# Calculus

## Applications of Integrals # Force, Work and Energy

## Solved Problems

Click or tap a problem to see the solution.

### 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.

### 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.

### Example 7

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

### 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.

### 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 $$\omega.$$

### Example 10

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

### 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.

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.

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

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.

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.

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.

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.

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