Examples, exercises and solutions - Classical Mechanics & Special Relativity for Starters

Updated: 05 feb 2026

Worked examples¶

Circular motion¶

Consider a particle of mass $m$ that is attached to a massless rope of length R. The other end of the rope is fixed in the origin and is free to rotate in a horizontal plane. The particle is made to move in a circular motion wit a velocity that has a constant magnitude $v$ . Show that on $m$ a force should act that is parallel to the rope at all times. Gravity is to be ignored.

Interpret the problem

Develop the solution

Evaluate the problem

Assess the problem

Let’s start with a drawing: $m$ is moving over a circle in a horizontal plane. We draw its position vector, $\vec{r}$ , its velocity vector, $\vec{v}$ and a force, $\vec{F}$ that can act on $m$ .

This is clearly a 2-dimensional problem: both $\vec{r}$ and $\vec{v}$ do change in the plane while $m$ moves along the circle. Moreover, since $\vec{v}$ is not a constant vector (it does have constant magnitude $v$ , but clearly its direction constantly changes), we can anticipate that a force must be acting on $m$ . After all, $\vec{p} = m \vec{v}$ is not a constant and, thus, according to N2 a force must act on $m$ .

We have at least two options to approach this problem: via momentum or via angular momentum. We opt for the latter, as we anticipate that the angular momentum may be a constant.

Unstable See-Saw¶

We have a seesaw as shown in the figure below. On the left side a mass $m$ is place, on the right side, $2m$ . Both arms of the seesaw have a length $L$ and zero mass. For now, we keep the seesaw horizontal. But at $t=0$ , we let go.

Is the seesaw stable for $t \gt 0$ ?
If not: what is the initial acceleration of the mass $2m$ (that is, its acceleration just after release)?

Interpret the problem

Develop the solution

Evaluate the problem

Assess the problem

As always, we start with a drawing. That is, in this case we complement the figure given with relevant information for our interpret-phase.

We have drawn: the two relevant forces of gravity acting om $m$ and $2m$ , respectively, as well as $F_p$ the force of the pivot acting on the seesaw. Moreover, we have (in blue) indicated our coordinate system. This is useful, as we anticipate that we will have to deal with torques and angular momentum. Furthermore, we have drawn the position vector (in green) of both masses with the pivot point as our origin.

The figure is made with the idea that the stability of the seesaw for $t \gt 0$ can be inspected by looking at the torques acting on it.

Exercises¶

Exercise 2 (Halley’s comet)

Halley’s comet orbits the sun in a very elongated ellipse: $e=0.967$ . With the origin of a Cartesian coordinate system in the sun, the comet’s orbit is given by:

\sqrt{x^2 + y^2} +e x = a(1-e^2)

(9)

with $e=0.967$ the eccentricity and $a=17.8 \text{AU}$ the half long axis of the ellipse. (Note: AU stands for ‘astronomical unit’ which is the averaged distance from the earth to the sun, i.e. about 149.6 million km.)

Since gravity is a central force (in this case we have fixed the sun in the origin), the angular momentum of Halley’s comet is constant. Find the ratio of the velocity of the comet when passing the sun at its closest distance (the so-called perihelion) and the velocity at the furthest distance (the aphelion). Compare this ratio to that of the earth.

Data: the ellipse for an object orbiting the sun is given by

(x^2 + y^2) + ex = a(1-e^2)

(10)

object	a (A.U.)	e (-)
earth	1.0	0.0167
Halley’s comet	17.8	0.967

Exercise 4 (Relative distance to the sun)

Measuring distances in astronomy is a difficult task: one can not use a standard measuring rule. However, there are plenty of other options. One is: use Kepler’s third law to find the relative distance from the planets to the sun. Here ‘relative’ means: express them in terms of the distance of the earth to the sun.

Data

planet	orbital period (year)
Mercury	0.241
Venus	0.615
Earth	1.000
Mars	1.881
Jupiter	11.86
Saturn	29.46
Uranus	84.0
Neptune	164.8

Exercise 6 (A Seesaw with mass 🌶 🌶 🌶)

So far, we have treated the seesaw as a weightless bar on which two masses are sitting. In reality the bar itself, of course, also has mass. In the symmetric case, the net torque of the part of the bar on the left of the pivot is equal and opposite to the right bar. Hence, it has no effect on the stability or rotation of the seesaw.

However, in general that is not the case. In this exercise we will learn how to deal with such a situation.

Consider a seesaw as shown in the figure below.

It has on the left of the pivot point(the origin) a length $L$ , while on the right side only $\frac{3}{4}L$ . The green bar that forms the seesaw is here formed by a weightless length with at intervals of $\frac{1}{4}L$ a mass $m$ . Thus on the left side: 4 masses and and the right side only three.

Show that this seesaw is unbalanced.
You can place one additional mass $2m$ at some point on the seesaw. Where should it be placed to balance the seesaw?

Next, we consider the bar made of homogeneous material (with density per unit length $\rho) and not of a weightless part with here and there a mass. Thus, the seesaw is now like in the figure.

The difference is, that the mass is now continuously distributed. To deal with this, we think of the green bar as made of tiny masses glued together. Each mass is $dm$ , has a width $dx$ and is at location $(x,x+dx)$ . The total mass of the bar is, like in the above simplified case, $7m$ and the bar is of uniform thickness and uniform material (hence its density per unit length is constant).

Show that this seesaw is also unbalanced.
Find the position where a mass $2m$ should be placed to balance it.

Answers¶

Solution to Exercise 1

We first complete the drawing. As we need to think about the angular momentum, we need the position vector of $m$ . Furthermore, we need to include all forces acting on $m$ in order to evaluate the effect of all torques on $m$ .

point particle sliding down a slope with forces and position vector.

Apart from gravity also a normal force from the slope on $m$ is present.

Note: in the figure we have now for the position vector, the velocity and both forces set the size of $m$ to zero. As a consequence, $\vec{r}$ and $\vec{v}$ are parallel during the entire motion

Next step, we develop our solution strategy. As this exercise is concerned with angular moment, we will use N2 for angular momentum as well as, of course, the definition of angular momentum $\vec{l} \equiv \vec{r} \times \vec{p}$ .

Now we evaluate our ideas:

\vec{l} \equiv \vec{r} \times \vec{p} = m\vec{r} \times \vec{v} = 0

(11)

since, in this case $\vec{r} // \vec{v}$ .

N2 for angular momentum reads as:

\frac{d\vec{l}}{dt} = \sum_i \vec{r}_i \times \vec{F}_i

(12)

We know that the left hand side is zero: $\vec{l}$ is constant. Thus, the sum of all torques on $m$ must also be zero. We have identified two forces acting on $m$ : gravity and the normal force. The latter does nothing but opposing the component of gravity perpendicular tot he slope (that is its $y$ -component). We split gravity in its $x$ and $y$ -component:

\vec{F}_g = F_{g,x}\hat{x} + F_{g,y} \hat{y}

(13)

with this we compute the total torque on $m$

\begin{split} \sum_i \vec{r}_i \times \vec{F}_i &= \vec{r} \times \left ( F_{g,x}\hat{x} + F_{g,y} \hat{y} \right ) + \vec{r} \times \vec{F}_n \\ &= \underbrace{\vec{r} \times F_{g,x}\hat{x}}_{=0} + \vec{r} \times \underbrace{\left ( F_{g,y}\hat{y} + \vec{F}_n \right )}_{=0} \\ &=0 \end{split}

(14)

The first term is zero because $\vec{r} // \hat{x}$ and the second one due to cancelling of the two forces in the brackets.

If we assess our answer, we see that this is perfectly in line with N2 for angular momentum: no net torque, hence no change of the angular momentum!

Solution to Exercise 2

We will exploit that the angular momentum of the earth and the comet are constant. The angular momentum is defined as $\vec{l} = \vec{r} \times \vec{p}$ . At the perihelion and at the aphelion, the position vector and the velocity are perpendicular. Thus, the angular momentum simplifies to $\vec{l} = x_hmv \hat{z}$ , with $x_h$ the $x$ -coordinate of the perihelion or aphelion.

We can find these coordinates by realizing that at the perihelion and the aphelion, the $y$ -coordinate is zero. Thus,

\begin{split} \sqrt{x_h^2 + 0^2} + ex_h = a(1-e^2) \rightarrow \\ \pm x_h +ex_h = a(1-e^2) \rightarrow \\ x_{ph} = a(1-e) \text{ or } x_{ah} = -a(1+e) \end{split}

(15)

where we have labelled the $x$ -coordinate of the perihelion as $x_{ph}$ and of the aphelion as $x_{ah}$ .

The velocity ratio of $v_{ph}$ over $v_{ah}$ is now easily found, by using that the angular momentum is a constant:

l = mx_h v \Rightarrow \frac{v_{ph}}{v_{ah}} = \frac{x_{ap}}{x_{ph}}

(16)

Putting in the given numbers, we find:

\begin{split} \left [ \frac{v_{ph}}{v_{ah}} \right ]_{Halley} &= \frac{1+e_H}{1-e_H} = 59.6 \\ \left [ \frac{v_{ph}}{v_{ah}} \right ]_e &= \frac{1+e_e}{1-e_e} = 1.03 \end{split}

(17)

Does this make sense? Yes: the earth orbits in an ellipse that is close to a circle. Hence we expect only a mild difference between the velocity at the perihelion and the aphelion. For Halley’s comet this is rather different. According to Kepler’s law of equal area, the comets sweeps through the same area close to the sun (i.e. around the perihelion) and far away from the sun (the aphelion). As the distance to the sun close to the perihelion is much smaller than that around the aphelion, the velocity much be much smaller around the aphelion to ensure equal areas in equal times. Note: our expressions are dimensionally correct.

Solution to Exercise 3

The energy of the object is given by:

\frac{1}{2}mv^2 + V(r) = E_0

(18)

We can find the potential from the force:

V(r) \equiv - \int_{\vec{r}_{ref}}^{\vec{r}} \vec{F} \cdot d\vec{r} = k\int_{\infty}^r \frac{dr}{r^2} = -\frac{k}{r}

(19)

where we have used that the force is central. Furthermore, we have taken as our reference point: infinity.

If the object follows a parabola, then its energy must be exactly zero. Thus:

\frac{1}{2}mv^2 - \frac{k}{r} = 0

(20)

a) This holds at any point of the trajectory. Thus for the point of closest distance to the force center we get:

\frac{1}{2}mv_{esc}^2 - \frac{k}{r_0} = 0 \Rightarrow v_{esc} = \sqrt{\frac{2k}{mr_0}}

(21)

b) Now that we have both the velocity and the position at one point on the parabola, we can compute the angular momentum. Since the force is central, we do know that the angular momentum is a constant. Hence, computing it for one particular point will give us the angular momentum for all points on the trajectory.

We use that at the point of closest approach the position vector and the velocity are perpendicular: $\vec{r}_0 \perp \vec{v}_{esc}$ . Thus the angular momentum is:

\vec{l} \equiv \vec{r} \times m\vec{v} = -mr_0v_{esc}\hat{z} = -\sqrt{2kmr_0}\hat{z}

(22)

Note: the minus sign is needed to have the angular momentum point into the drawing as $\hat{z}$ points towards us.

Solution to Exercise 4

According to Kepler 3, the orbital period, $T$ , and the length of the semi major axis, $a$ are related as:

T^2 = k a^3

(23)

with k a know constant, the same for all planets orbiting the sun. From this law, we immediately get that

\frac{a_{planet}}{a_{earth}} = \left ( \frac{T_{planet}}{T_{earth}} \right ) ^{2/3}

(24)

Thus, if we want to find the relative distance to the sun of each planet (that is, we express $a$ in terms of $a_{earth}$ which is 1 A.U.), we get:

Data

planet	orbital period (year)	semi-major axis (A.U.)
Mercury	0.241	0.39
Venus	0.615	0.72
Earth	1.000	1.00
Mars	1.881	1.52
Jupiter	11.86	5.20
Saturn	29.46	9.58
Uranus	84.0	19.2
Neptune	164.8	30.1

Fun fact: Neptune’s orbital period is so large, that since its discovery in 1846, it has only completed 1 Neptune-year (first full orbit in 2011!).

Solution to Exercise 5

First we make a sketch.

We have drawn gravity and the tension in the rod connecting the mass m tot the pivot point. From a physics point of view, the situation is as follows: the swing initially has only potential energy and it will gain kinetic energy on its way down, until all potential energy is converted into kinetic energy.

Two forces are acting on the swing, giving rise to two torques that cause the angular momentum to change. The velocity of the mass is always perpendicular to its position vector (a consequence of using a ‘clever’ choice for the origin of our coordinate system).

Now, we can set up the modeling for this problem.

Conservation of energy:

\frac{1}{2}mv^2 + mgy = const

(25)

Angular momentum:

\vec{l} = \vec{r} \times \vec{p} \Rightarrow l = mLv

(26)

Change of angular momentum:

\frac{d\vec{l}}{dt} = \sum_i \vec{r}_i \times \vec{F}_i

(27)

With this, we are ready for the develop phase. Conservation of energy gives us for the velocity at the lowest point:

\frac{1}{2}mv_{max}^2 + 0 = mgH \Rightarrow v_{max} = \sqrt{2gL} = 22 m/s

(28)

The minimum angular momentum is of course $l_{min} = 0$ at the highest point where the swing has zero velocity and changes direction.

Tha maximum angular momentum is at the lowest point: $l_{max} = mL\sqrt{2gL}$

Finally, the torque comes from gravity: $\vec{\Gamma} = \vec{r} \times -mg\hat{y} = mgL\sin \theta \hat{z}$ . The tension in the rod is always parallel to the position vector of $m$ and thus has zero torque.

Solution to Exercise 6

a) The seesaw is balanced if: $\sum_i \vec{r}_i \times \vec{F}_i = 0$ . Hence, we compute the net torque:

\begin{split} \sum_i \vec{r}_i \times \vec{F}_i &= -Lmg\hat{y} -\frac{3}{4}Lmg\hat{y} -\frac{2}{4}Lmg\hat{y} -\frac{1}{4}Lmg\hat{y} \\ &\;\;\;\;+ \frac{1}{4}Lmg\hat{y} +\frac{2}{4}Lmg\hat{y} -\frac{3}{4}Lmg\hat{y} \\ &= -mgL\hat{y} \neq 0 \end{split}

(29)

Thus the net torque is not zero and, indeed, the seesaw is unbalanced.

b) Where to place on the right hand side an additional mass $m$ ? We need to compensate $-mgL\hat{y}$ by adding $2m$ at some $x\gt0$ . Thus, we require:

\sum_i \vec{r}_i \times \vec{F}_i = -mgL\hat{y} +2mgx\hat{y} = 0

(30)

and we find that we need to place $2m$ at $x=\frac{1}{2}L$ .

c) In the second situation, we need to rethink how to compute the total torque. It still requires to calculate the contribution of all small masses $dm$ . We will label each small mass with index $i$ . It has a distance to the origin of $x_i$ . (Actually it does not have a single distance, but a varying one between $x_i$ and $x_i+dx$ . But since $dx$ is very small (and the same for all masses $dm_i$ ), it is accurate enough to use $x_i$ . In your calculus classes you will get the proof of this.)

\sum_i \vec{r}_i \times \vec{F}_i = \sum_i dm_i g x_i \hat{y}

(31)

Next, we realize that each mass $dm_i$ is given by

dm_i = \rho dx

(32)

Thus, the net torque can be written as

\sum_i \vec{r}_i \times \vec{F}_i = \sum_i \rho g x_i dx\hat{y}

(33)

This is a Riemann sum and if we let $dx \rightarrow 0$ , we arrive at an integral:

\sum_i \vec{r}_i \times \vec{F}_i = \int_{-L}^{\frac{3}{4}L} \rho g x dx = \left [ \frac{1}{2} \rho g x^2 \right ]_{-L}^{\frac{3}{4}L} = -\frac{7}{32}\rho gL^2

(34)

In the final step, we will replace $\rho$ in terms of the mass of the bar. We do that by noting that the total mass of the bar is the sum of all small parts $dm_i$ :

M_{bar} = \sum_i dm_i = \sum_i \rho dx = \int_{-L}^{\frac{3}{4}L} \rho dx = \rho \left [ x \right ]_{-L}^{\frac{3}{4}L} = \frac{7}{4} \rho L^2

(35)

This is to be expected: the total length of the bar is $\frac{7}{4}L$ and with a constant density per unit length we get the above result.

Thus in terms of $m$ we find:

7m = \frac{7}{4} \rho L \Rightarrow \rho = 4\frac{m}{L}

(36)

If we substitute this in the equation for the net torque, we find:

\sum_i \vec{r}_i \times \vec{F}_i = -\frac{7}{8}mgL \hat{y}

(37)

Indeed, the seesaw is unbalanced.
Note: we have an expression for the torque with the right dimensions!

d) Now we can find the position where we should place a mass $2m$ to balance the seesaw: we need to do that such that the net torque is zero.

0 = -\frac{7}{8}mgL + 2mgx \Rightarrow x = -\frac{7}{16}L

(38)

Conclusion: by splitting up the real bar in infinitesimally small parts, we can still compute the net torque (with many thanks to Bernhard Riemann, 1826-1866, a German mathematician).

Solution to Exercise 7

First, we complete the sketch by adding the force.

We note that the masses move under the influence of a central force. Hence, no torque is exerted and we conclude from N2 for angular momentum that the angular momentum is constant. From this, we can find the momentum and, hence, the velocity of each mass.
Subsequently, we can compute the kinetic energy in the initial and final situation. From that, we easily find the work done as $W_{12} = E_{kin,2} - E_{kin,1}$ .
Our final step will then be: compute the force from the definition of work.

Here we go:

\vec{F} // \vec{r} \Rightarrow \vec{r} \times \vec{F} = 0 \Rightarrow \frac{d\vec{l}}{dt} = 0 \Rightarrow l = const

(39)

If we substitute the initial and final conditions, we get:

l=Rmv + Rmv = 2mRv = 2 \left ( m\frac{R}{2}v_f \right ) \Rightarrow v_f = 2v

(40)

The work done is:

\begin{split} W_{12} &= E_{kin,2} - E_{kin,1} \\ &= 2 \left ( \frac{1}{2}m(2v)^2 \right ) - 2 \left ( \frac{1}{2}mv^2 \right ) \\ &= 3mv^2 \end{split}

(41)

From the definition of work, we can find the force, using that it is central and of constant magnitude:

\begin{split} W_{12} &\equiv \int_1^2 \vec{F} \cdot d\vec{r} \\ &= \int_R^{R/2} Fdr \\ &= -\frac{R}{2}F \end{split}

(42)

Thus, we get for $F$ :

-\frac{R}{2}F = 3mv^2 \Rightarrow F = - 6\frac{mv^2}{R}

(43)

Assess: we have the right units. Moreover, the value we have found is negative. This makes sense: to get $m$ from a distance $R$ to $\frac{R}{2}$ we need an inwards pointing force. Thus, $F$ must be negative as the central force is written as $\vec{F} = F\hat{r}$ . The direction of the force follows from the sign of $F$ , negative to make it attracting to the origin!