is computed by integrating the work of lifting each thin layer.
When a quantity is spread continuously (mass along a rod, force along a path), it is cut into small pieces, the contribution of each piece is approximated, and the sum becomes an integral. This guide shows the method for work, fluid pumping, spring and gravity problems, center of mass and centroids, moments of inertia, average values, and the force of a fluid on a dam, with worked examples and a table of formulas.
How do you compute work?
The work done by a constant force F over a distance d is Fd. If the force depends on position, F(x), the work from a to b is W = ∫ab F(x) dx, by the fundamental theorem of calculus. The unit of work is the joule (1 J = 1 N · m).
Example 1: a spring (Hooke’s law)
F = kx with k = 200 N/m. Stretching the spring from its natural length by 0.3 m: W = ∫00.3 200x dx = 100x2|00.3 = 9 J. Stretching it from 0.3 to 0.5 m: 100(0.25 − 0.09) = 16 J.
Example 2: finding the spring constant
A force of 12 N stretches a spring by 0.2 m, so k = 60 N/m. The work for stretching it by 0.5 m: 30 · 0.25 = 7.5 J.
Example 3: a cable
A 20 m cable of mass 40 kg hangs from the top of a building. Lifting it to the top: a piece of length dy at depth y has the mass 2 dy kg (2 kg/m) and is lifted by y, requiring the work 2 g y dy. W = ∫020 2 · 9.8 y dy = 9.8 · 400 = 3,920 J.
Example 4: gravity at a large distance
The work to lift a mass m from the Earth’s surface (radius R) to infinity against gravity F = GMm/r2: W = ∫R∞ GMm/r2 dr = GMm/R, an improper integral. Setting (1/2)mv2 = GMm/R gives the escape velocity v = √(2GM/R) ≈ 11.2 km/s.
How do you compute the work of pumping?
Slice the liquid into thin horizontal layers. The weight of a layer is density × g × volume; it must be lifted through the distance from its depth to the outlet. Add up with an integral.
Example 5: a cylindrical tank
A tank is a vertical cylinder with radius 2 m and height 5 m, full of water (density 1000 kg/m3). Pumping everything out over the top: a layer at depth y (from the top) has the volume 4π dy, mass 4000π dy kg, weight 4000π · 9.8 dy N and is lifted y m. W = ∫05 39,200π y dy = 39,200π · 12.5 = 490,000π ≈ 1.54 × 106 J.
Example 6: a conical tank
A cone with the apex down, height 6 m and top radius 3 m is full of water. At depth y from the top, the radius is r = 3 − y/2 (the top has r = 3 at y = 0 and the apex r = 0 at y = 6). The layer volume is π(3 − y/2)2 dy. The work: ∫06 1000 · 9.8 · π (3 − y/2)2 y dy. Expand (3 − y/2)2 y = 9y − 3y2 + y3/4 and integrate: [4.5y2 − y3 + y4/16]06 = 162 − 216 + 81 = 27. W = 9800π · 27 = 264,600π ≈ 8.31 × 105 J.
How do you find mass and center of mass?
For a rod with linear density ρ(x), the mass is m = ∫ ρ(x) dx and the center of mass is x̄ = (1/m) ∫ x ρ(x) dx. For a flat plate (lamina) under y = f(x) with a constant density, the centroid has x̄ = (1/A) ∫ x f(x) dx and ȳ = (1/A) ∫ (1/2) f(x)2 dx, where A = ∫ f(x) dx.
Example 7: a rod with variable density
ρ(x) = 2 + x (kg/m) on [0, 4]. m = ∫04 (2 + x) dx = 8 + 8 = 16 kg. ∫ x(2 + x) dx = [x2 + x3/3]04 = 16 + 64/3 = 112/3. x̄ = (112/3)/16 = 7/3 ≈ 2.33 m, shifted towards the heavy end from the midpoint 2.
Example 8: a parabolic plate
The region under y = x2 on [0, 1]. A = 1/3. x̄ = 3 ∫01 x3 dx = 3/4. ȳ = 3 · (1/2) ∫01 x4 dx = 3/10. The centroid is (0.75, 0.3).
Example 9: a semicircle
The centroid of the upper half disc of radius R lies on the axis of symmetry at ȳ = (1/A) ∫ (1/2)(R2 − x2) dx = (2/(πR2))(1/2)(4R3/3) = 4R/(3π) ≈ 0.424R.
How do you compute moments of inertia?
For a body rotating about an axis, I = ∫ r2 dm, where r is the distance from the axis. The kinetic energy of rotation is (1/2)Iω2.
Example 10: a rod
A uniform rod of mass M and length L (linear density M/L), axis perpendicular through the end: I = ∫0L x2(M/L) dx = ML2/3. Through the center: I = ∫−L/2L/2 x2(M/L) dx = ML2/12. The two are related by the parallel axis theorem: I = Ic + M d2 = ML2/12 + M(L/2)2 = ML2/3 ✓.
Example 11: a disc
A uniform disc of mass M and radius R, about the axis through the center perpendicular to the disc. Use rings of radius r and width dr, with mass dm = M · 2πr dr/(πR2) = 2Mr dr/R2. I = ∫0R r2 · 2Mr/R2 dr = (2M/R2)(R4/4) = MR2/2.
Example 12: a solid sphere
I = (2/5)MR2 about a diameter, computed by integrating the discs’ moments (1/2)dm y2. For a sphere of M = 5 kg, R = 0.1 m: I = 0.4 · 5 · 0.01 = 0.02 kg · m2.
What about fluid force and average values?
The pressure at depth y is p = ρ g y, and the force on a vertical plate is F = ∫ ρ g y w(y) dy, where w(y) is the width at depth y.
Example 13: a dam
A rectangular dam wall 20 m wide, with water 10 m deep: F = ∫010 1000 · 9.8 · y · 20 dy = 196,000 · 50 = 9.8 × 106 N.
Example 14: average value
The temperature in a room varies as T(t) = 20 + 4 sin(πt/12) for 0 ≤ t ≤ 24 (hours). The average temperature: (1/24) ∫024 T dt = 20 (the sine integrates to 0 over a full period). The average over 0 ≤ t ≤ 12: 20 + (4/12)[−(12/π)cos(πt/12)]012 = 20 + (4/12)(24/π) = 20 + 8/π ≈ 22.55 °C.
Table of formulas
| Quantity | Formula | Example | Result |
|---|---|---|---|
| work of a variable force | ∫ F(x) dx | spring k = 200, 0.3 m | 9 J |
| mass of a rod | ∫ ρ(x) dx | ρ = 2 + x on [0, 4] | 16 kg |
| center of mass | ∫ xρ dx / ∫ ρ dx | same rod | 7/3 m |
| moment of inertia | ∫ r2 dm | rod about the end | ML2/3 |
| moment of inertia | ∫ r2 dm | disc about its axis | MR2/2 |
| fluid force | ∫ ρ g y w(y) dy | dam 20 m × 10 m | 9.8 × 106 N |
Where do these ideas come from?
Archimedes used the method of balancing (the law of the lever) to find centers of gravity and areas, an early physical use of integration ideas. The concepts of work and moments were developed by Newton, Euler and others in the eighteenth century. The moment of inertia was introduced by Euler in his study of rigid body rotation (1730s), and the parallel axis theorem is associated with Jakob Steiner (nineteenth century), which is why it is called Steiner’s theorem in some countries.
Where are they used?
Engineers compute the energy needed for pumping and lifting loads, the stability of structures from centers of mass, the inertia of rotating parts (flywheels, turbines) and forces on dams and ship hulls. In robotics and sports biomechanics, the location of the center of mass of the body determines balance. Average values appear in electrical engineering as rms values.
Which mistakes are common?
Using the total weight instead of integrating
When different parts are lifted different distances, you must integrate; the total weight times a single distance is wrong.
Wrong distance in pumping problems
Measure the distance from each layer to the outlet, not to the surface of the liquid.
Forgetting units
Convert to SI: kg, m, N, J. Pounds and feet require the corresponding conversions of density and g.
Mixing up the density and the mass
dm = ρ dx for a rod and ρ dA for a plate. Integrate the density, not the mass.