# 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.

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

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 \(\theta = \pi:\)

### 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

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

The total work done in building the pyramid is

Substituting the values for the Great Pyramid yields

### 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:

The mass of the slice is given by

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

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 \(z = H:\)

### 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:

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

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

then

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

Recall from trigonometry that

As a result, we get the following simple expression:

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

### 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

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

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

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

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 \(9\))

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

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: