Angular momentum and central forces
A satellite, the ball in a track-and-field hammer throw and a wrecking ball swinging on a cable have something in common that is neither mass nor speed on its own, but how their motion turns around a point. That quantity, which combines position and linear momentum, is angular momentum.
Angular momentum of a particle
Making a stone on a short string spin takes little effort, but with a 2-meter string at the same speed it is much harder to slow down or deflect the motion: which physical quantity combines mass, speed and distance to measure this "rotational inertia"?
In a hospital lab, 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…
In a hospital lab, 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:
- 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.
- b) Calculate the angular momentum of the tube about the axis of rotation, expressed in $\text{kg} \times \text{m}^2/\text{s}$.
An athlete whirls a sling: a stone of $0.1\ \text{kg}$ describes a horizontal circle of radius $0.8\ \text{m}$ with a speed of $12\ \text{m/s}$. a) Calculate the linear…
An athlete whirls a sling: a stone of $0.1\ \text{kg}$ describes a horizontal circle of radius $0.8\ \text{m}$ with a speed of $12\ \text{m/s}$.
- a) Calculate the linear momentum of the stone, $p = m \times v$.
- b) Calculate its angular momentum about the center of the circle, $L = m \times v \times r$.
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 $L = m \times v \times r$.
- a) From the units of mass, velocity, and distance, deduce the SI unit of angular momentum.
- b) Verify that this unit is equivalent to $\text{J} \times \text{s}$ (recall that $1\ \text{J} = 1\ \text{kg} \times \text{m}^2/\text{s}^2$).
- c) From the formula, explain what happens to the angular momentum of a particle if its speed is doubled without changing its trajectory.
Orbits and spinning objects: period, ω and L
The Moon, a satellite or a Ferris wheel is described by its period, not by its speed. In one period $T$ the object travels once around a circle, a distance $2\pi r$, so its speed is $2\pi r / T$.
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 $24\ \text{h}$.
- a) Express the period in seconds and calculate the orbital speed, $v = \frac{2 \times \pi \times r}{T}$.
- b) Calculate the angular momentum of the satellite about the center of the Earth, in scientific notation.
- c) How would the result change if the satellite were twice as massive? Justify your answer without repeating the full calculation.
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: $7.35 \times 10^{22}\ \text{kg}$).
- a) Express the period in seconds and calculate the orbital speed of the Moon, $v = \frac{2 \times \pi \times r}{T}$.
- b) Calculate the orbital angular momentum of the Moon about the center of the Earth, in scientific notation and to three significant figures.
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:
- a) Express the mass and radius in SI units.
- b) Calculate the linear speed of the particle, $v = \omega \times r$.
- c) Calculate its angular momentum about the axis of the disc.
Angular momentum when the velocity is not perpendicular
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:
- 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.
- b) Calculate the angular momentum of the spacecraft about the center of the Moon.
- c) If the spacecraft were to pass through the same point with its velocity pointing directly toward the Moon, what would its angular momentum be? Reason your answer.
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 $290\ \text{kg} \times \text{m}^2/\text{s}$.
- a) Rearrange the expression $L = m \times v \times r$ to solve for the radius of rotation.
- b) Calculate the radius of rotation of the ball, to two significant figures.
- 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.
Two test bodies, A and B, rotate in the laboratory about the same axis. Body A has a mass of $0.3\ \text{kg}$, a speed of $4\ \text{m/s}$, and a radius of rotation of…
Two test bodies, A and B, rotate in the laboratory about the same axis. Body A has a mass of $0.3\ \text{kg}$, a speed of $4\ \text{m/s}$, and a radius of rotation of $0.5\ \text{m}$; body B has a mass of $0.6\ \text{kg}$, a speed of $2\ \text{m/s}$, and a radius of $1\ \text{m}$. The diagram shows both bodies viewed from above:
- a) Calculate the angular momentum of each body about the axis.
- b) Calculate the ratio $\frac{L_B}{L_A}$ and interpret it: how many times greater is the angular momentum of B?
- c) Without repeating calculations, reason how this ratio would change if the speed of A were doubled.
Central forces
If you swing a ball on a string, the tension in the string always pulls toward your hand; if you look at the Moon, gravity pulls it toward the center of the Earth: what geometric property do all forces that permanently point towards a fixed point have in common?
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:
- a) Calculate the torque of each force about the axis, $M = r \times F \times \sen\theta$.
- 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.
- c) Relate the result to the behavior of central forces: which of the two forces does a central force resemble?
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:
- a) Calculate its initial angular momentum about the center of the Earth.
- 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.
- 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.
Why a central force conserves angular momentum
When an ice skater spinning on the spot pulls her arms in towards her chest, her spin rate shoots up without anyone pushing her: why does the speed change when the distance from the axis changes if no external force produces rotation?
The simulation below shows a planet on an elliptical orbit: the force always points toward the Sun, so the angular momentum stays constant and the planet moves faster when it is closer.
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 $M_{Tierra} = 5.98 \times 10^{24}\ \text{kg}$.
- 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 $v = \sqrt{\frac{G \times M}{r}}$.
- b) Calculate the orbital speed of the satellite.
- c) Calculate its angular momentum about the center of the Earth.
- d) Justify, based on the central nature of the gravitational force, that this angular momentum remains constant throughout the orbit.
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:
- a) Calculate the initial angular momentum of the small ball.
- 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.
- c) Calculate the speed of the small ball when the radius is $r_2$.
- d) Compare the initial and final kinetic energies and explain where the increase comes from.
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.
- a) Calculate the angular momentum of the probe at that point.
- 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.
- 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.
Systems and spinning bodies: L = Iω
A merry-go-round, a disc or a pair of skaters is made of many particles that all turn with the same angular speed $\omega$. The angular momentum of the whole system is the sum of the angular momenta of its parts, using the same sign convention for all of them.
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 $\omega = 3\ \text{rad/s}$.
- a) Calculate the linear speed of each skater, $v = \omega \times r$.
- b) Calculate the angular momentum of each skater about the center of rotation.
- c) The total angular momentum of the system is the sum of both: calculate it and justify that, if external forces have zero torque, this total is conserved even if the skaters push against each other (internal forces do not change the total angular momentum).
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 $2.98 \times 10^{4}\ \text{m/s}$.
- a) Calculate the moment of inertia of the Earth, in scientific notation.
- b) Express the rotational period in seconds and calculate the rotational angular momentum of the Earth.
- c) Calculate its orbital angular momentum about the Sun and find the ratio between both: which one dominates and by approximately what factor?
Summary
Back to the start: the satellite, the hammer ball and the wrecking ball all have angular momentum $L = m v r \sin\theta$ about a point, and it stays constant when the only force is central.