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Gravitational field · Angular momentum

Angular momentum and central forces

A communications satellite over the Caribbean, an electron in its orbit, and a stone spinning in a sling have something in common that is not just mass or speed, but how their motion rotates around a point. That quantity, which combines position and momentum, is the angular momentum of the particle. By the end you will see why a planet moves faster when it passes close to the Sun.

Angular momentum of a particle

Spinning a stone tied to a short string takes little effort, but with a two-meter string, at the same speed, it is much harder to stop or deflect the rotation. What quantity combines mass, speed, and distance to measure that "rotational inertia"?

Angular momentum of a particle about point O: the position vector r and the momentum p lie in the plane, and L = r × p is perpendicular to it, pointing out of the plane. plane of r and p O r θ p = m·v L = r × p out of the plane

Direction and sense. With your right hand, point your fingers in the direction of $\vec{r}$ and curl them toward $\vec{v}$: your thumb indicates $\vec{L}$. If a particle rotates counterclockwise as seen from above, $\vec{L}$ points out of the page toward you; if it rotates clockwise, it points into the page.

Units. $[m][v][r] = \text{kg}\cdot\text{m/s}\cdot\text{m} = \text{kg}\cdot\text{m}^2/\text{s}$. Since $1\ \text{J} = 1\ \text{kg}\cdot\text{m}^2/\text{s}^2$, it follows that $\text{kg}\cdot\text{m}^2/\text{s} = \text{J}\cdot\text{s}$.

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In a hospital laboratory, a centrifuge spins a test tube of $0.02\ \text{kg}$ that describes a circle of radius $0.15\ \text{m}$ with a speed of $9\ \text{m/s}$. The…

In a hospital laboratory, a centrifuge spins a test tube of $0.02\ \text{kg}$ that describes a circle of radius $0.15\ \text{m}$ with a speed of $9\ \text{m/s}$. The diagram shows the rotation viewed from above:

rotation r = 0.15 m axis tube 0.02 kg v = 9 m/s
  1. a) Write down the expression for the angular momentum of a particle about a point, $L = m \times v \times r$, valid when the velocity is perpendicular to the radius.
  2. b) Calculate the angular momentum of the tube about the axis of rotation, expressed in $\text{kg} \times \text{m}^2/\text{s}$.
Solve
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When the velocity of a particle is perpendicular to its position vector (for example, in circular motion about the center), the magnitude of its angular momentum is…

When the velocity of a particle is perpendicular to its position vector (for example, in circular motion about the center), the magnitude of its angular momentum is calculated as $L = m \times v \times r$.

  1. a) From the units of mass, velocity, and distance, deduce the SI unit of angular momentum.
  2. b) Verify that this unit is equivalent to $\text{J} \times \text{s}$ (recall that $1\ \text{J} = 1\ \text{kg} \times \text{m}^2/\text{s}^2$).
  3. c) From the formula, explain what happens to the angular momentum of a particle if its speed is doubled without changing its trajectory.
Solve
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The Earth describes a practically circular orbit of radius $1.5 \times 10^{11}\ \text{m}$ around the Sun with a speed of $29.8\ \text{km/s}$ (mass of the Earth:…

The Earth describes a practically circular orbit of radius $1.5 \times 10^{11}\ \text{m}$ around the Sun with a speed of $29.8\ \text{km/s}$ (mass of the Earth: $5.98 \times 10^{24}\ \text{kg}$).

  1. a) Express the speed in $\text{m/s}$, in standard form.
  2. b) Calculate the orbital angular momentum of the Earth about the Sun, to three significant figures.
  3. c) State the direction of the vector $\vv{L}$ relative to the orbital plane and whether it changes throughout the year.
Solve
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An abrasive particle of mass $1\ \text{g}$ is attached to the edge of an angle grinder disc, which rotates with angular velocity $\omega = 250\ \text{rad/s}$ at…

An abrasive particle of mass $1\ \text{g}$ is attached to the edge of an angle grinder disc, which rotates with angular velocity $\omega = 250\ \text{rad/s}$ at $5\ \text{cm}$ from the axis. The diagram shows the disc viewed from above:

axis r = 5 cm particle (1 g) v = ? rotation: 250 rad/s
  1. a) Express the mass and radius in SI units.
  2. b) Calculate the linear speed of the particle, $v = \omega \times r$.
  3. c) Calculate its angular momentum about the axis of the disc.
Solve

When the velocity is not perpendicular to r

Position vector r from point O to a spacecraft and velocity vector v at the spacecraft, forming an angle θ between them that is not a right angle. O r v θ spacecraft

When $\vec{v}$ makes any angle $\theta$ with $\vec{r}$, the full magnitude with $\sin\theta$ is used. If $\vec{v}$ points directly toward $O$ or directly away from it, $\theta = 0^\circ$ or $180^\circ$ and $L = 0$.

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A particle in a flight simulator describes a circle on the screen in an anticlockwise direction around point O. a) Write down the vector definition of angular momentum,…

A particle in a flight simulator describes a circle on the screen in an anticlockwise direction around point O.

  1. a) Write down the vector definition of angular momentum, $\vv{L} = \vv{r} \times \vv{p}$, and state what $\vv{r}$ and $\vv{p} = m \times \vv{v}$ represent.
  2. b) Draw in your notebook the circle and the vectors $\vv{r}$, $\vv{v}$ and $\vv{p}$ at any arbitrary point and, applying the right-hand rule to the vector product, deduce the direction and sense of $\vv{L}$ relative to the plane of the screen.

You can do it on paper and send a photo with your answer.

Solve
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The Orion spacecraft, with a mass of $2.5 \times 10^{4}\ \text{kg}$, passes through a point $2 \times 10^{6}\ \text{m}$ from the center of the Moon with a speed of…

The Orion spacecraft, with a mass of $2.5 \times 10^{4}\ \text{kg}$, passes through a point $2 \times 10^{6}\ \text{m}$ from the center of the Moon with a speed of $1.5 \times 10^{3}\ \text{m/s}$, in a direction making an angle of $30^\circ$ with the line joining it to the Moon ($\sen 30^\circ = 0.5$). The figure shows that instant:

Moon r = 2 000 km Orion spacecraft v = 1.5 km/s 30°
  1. a) Write down the general expression for the magnitude of angular momentum, $L = m \times v \times r \times \sen\theta$, and explain the meaning of each term.
  2. b) Calculate the angular momentum of the spacecraft about the center of the Moon.
  3. c) If the spacecraft were to pass through the same point with its velocity pointing directly towards the Moon, what would its angular momentum be? Reason your answer.
Solve
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In the hammer throw event, the ball of $7.26\ \text{kg}$ rotates with a speed of $25\ \text{m/s}$ and its angular momentum about the axis of rotation is…

In the hammer throw event, the ball of $7.26\ \text{kg}$ rotates with a speed of $25\ \text{m/s}$ and its angular momentum about the axis of rotation is $290\ \text{kg} \times \text{m}^2/\text{s}$.

  1. a) Rearrange the expression $L = m \times v \times r$ to solve for the radius of rotation.
  2. b) Calculate the radius of rotation of the ball, to two significant figures.
  3. c) Compare the result with the regulation length of the wire, which is about $1.2\ \text{m}$, and explain the reason for the difference.
Solve

Calculations with orbital and rotation data

In astronomy and rotation problems you are sometimes given the period $T$ or the angular velocity $\omega$ instead of the speed. First you convert to $v$ and then apply $L = m\,v\,r$:

  • With the period: $v = \dfrac{2\pi r}{T}$ (with $T$ in seconds).

  • With the angular velocity: $v = \omega\, r$ (with $\omega$ in rad/s).

  • For several bodies rotating together, the total $L$ is the sum of the $L$ of each one (same axis and same direction).

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A communications satellite of $300\ \text{kg}$ occupies a circular geostationary orbit of radius $4.22 \times 10^{7}\ \text{m}$ and completes one revolution every…

A communications satellite of $300\ \text{kg}$ occupies a circular geostationary orbit of radius $4.22 \times 10^{7}\ \text{m}$ and completes one revolution every $24\ \text{h}$.

  1. a) Express the period in seconds and calculate the orbital speed, $v = \frac{2 \times \pi \times r}{T}$.
  2. b) Calculate the angular momentum of the satellite about the center of the Earth, in scientific notation.
  3. c) How would the result change if the satellite were twice as massive? Justify your answer without repeating the full calculation.
Solve
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The Moon describes an almost circular orbit of radius $3.84 \times 10^{8}\ \text{m}$ around the Earth with a period of $27.3$ days (mass of the Moon:…

The Moon describes an almost circular orbit of radius $3.84 \times 10^{8}\ \text{m}$ around the Earth with a period of $27.3$ days (mass of the Moon: $7.35 \times 10^{22}\ \text{kg}$).

  1. a) Express the period in seconds and calculate the orbital speed of the Moon, $v = \frac{2 \times \pi \times r}{T}$.
  2. b) Calculate the orbital angular momentum of the Moon about the center of the Earth, in scientific notation and to three significant figures.
Solve
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A pair of figure skaters perform a routine in which both spin joined by their arms around the midpoint between them. Model each skater as a point mass:…

A pair of figure skaters perform a routine in which both spin joined by their arms around the midpoint between them. Model each skater as a point mass: $m_1 = 60\ \text{kg}$ at $0.9\ \text{m}$ from the center of rotation and $m_2 = 50\ \text{kg}$ at $1.1\ \text{m}$, both with the same angular velocity $\omega = 3\ \text{rad/s}$.

  1. a) Calculate the linear speed of each skater, $v = \omega \times r$.
  2. b) Calculate the angular momentum of each skater about the center of rotation.
  3. c) The total angular momentum of the system is the sum of both: calculate it and justify that, if external forces have zero moment, this total is conserved even if the skaters push against each other (internal forces do not change the total angular momentum).
Solve

Central forces

If you spin a ball tied to a string, the tension always pulls toward your hand; gravity pulls the Moon always toward the center of the Earth. What do these forces have in common?

Three particles at different positions with forces whose lines of action all pass through the same center of force O. all forces point toward O F₁ F₂ F₃ O center of force

Notice the opposite case: a force perpendicular to a bar, applied far from the axis, has $\theta = 90^\circ$ and the maximum moment, $M = rF$. A force pointing toward the axis does not make the bar rotate.

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A force is central when its line of action always passes through a fixed point called the center of force. State which of the following forces are central and which are…

A force is central when its line of action always passes through a fixed point called the center of force. State which of the following forces are central and which are not, justifying each answer:

  1. a) The gravitational attraction of the Sun on a planet.
  2. b) The tension in the string of a ball spinning attached to a pole.
  3. c) The electric force exerted by the nucleus of an atom on an electron.
  4. d) Air resistance acting on an artificial satellite in low orbit.
  5. e) The thrust of a rocket engine accelerating in the direction of motion.
Solve
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In the Bohr model of the atom, the electron of a hydrogen atom describes a circular orbit of radius $5.3 \times 10^{-11}\ \text{m}$ with a speed of…

In the Bohr model of the atom, the electron of a hydrogen atom describes a circular orbit of radius $5.3 \times 10^{-11}\ \text{m}$ with a speed of $2.2 \times 10^{6}\ \text{m/s}$ (mass of the electron: $9.1 \times 10^{-31}\ \text{kg}$). The diagram represents the model:

nucleus r = 53 pm electron v = 2 200 km/s
  1. a) Calculate the angular momentum of the electron about the nucleus, in standard form and to two significant figures.
  2. b) State which force acts as the centripetal force in this case and justify that it is a central force.
Solve
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The moment of a force about a point is $\vv{M} = \vv{r} \times \vv{F}$, and its magnitude is $M = r \times F \times \sen\theta$, where $\theta$ is the angle between…

The moment of a force about a point is $\vv{M} = \vv{r} \times \vv{F}$, and its magnitude is $M = r \times F \times \sen\theta$, where $\theta$ is the angle between $\vv{r}$ and $\vv{F}$. The figure shows the vectors and the angle:

r F θ O
  1. a) Reason what the moment of a central force about the center of force is: what is the angle $\theta$ between the position vector and the force?
  2. b) The rate of change of angular momentum satisfies $\frac{\Delta \vv{L}}{\Delta t} = \vv{M}$. Deduce what happens to the angular momentum of a body subjected solely to a central force.
  3. c) Give two examples of real-world motions in which the above holds true.
Solve
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Two forces of $20\ \text{N}$ are applied to a rod that can rotate about an axis, at points located $0.4\ \text{m}$ from the axis: force A points radially towards the…

Two forces of $20\ \text{N}$ are applied to a rod that can rotate about an axis, at points located $0.4\ \text{m}$ from the axis: force A points radially towards the axis and force B is perpendicular to the rod ($\sen 90^\circ = 1$, $\sen 0^\circ = 0$). The diagram shows the rod and the two forces:

axis force A: 20 N radial force B: 20 N perpendicular 0.4 m
  1. a) Calculate the moment of each force about the axis, $M = r \times F \times \sen\theta$.
  2. b) From $\frac{\Delta \vv{L}}{\Delta t} = \vv{M}$, state which of the two forces changes the angular momentum of the rod and which does not.
  3. c) Relate the result to the behavior of central forces: which of the two forces does a central force resemble?
Solve

Why a central force conserves L

When a merengue dancer spins and pulls in her arms, she spins faster without anyone pushing her. In the sky something similar happens: why does a planet go faster when it is close to the Sun?

$L$ being constant means that $\vec{L}$ changes in neither magnitude, direction, nor sense: the orbit stays in one plane and $m\,v\,r\sin\theta$ does not vary.

Go back to the beginning: the planet speeds up near the Sun because gravity is a central force and conserves $L$; the same thing happens to the dancer when she pulls in her arms.

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It is desired to place an Earth observation satellite of $500\ \text{kg}$ into a circular orbit of radius $7 \times 10^{6}\ \text{m}$ measured from the center of the…

It is desired to place an Earth observation satellite of $500\ \text{kg}$ into a circular orbit of radius $7 \times 10^{6}\ \text{m}$ measured from the center of the Earth. Data: $G = 6.67 \times 10^{-11}\ \text{N} \times \text{m}^2/\text{kg}^2$ and $M_{Tierra} = 5.98 \times 10^{24}\ \text{kg}$.

  1. a) The only force acting on the satellite is the gravitational force, which acts as a centripetal force. Equate $G \times \frac{M \times m}{r^2} = m \times \frac{v^2}{r}$ and deduce the expression $v = \sqrt{\frac{G \times M}{r}}$.
  2. b) Calculate the orbital speed of the satellite.
  3. c) Calculate its angular momentum about the center of the Earth.
  4. d) Justify, based on the central nature of the gravitational force, that this angular momentum remains constant throughout the orbit.
Solve
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In a laboratory experiment, a small ball of $0.1\ \text{kg}$ rotates on a frictionless horizontal table attached to a string passing through a vertical tube fixed to the…

In a laboratory experiment, a small ball of $0.1\ \text{kg}$ rotates on a frictionless horizontal table attached to a string passing through a vertical tube fixed to the table. Initially it rotates with a radius $r_1 = 0.6\ \text{m}$ and a speed $v_1 = 3\ \text{m/s}$. Pulling the string from below the tube reduces the radius to $r_2 = 0.3\ \text{m}$. The diagram shows the set-up viewed from above:

tube r₁ = 0.6 m v₁ = 3 m/s r₂ = 0.3 m v₂ = ?
  1. a) Calculate the initial angular momentum of the small ball.
  2. b) The tension in the string is a central force, as it always points towards the tube: justify, from its moment, that the angular momentum of the small ball is conserved during the process.
  3. c) Calculate the speed of the small ball when the radius is $r_2$.
  4. d) Compare the initial and final kinetic energies and explain where the increase comes from.
Solve
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A proton ($m = 1.67 \times 10^{-27}\ \text{kg}$) moves in a straight line with constant speed $2 \times 10^{6}\ \text{m/s}$ inside the chamber of a particle accelerator.…

A proton ($m = 1.67 \times 10^{-27}\ \text{kg}$) moves in a straight line with constant speed $2 \times 10^{6}\ \text{m/s}$ inside the chamber of a particle accelerator. Its trajectory passes at a minimum distance $d = 3 \times 10^{-9}\ \text{m}$ from a sensor located at point O.

  1. a) The angular momentum about O is $L = m \times v \times r \times \sen\theta$. Draw the situation in your notebook and reason that, although $r$ and $\theta$ change along the path, the product $r \times \sen\theta$ is always equal to $d$.
  2. b) Calculate the angular momentum of the proton about O, in standard form.
  3. c) The proton moves freely, without forces: is it consistent for its angular momentum to be constant even though it does not describe a curve? Justify your answer.

You can do it on paper and send a photo with your answer.

Solve
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A probe of $800\ \text{kg}$ with its engines turned off approaches Mars. At the point of closest approach, it is at $4 \times 10^{6}\ \text{m}$ from the center of the…

A probe of $800\ \text{kg}$ with its engines turned off approaches Mars. At the point of closest approach, it is at $4 \times 10^{6}\ \text{m}$ from the center of the planet and travels with a speed of $5 \times 10^{3}\ \text{m/s}$, perpendicular to the line joining it to Mars.

  1. a) Calculate the angular momentum of the probe at that point.
  2. b) While the engines are turned off, the only force acting on the probe is the gravitational attraction of Mars: justify why the angular momentum remains constant throughout the maneuver.
  3. c) When the probe moves away again and is at $1.6 \times 10^{7}\ \text{m}$ from the center of Mars, moving perpendicular to the line joining them, calculate its speed.
Solve

A glimpse of the atom and the rotation of the Earth

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In the Bohr model of the hydrogen atom, the electron can only have certain allowed values of angular momentum. The first allowed value is $L_1 = \frac{h}{2 \times \pi}$,…

In the Bohr model of the hydrogen atom, the electron can only have certain allowed values of angular momentum. The first allowed value is $L_1 = \frac{h}{2 \times \pi}$, with $h = 6.63 \times 10^{-34}\ \text{J} \times \text{s}$.

  1. a) Calculate the value of $L_1$, in standard form and to three significant figures.
  2. b) Using the data for the ground orbit (radius $5.3 \times 10^{-11}\ \text{m}$, speed $2.2 \times 10^{6}\ \text{m/s}$, mass of the electron $9.1 \times 10^{-31}\ \text{kg}$), calculate the angular momentum of the electron with $L = m \times v \times r$ and compare it with $L_1$.
  3. c) The second allowed level is $L_2 = 2 \times L_1$. If the radius of that orbit is $2.12 \times 10^{-10}\ \text{m}$, calculate the speed the electron must have in it.
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A satellite of $200\ \text{kg}$ in low orbit ($r = 6.8 \times 10^{6}\ \text{m}$, $v = 7.7 \times 10^{3}\ \text{m/s}$) experiences slight friction with the upper layers…

A satellite of $200\ \text{kg}$ in low orbit ($r = 6.8 \times 10^{6}\ \text{m}$, $v = 7.7 \times 10^{3}\ \text{m/s}$) experiences slight friction with the upper layers of the atmosphere. The figure shows the orbit and the direction of the friction:

Earth r = 6.800 km satellite v friction
  1. a) Calculate its initial angular momentum about the center of the Earth.
  2. b) Friction is a tangential force, opposite to velocity: is its moment about the center of the Earth zero? Is friction a central force? Reason both answers.
  3. c) In light of the above, predict what will happen to the angular momentum and the orbital radius over the months: will the satellite maintain its orbit indefinitely? Justify your answer.
Solve
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The rotational angular momentum of a spherical planet is given by $L = I \times \omega$, with $I = \frac{2}{5} \times M \times R^2$ and…

The rotational angular momentum of a spherical planet is given by $L = I \times \omega$, with $I = \frac{2}{5} \times M \times R^2$ and $\omega = \frac{2 \times \pi}{T}$. Earth data: $M = 5.98 \times 10^{24}\ \text{kg}$, $R = 6.37 \times 10^{6}\ \text{m}$, rotational period $T = 24\ \text{h}$; in its orbit, radius $1.5 \times 10^{11}\ \text{m}$ and speed $2.98 \times 10^{4}\ \text{m/s}$.

  1. a) Calculate the moment of inertia of the Earth, in standard form.
  2. b) Express the rotational period in seconds and calculate the rotational angular momentum of the Earth.
  3. c) Calculate its orbital angular momentum about the Sun and find the ratio between both: which one dominates and by approximately what factor?
Solve

Summary

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