Calculus

Applications of Integrals

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Force, Work and Energy

Solved Problems

Example 5.

Find the force of gravity exerted by a semicircular ring of radius \(R\) and mass \(m\) on a particle of mass \({m_0}\) located at the centre of the ring.

Solution.

Force of gravity exerted by a semicircular ring on a point mass.
Figure 3.

Consider a small arc of the ring cut off by an angle \(d\theta.\) The mass of this arc is \(dm = \frac{{md\theta }}{\pi }.\) The force of gravity between the arc element \(dm\) and the central mass \({m_0}\) is given by the equation

\[dF = G\frac{{{m_0}dm}}{{{R^2}}} = G\frac{{{m_0}m}}{{\pi {R^2}}}d\theta ,\]

where \(G\) is the gravitational constant.

Due to symmetry, the horizontal components \({dF_x}\) will cancel each other, so we will consider only the vertical components \({dF_y} = dF\sin\theta.\) The resulting force is found by integration from \(\theta = 0\) to \(\theta = \pi:\)

\[F = {F_y} = \int\limits_0^\pi {d{F_y}} = G\frac{{{m_0}m}}{{\pi {R^2}}}\int\limits_0^\pi {\sin \theta d\theta } = G\frac{{{m_0}m}}{{\pi {R^2}}}\left. {\left( { - \cos \theta } \right)} \right|_0^\pi = \frac{{2G{m_0}m}}{{\pi {R^2}}}.\]

Example 6.

The dimensions of the Great Pyramid of Giza (also known as the Pyramid of Cheops) are as follows: height \(H = 140\,\text{m},\) side of the square base \(a = 200\,\text{m}.\) Assuming that the pyramid was built of limestone with density of \(\rho = 2500\,\frac{{kg}}{{{m^3}}},\) estimate the total work done in its building.

Solution.

Pyramid of Cheops
Figure 4.

Consider a thin slice in the pyramid with thickness \(dz\) drawn \(z\) units from the vertex.

Calculation of total work done in building the Great Pyramid.
Figure 5.

The side length of the slice is \(\frac{{az}}{H}.\) Hence, the mass of the slice is

\[dm = \rho dV = \rho {\left( {\frac{{az}}{H}} \right)^2}dz = \frac{{\rho {a^2}}}{{{H^2}}}{z^2}dz.\]

We assume that the work is done by gravity, so for our representative slice, the work is given by

\[dW = dm \cdot g\left( {H - z} \right) = \frac{{\rho g{a^2}}}{{{H^2}}}\left( {H - z} \right){z^2}dz.\]

The total work done in building the pyramid is

\[W = \int\limits_0^H {dW} = \int\limits_0^H {\frac{{\rho g{a^2}}}{{{H^2}}}\left( {H - z} \right){z^2}dz} = \frac{{\rho g{a^2}}}{{{H^2}}}\int\limits_0^H {\left( {H{z^2} - {z^3}} \right)dz} = \frac{{\rho g{a^2}}}{{{H^2}}}\left. {\left( {\frac{{H{z^3}}}{3} - \frac{{{z^4}}}{4}} \right)} \right|_0^H = \frac{{\rho g{a^2}}}{{{H^2}}}\left( {\frac{{{H^4}}}{3} - \frac{{{H^4}}}{4}} \right) = \frac{{\rho g{a^2}{H^2}}}{{12}}.\]

Substituting the values for the Great Pyramid yields

\[{W_{Cheops}} = \frac{{2500 \cdot 9.8 \cdot {{200}^2} \cdot {{140}^2}}}{{12}} \approx 1.6 \times {10^{12}}\,\left( J \right).\]

Example 7.

Find the work required to produce a conical pile of sand with base radius \(R\) and altitude \(H.\)

Solution.

Calculating work required to produce a conical pile of sand.
Figure 6.

Take a thin cylindrical slice with thickness \(dz\) at a height \(z\) from the base of the cone. Using the similarity of triangles, we can write the following proportion:

\[\frac{r}{R} = \frac{{H - z}}{H},\;\; \Rightarrow r = \frac{{R\left( {H - z} \right)}}{H}.\]

The mass of the slice is given by

\[dm = \rho dV = \pi \rho {r^2}dz = \frac{{\pi \rho {R^2}{{\left( {H - z} \right)}^2}}}{{{H^2}}}dz.\]

Assuming that the work is required to overcome gravity, we have

\[dW = dm \cdot gz = \frac{{\pi \rho g{R^2}}}{{{H^2}}}{\left( {H - z} \right)^2}zdz,\]

where \(dW\) denotes the elementary work necessary to pour the layer of sand at height \(z.\)

To calculate the total work, we integrate from \(z = 0\) to \(z = H:\)

\[W = \int\limits_0^H {dW} = \frac{{\pi \rho g{R^2}}}{{{H^2}}}\int\limits_0^H {{{\left( {H - z} \right)}^2}zdz} = \frac{{\pi \rho g{R^2}}}{{{H^2}}}\int\limits_0^H {\left( {{H^2} - 2Hz + {z^2}} \right)zdz} = \frac{{\pi \rho g{R^2}}}{{{H^2}}}\int\limits_0^H {\left( {{H^2}z - 2H{z^2} + {z^3}} \right)dz} = \frac{{\pi \rho g{R^2}}}{{{H^2}}}\left. {\left( {\frac{{{H^2}{z^2}}}{2} - \frac{{2H{z^3}}}{3} + \frac{{{z^4}}}{4}} \right)} \right|_0^H = \pi \rho g{R^2}{H^2}\left( {\frac{1}{2} - \frac{2}{3} + \frac{1}{4}} \right) = \frac{{\pi \rho g{R^2}{H^2}}}{{12}}.\]

Example 8.

An infinite straight line is uniformly charged with positive electricity. The linear density of electricity is \(\mu.\) Calculate the force exerted by the straight line on a unit charge located in air at a distance \(\ell\) from the line.

Solution.

Electrostatic force between an infinite charged straight line and a unit charge
Figure 7.

Let's take a small segment of length \(dx\) on the straight line. It carries the charge \(dq=\mu dx.\) The force \(dF\) exerted by the elementary charge \(dq\) on a unit test charge is determined by Coulomb's law:

\[dF = k\frac{{{q_1}{q_2}}}{{{r^2}}} = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{1 \cdot \mu dx}}{{{r^2}}} = \frac{1}{{4\pi {\varepsilon _0}}}\frac{\mu }{{{r^2}}}dx.\]

Using the Pythagorean theorem, the distance \(r\) between the charges can be written as \(r = \sqrt {{l^2} + {x^2}} ,\) so

\[dF = \frac{1}{{4\pi {\varepsilon _0}}}\frac{\mu }{{{l^2} + {x^2}}}dx.\]

For integration over the infinite line it is convenient to change variable. We rewrite the expression for \(dF\) in terms of the angle \(\theta.\) Since

\[x = l\tan \theta,\]

then

\[dx = l\,{\sec ^2}\theta \,d\theta.\]

Thus, we obtain \(dF\) in the form

\[dF = \frac{1}{{4\pi {\varepsilon _0}}}\frac{\mu }{{{l^2} + {x^2}}}dx = \frac{1}{{4\pi {\varepsilon _0}}}\frac{\mu }{{{l^2} + {l^2}\,{{\tan }^2}\theta }}\,l\,{\sec ^2}\theta d\theta = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{\mu \cancel{l}\,{{\sec }^{2}}\theta }}{{{l^{\cancel{2}}}\left( {1 + {{\tan }^2}\theta } \right)}}d\theta = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{\mu \,{{\sec }^2}\theta }}{{l\left( {1 + {{\tan }^2}\theta } \right)}}d\theta .\]

Recall from trigonometry that

\[1 + {\tan ^2}\theta = {\sec ^2}\theta .\]

As a result, we get the following simple expression:

\[dF = \frac{1}{{4\pi {\varepsilon _0}}}\frac{\mu }{l}d\theta .\]

By symmetry, all horizontal components of the force cancel, so further we can consider only vertical components \(d{F_y} = {dF}\cos\theta.\) Hence, the total electrostatic force exerted on a unit test charge is given by

\[F = {F_y} = \int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {d{F_y}} = \int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\frac{1}{{4\pi {\varepsilon _0}}}\frac{\mu }{l}\cos \theta d\theta } = \frac{\mu }{{4\pi {\varepsilon _0}l}}\int\limits_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} {\cos \theta d\theta } = \frac{\mu }{{4\pi {\varepsilon _0}l}}\left. {\left( {\sin \theta } \right)} \right|_{ - \frac{\pi }{2}}^{\frac{\pi }{2}} = \frac{\mu }{{4\pi {\varepsilon _0}l}}\left[ {1 - \left( { - 1} \right)} \right] = \frac{\mu }{{2\pi {\varepsilon _0}l}}.\]

Example 9.

Find the kinetic energy of a homogeneous circular cylinder of density \(\rho\) with base radius \(R\) and altitude \(H\) rotating about its axis with angular velocity \(\omega.\)

Solution.

Consider a thin hollow cylinder with inner radius \(r\) and wall thickness \(dr\) inside the initial cylinder.

Homogeneous circular cylinder rotating around its axis with angular velocity omega.
Figure 8.

The mass of the the hollow cylinder is

\[dm = \rho dV = 2\pi \rho Hrdr.\]

The linear velocity of the layer is \(v = wr,\) so the kinetic energy of the elementary cylinder is given by

\[dE = \frac{{dm{v^2}}}{2} = \frac{{\cancel{2}\pi \rho Hrdr \cdot {{\left( {wr} \right)}^2}}}{\cancel{2}} = \pi \rho {w^2}H{r^3}dr.\]

The total kinetic energy of the initial cylinder can be found through integration:

\[E = \int\limits_0^R {dE\left( r \right)} = \int\limits_0^R {\pi \rho {w^2}H{r^3}dr} = \pi \rho {w^2}H\int\limits_0^R {{r^3}dr} = \left. {\frac{{\pi \rho {w^2}H{r^4}}}{4}} \right|_0^R = \frac{{\pi \rho {w^2}H{R^4}}}{4}.\]

Notice that the energy of rotation of the body can be expressed in terms of the moment of inertia. In our case,

\[E = \frac{{\pi \rho {w^2}H{R^4}}}{4} = \pi \rho {R^2}H \cdot \frac{{{R^2}}}{2} \cdot \frac{{{w^2}}}{2} = \frac{{m{R^2}}}{2} \cdot \frac{{{w^2}}}{2} = \frac{{I{w^2}}}{2},\]

where \(I = \frac{{m{R^2}}}{2}\) is the moment of inertia of the solid circular cylinder of mass \(m\) and radius \(R.\)

Example 10.

Calculate the kinetic energy of a homogeneous ball of density \(\rho\) and radius \(R\) rotating about its axis with angular velocity \(\omega.\)

Solution.

Consider a thin disk of radius \(r\) and thickness \(dz\) inside the ball. The bottom side of the disk crosses the vertical axis at a distance of \(z\) units from the center of the ball.

A homogeneous ball of radius R rotating about its axis with angular velocity omega.
Figure 9.

The disk has the shape of a cylinder. Therefore, its kinetic energy of rotation about the vertical axis is written as (see the previous Example \(9\))

\[dE = \frac{{\pi \rho {\omega ^2}{r^4}}}{4}dz.\]

The inner radius \(r\) is expressed in the form

\[r = \sqrt {{R^2} - {z^2}} ,\;\; \Rightarrow {r^4} = {\left( {{R^2} - {z^2}} \right)^2}.\]

To find the total kinetic energy of the ball we integrate the function \({dE\left( z \right)}\) from \(0\) to \(R\) and multiply the result by \(2.\) This yields:

\[E = 2\int\limits_0^R {dE\left( z \right)} = 2\int\limits_0^R {\frac{{\pi \rho {\omega ^2}{r^4}}}{4}dz} = \frac{{\pi \rho {\omega ^2}}}{2}\int\limits_0^R {{{\left( {{R^2} - {z^2}} \right)}^2}dz} = \frac{{\pi \rho {\omega ^2}}}{2}\int\limits_0^R {\left( {{R^4} - 2{R^2}{z^2} + {z^4}} \right)dz} = \frac{{\pi \rho {\omega ^2}}}{2}\left. {\left[ {{R^4}z - \frac{{2{R^2}{z^3}}}{3} + \frac{{{z^5}}}{5}} \right]} \right|_0^R = \frac{{\pi \rho {\omega ^2}}}{2}\left( {{R^5} - \frac{{2{R^5}}}{3} + \frac{{{R^5}}}{5}} \right) = \frac{{\pi \rho {\omega ^2}{R^5}}}{2}\left( {1 - \frac{2}{3} + \frac{1}{5}} \right) = \frac{{4\pi \rho {\omega ^2}{R^5}}}{{15}}.\]
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