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I wrote a quickie answer where I estimate the total rotational angular momentum of Jupiter as about 7E+38 kg m^2/s. However it bothers me because one of the sources gets this number assuming uniform density, and the other two cite 6.9E+38 kg m^2/s but it's not clear to me where this comes from.

Are there better estimates that use reasonable models for Jupiter's density distribution?

**update:** If not answerable due to uncertainties in density distribution and/or gradients in rotation rate (i.e. non rigid-body rotation) which I sort-of remember reading about recently, then an explanation why a better value isn't available would be an acceptable answer.

A better estimate might be to use the moment of inertia of a $n=3/2$ (fully convective) polytrope, which will be a good approximation in an object like Jupiter, even when it approaches electron degeneracy in its interior. You can look this up and it is given by $I = kMR^2$, with $k=0.205$ (almost exactly a factor of two smaller than a uniform sphere because the mass is concentrated towards the centre).

If we use a mass $M$ of $1.898 imes 10^{27}$ kg, the equatorial radius of Jupiter ($R=71500$ km) and the rotation rate determined from its magnetosphere ($P= 9.93$ hours), then the angular momentum ($2pi I/P$) is $3.496 imes 10^{38}$ kg m$^2$ s$^{-1}$.

If instead, we use the volumetric average radius of $R=69900$ km, this reduces to $3.34 imes 10^{38}$ kg m$^2$ s$^{-1}$.

This simple approach neglects the more complicated equation of state of Jupiter, it's non-sphericity and the probable presence of a solid core. Consideration of these, together with constraints provided by the measured even harmonics of the gravitational field (referred to as $J_2$ and $J_4$, led Helled et al. (2011) to suggest that $k=0.264$, with an uncertainty below 1%. Combined with the average radius, this gives $4.30 imes 10^{38}$ kg m$^2$ s$^{-1}$.

More recent work by Ni (2018) has used refinements to the gravitational harmonics from Juno measurements, along with a more sophisticated interior model to suggest $k=0.274$ with 0.5% uncertainty.

## 11 Chapter Review

Can a round object released from rest at the top of a frictionless incline undergo rolling motion?

A cylindrical can of radius *R* is rolling across a horizontal surface without slipping. (a) After one complete revolution of the can, what is the distance that its center of mass has moved? (b) Would this distance be greater or smaller if slipping occurred?

A wheel is released from the top on an incline. Is the wheel most likely to slip if the incline is steep or gently sloped?

Which rolls down an inclined plane faster, a hollow cylinder or a solid sphere? Both have the same mass and radius.

A hollow sphere and a hollow cylinder of the same radius and mass roll up an incline without slipping and have the same initial center of mass velocity. Which object reaches a greater height before stopping?

#### 11.2 Angular Momentum

Can you assign an angular momentum to a particle without first defining a reference point?

For a particle traveling in a straight line, are there any points about which the angular momentum is zero? Assume the line intersects the origin.

Under what conditions does a rigid body have angular momentum but not linear momentum?

If a particle is moving with respect to a chosen origin it has linear momentum. What conditions must exist for this particle’s angular momentum to be zero about the chosen origin?

If you know the velocity of a particle, can you say anything about the particle’s angular momentum?

#### 11.3 Conservation of Angular Momentum

What is the purpose of the small propeller at the back of a helicopter that rotates in the plane perpendicular to the large propeller?

Suppose a child walks from the outer edge of a rotating merry-go-round to the inside. Does the angular velocity of the merry-go-round increase, decrease, or remain the same? Explain your answer. Assume the merry-go-round is spinning without friction.

As the rope of a tethered ball winds around a pole, what happens to the angular velocity of the ball?

Suppose the polar ice sheets broke free and floated toward Earth’s equator without melting. What would happen to Earth’s angular velocity?

Explain why stars spin faster when they collapse.

Competitive divers pull their limbs in and curl up their bodies when they do flips. Just before entering the water, they fully extend their limbs to enter straight down (see below). Explain the effect of both actions on their angular velocities. Also explain the effect on their angular momentum.

#### 11.4 Precession of a Gyroscope

Gyroscopes used in guidance systems to indicate directions in space must have an angular momentum that does not change in direction. When placed in the vehicle, they are put in a compartment that is separated from the main fuselage, such that changes in the orientation of the fuselage does not affect the orientation of the gyroscope. If the space vehicle is subjected to large forces and accelerations how can the direction of the gyroscopes angular momentum be constant at all times?

Earth precesses about its vertical axis with a period of 26,000 years. Discuss whether Equation 11.12 can be used to calculate the precessional angular velocity of Earth.

### Problems

#### 11.1 Rolling Motion

What is the angular velocity of a 75.0-cm-diameter tire on an automobile traveling at 90.0 km/h?

A boy rides his bicycle 2.00 km. The wheels have radius 30.0 cm. What is the total angle the tires rotate through during his trip?

If the boy on the bicycle in the preceding problem accelerates from rest to a speed of 10.0 m/s in 10.0 s, what is the angular acceleration of the tires?

Formula One race cars have 66-cm-diameter tires. If a Formula One averages a speed of 300 km/h during a race, what is the angular displacement in revolutions of the wheels if the race car maintains this speed for 1.5 hours?

A marble rolls down an incline at 30 ° 30° from rest. (a) What is its acceleration? (b) How far does it go in 3.0 s?

Repeat the preceding problem replacing the marble with a solid cylinder. Explain the new result.

A rigid body with a cylindrical cross-section is released from the top of a 30 ° 30° incline. It rolls 10.0 m to the bottom in 2.60 s. Find the moment of inertia of the body in terms of its mass *m* and radius *r.*

A yo-yo can be thought of a solid cylinder of mass *m* and radius *r* that has a light string wrapped around its circumference (see below). One end of the string is held fixed in space. If the cylinder falls as the string unwinds without slipping, what is the acceleration of the cylinder?

A solid cylinder of radius 10.0 cm rolls down an incline with slipping. The angle of the incline is 30 ° . 30°. The coefficient of kinetic friction on the surface is 0.400. What is the angular acceleration of the solid cylinder? What is the linear acceleration?

A bowling ball rolls up a ramp 0.5 m high without slipping to storage. It has an initial velocity of its center of mass of 3.0 m/s. (a) What is its velocity at the top of the ramp? (b) If the ramp is 1 m high does it make it to the top?

A 40.0-kg solid cylinder is rolling across a horizontal surface at a speed of 6.0 m/s. How much work is required to stop it?

A 40.0-kg solid sphere is rolling across a horizontal surface with a speed of 6.0 m/s. How much work is required to stop it? Compare results with the preceding problem.

A solid cylinder rolls up an incline at an angle of 20 ° . 20°. If it starts at the bottom with a speed of 10 m/s, how far up the incline does it travel?

A solid cylindrical wheel of mass *M* and radius *R* is pulled by a force F ⃗ F→ applied to the center of the wheel at 37 ° 37° to the horizontal (see the following figure). If the wheel is to roll without slipping, what is the maximum value of ∣∣∣ F ⃗ ∣∣∣ ? |F→|? The coefficients of static and kinetic friction are μ S = 0.40 and μ k = 0.30 . μS=0.40andμk=0.30.

A hollow cylinder is given a velocity of 5.0 m/s and rolls up an incline to a height of 1.0 m. If a hollow sphere of the same mass and radius is given the same initial velocity, how high does it roll up the incline?

#### 11.2 Angular Momentum

A 0.2-kg particle is travelling along the line y = 2.0 m y=2.0m with a velocity 5.0 m / s 5.0m/s . What is the angular momentum of the particle about the origin?

A bird flies overhead from where you stand at an altitude of 300.0 m and at a speed horizontal to the ground of 20.0 m/s. The bird has a mass of 2.0 kg. The radius vector to the bird makes an angle θ θ with respect to the ground. The radius vector to the bird and its momentum vector lie in the *xy*-plane. What is the bird’s angular momentum about the point where you are standing?

A Formula One race car with mass 750.0 kg is speeding through a course in Monaco and enters a circular turn at 220.0 km/h in the counterclockwise direction about the origin of the circle. At another part of the course, the car enters a second circular turn at 180 km/h also in the counterclockwise direction. If the radius of curvature of the first turn is 130.0 m and that of the second is 100.0 m, compare the angular momenta of the race car in each turn taken about the origin of the circular turn.

A particle of mass 5.0 kg has position vector r ⃗ = ( 2.0 i ˆ − 3.0 j ˆ ) m r→=(2.0i^−3.0j^)m at a particular instant of time when its velocity is v ⃗ = ( 3.0 i ˆ ) m / s v→=(3.0i^)m/s with respect to the origin. (a) What is the angular momentum of the particle? (b) If a force F ⃗ = 5.0 j ˆ N F→=5.0j^N acts on the particle at this instant, what is the torque about the origin?

Use the right-hand rule to determine the directions of the angular momenta about the origin of the particles as shown below. The *z-*axis is out of the page.

Suppose the particles in the preceding problem have masses m 1 = 0.10 kg, m 2 = 0.20 kg, m 3 = 0.30 kg, m1=0.10kg,m2=0.20kg,m3=0.30kg, m 4 = 0.40 kg m4=0.40kg . The velocities of the particles are v 1 = 2.0 i ˆ m / s v1=2.0i^m/s , v 2 = ( 3.0 i ˆ − 3.0 j ˆ ) m / s v2=(3.0i^−3.0j^)m/s , v 3 = −1.5 j ˆ m / s v3=−1.5j^m/s , v 4 = −4.0 i ˆ m / s v4=−4.0i^m/s . (a) Calculate the angular momentum of each particle about the origin. (b) What is the total angular momentum of the four-particle system about the origin?

Two particles of equal mass travel with the same speed in opposite directions along parallel lines separated by a distance*d*. Show that the angular momentum of this two-particle system is the same no matter what point is used as the reference for calculating the angular momentum.

An airplane of mass 4.0 × 10 4 kg 4.0×104kg flies horizontally at an altitude of 10 km with a constant speed of 250 m/s relative to Earth. (a) What is the magnitude of the airplane’s angular momentum relative to a ground observer directly below the plane? (b) Does the angular momentum change as the airplane flies along its path?

At a particular instant, a 1.0-kg particle’s position is r ⃗ = ( 2.0 i ˆ − 4.0 j ˆ + 6.0 k ˆ ) m r→=(2.0i^−4.0j^+6.0k^)m , its velocity is v ⃗ = ( −1.0 i ˆ + 4.0 j ˆ + 1.0 k ˆ ) m / s v→=(−1.0i^+4.0j^+1.0k^)m/s , and the force on it is F ⃗ = ( 10.0 i ˆ + 15.0 j ˆ ) N F→=(10.0i^+15.0j^)N . (a) What is the angular momentum of the particle about the origin? (b) What is the torque on the particle about the origin? (c) What is the time rate of change of the particle’s angular momentum at this instant?

A particle of mass *m* is dropped at the point ( − d , 0 ) (−d,0) and falls vertically in Earth’s gravitational field − g j ˆ . −gj^. (a) What is the expression for the angular momentum of the particle around the *z*-axis, which points directly out of the page as shown below? (b) Calculate the torque on the particle around the *z*-axis. (c) Is the torque equal to the time rate of change of the angular momentum?

(a) Calculate the angular momentum of Earth in its orbit around the Sun. (b) Compare this angular momentum with the angular momentum of Earth about its axis.

A boulder of mass 20 kg and radius 20 cm rolls down a hill 15 m high from rest. What is its angular momentum when it is half way down the hill? (b) At the bottom?

A satellite is spinning at 6.0 rev/s. The satellite consists of a main body in the shape of a sphere of radius 2.0 m and mass 10,000 kg, and two antennas projecting out from the center of mass of the main body that can be approximated with rods of length 3.0 m each and mass 10 kg. The antenna’s lie in the plane of rotation. What is the angular momentum of the satellite?

A propeller consists of two blades each 3.0 m in length and mass 120 kg each. The propeller can be approximated by a single rod rotating about its center of mass. The propeller starts from rest and rotates up to 1200 rpm in 30 seconds at a constant rate. (a) What is the angular momentum of the propeller at t = 10 s t = 20 s? t=10st=20s? (b) What is the torque on the propeller?

A pulsar is a rapidly rotating neutron star. The Crab nebula pulsar in the constellation Taurus has a period of 33.5 × 10 −3 s 33.5×10−3s , radius 10.0 km, and mass 2.8 × 10 30 kg . 2.8×1030kg. The pulsar’s rotational period will increase over time due to the release of electromagnetic radiation, which doesn’t change its radius but reduces its rotational energy. (a) What is the angular momentum of the pulsar? (b) Suppose the angular velocity decreases at a rate of 10 −14 rad / s 2 10−14rad/s2 . What is the torque on the pulsar?

The blades of a wind turbine are 30 m in length and rotate at a maximum rotation rate of 20 rev/min. (a) If the blades are 6000 kg each and the rotor assembly has three blades, calculate the angular momentum of the turbine at this rotation rate. (b) What is the torque require to rotate the blades up to the maximum rotation rate in 5 minutes?

A roller coaster has mass 3000.0 kg and needs to make it safely through a vertical circular loop of radius 50.0 m. What is the minimum angular momentum of the coaster at the bottom of the loop to make it safely through? Neglect friction on the track. Take the coaster to be a point particle.

A mountain biker takes a jump in a race and goes airborne. The mountain bike is travelling at 10.0 m/s before it goes airborne. If the mass of the front wheel on the bike is 750 g and has radius 35 cm, what is the angular momentum of the spinning wheel in the air the moment the bike leaves the ground?

#### 11.3 Conservation of Angular Momentum

A disk of mass 2.0 kg and radius 60 cm with a small mass of 0.05 kg attached at the edge is rotating at 2.0 rev/s. The small mass suddenly separates from the disk. What is the disk’s final rotation rate?

The Sun’s mass is 2.0 × 10 30 kg, 2.0×1030kg, its radius is 7.0 × 10 5 km, 7.0×105km, and it has a rotational period of approximately 28 days. If the Sun should collapse into a white dwarf of radius 3.5 × 10 3 km, 3.5×103km, what would its period be if no mass were ejected and a sphere of uniform density can model the Sun both before and after?

A cylinder with rotational inertia I 1 = 2.0 kg · m 2 I1=2.0kg·m2 rotates clockwise about a vertical axis through its center with angular speed ω 1 = 5.0 rad / s . ω1=5.0rad/s. A second cylinder with rotational inertia I 2 = 1.0 kg · m 2 I2=1.0kg·m2 rotates counterclockwise about the same axis with angular speed ω 2 = 8.0 rad / s ω2=8.0rad/s . If the cylinders couple so they have the same rotational axis what is the angular speed of the combination? What percentage of the original kinetic energy is lost to friction?

A diver off the high board imparts an initial rotation with his body fully extended before going into a tuck and executing three back somersaults before hitting the water. If his moment of inertia before the tuck is 16.9 kg · m 2 16.9kg·m2 and after the tuck during the somersaults is 4.2 kg · m 2 4.2kg·m2 , what rotation rate must he impart to his body directly off the board and before the tuck if he takes 1.4 s to execute the somersaults before hitting the water?

An Earth satellite has its apogee at 2500 km above the surface of Earth and perigee at 500 km above the surface of Earth. At apogee its speed is 730 m/s. What is its speed at perigee? Earth’s radius is 6370 km (see below).

A Molniya orbit is a highly eccentric orbit of a communication satellite so as to provide continuous communications coverage for Scandinavian countries and adjacent Russia. The orbit is positioned so that these countries have the satellite in view for extended periods in time (see below). If a satellite in such an orbit has an apogee at 40,000.0 km as measured from the center of Earth and a velocity of 3.0 km/s, what would be its velocity at perigee measured at 200.0 km altitude?

Shown below is a small particle of mass 20 g that is moving at a speed of 10.0 m/s when it collides and sticks to the edge of a uniform solid cylinder. The cylinder is free to rotate about its axis through its center and is perpendicular to the page. The cylinder has a mass of 0.5 kg and a radius of 10 cm, and is initially at rest. (a) What is the angular velocity of the system after the collision? (b) How much kinetic energy is lost in the collision?

A bug of mass 0.020 kg is at rest on the edge of a solid cylindrical disk ( M = 0.10 kg, R = 0.10 m ) (M=0.10kg,R=0.10m) rotating in a horizontal plane around the vertical axis through its center. The disk is rotating at 10.0 rad/s. The bug crawls to the center of the disk. (a) What is the new angular velocity of the disk? (b) What is the change in the kinetic energy of the system? (c) If the bug crawls back to the outer edge of the disk, what is the angular velocity of the disk then? (d) What is the new kinetic energy of the system? (e) What is the cause of the increase and decrease of kinetic energy?

A uniform rod of mass 200 g and length 100 cm is free to rotate in a horizontal plane around a fixed vertical axis through its center, perpendicular to its length. Two small beads, each of mass 20 g, are mounted in grooves along the rod. Initially, the two beads are held by catches on opposite sides of the rod’s center, 10 cm from the axis of rotation. With the beads in this position, the rod is rotating with an angular velocity of 10.0 rad/s. When the catches are released, the beads slide outward along the rod. (a) What is the rod’s angular velocity when the beads reach the ends of the rod? (b) What is the rod’s angular velocity if the beads fly off the rod?

A merry-go-round has a radius of 2.0 m and a moment of inertia 300 kg · m 2 . 300kg·m2. A boy of mass 50 kg runs tangent to the rim at a speed of 4.0 m/s and jumps on. If the merry-go-round is initially at rest, what is the angular velocity after the boy jumps on?

A playground merry-go-round has a mass of 120 kg and a radius of 1.80 m and it is rotating with an angular velocity of 0.500 rev/s. What is its angular velocity after a 22.0-kg child gets onto it by grabbing its outer edge? The child is initially at rest.

Three children are riding on the edge of a merry-go-round that is 100 kg, has a 1.60-m radius, and is spinning at 20.0 rpm. The children have masses of 22.0, 28.0, and 33.0 kg. If the child who has a mass of 28.0 kg moves to the center of the merry-go-round, what is the new angular velocity in rpm?

(a) Calculate the angular momentum of an ice skater spinning at 6.00 rev/s given his moment of inertia is 0.400 kg · m 2 0.400kg·m2 . (b) He reduces his rate of spin (his angular velocity) by extending his arms and increasing his moment of inertia. Find the value of his moment of inertia if his angular velocity decreases to 1.25 rev/s. (c) Suppose instead he keeps his arms in and allows friction of the ice to slow him to 3.00 rev/s. What average torque was exerted if this takes 15.0 s?

Twin skaters approach one another as shown below and lock hands. (a) Calculate their final angular velocity, given each had an initial speed of 2.50 m/s relative to the ice. Each has a mass of 70.0 kg, and each has a center of mass located 0.800 m from their locked hands. You may approximate their moments of inertia to be that of point masses at this radius. (b) Compare the initial kinetic energy and final kinetic energy.

A baseball catcher extends his arm straight up to catch a fast ball with a speed of 40 m/s. The baseball is 0.145 kg and the catcher’s arm length is 0.5 m and mass 4.0 kg. (a) What is the angular velocity of the arm immediately after catching the ball as measured from the arm socket? (b) What is the torque applied if the catcher stops the rotation of his arm 0.3 s after catching the ball?

In 2015, in Warsaw, Poland, Olivia Oliver of Nova Scotia broke the world record for being the fastest spinner on ice skates. She achieved a record 342 rev/min, beating the existing Guinness World Record by 34 rotations. If an ice skater extends her arms at that rotation rate, what would be her new rotation rate? Assume she can be approximated by a 45-kg rod that is 1.7 m tall with a radius of 15 cm in the record spin. With her arms stretched take the approximation of a rod of length 130 cm with 10 % 10% of her body mass aligned perpendicular to the spin axis. Neglect frictional forces.

A satellite in a geosynchronous circular orbit is 42,164.0 km from the center of Earth. A small asteroid collides with the satellite sending it into an elliptical orbit of apogee 45,000.0 km. What is the speed of the satellite at apogee? Assume its angular momentum is conserved.

A gymnast does cartwheels along the floor and then launches herself into the air and executes several flips in a tuck while she is airborne. If her moment of inertia when executing the cartwheels is 13.5 kg · m 2 13.5kg·m2 and her spin rate is 0.5 rev/s, how many revolutions does she do in the air if her moment of inertia in the tuck is 3.4 kg · m 2 3.4kg·m2 and she has 2.0 s to do the flips in the air?

The centrifuge at NASA Ames Research Center has a radius of 8.8 m and can produce forces on its payload of 20 *g*s or 20 times the force of gravity on Earth. (a) What is the angular momentum of a 20-kg payload that experiences 10 *g*s in the centrifuge? (b) If the driver motor was turned off in (a) and the payload lost 10 kg, what would be its new spin rate, taking into account there are no frictional forces present?

A ride at a carnival has four spokes to which pods are attached that can hold two people. The spokes are each 15 m long and are attached to a central axis. Each spoke has mass 200.0 kg, and the pods each have mass 100.0 kg. If the ride spins at 0.2 rev/s with each pod containing two 50.0-kg children, what is the new spin rate if all the children jump off the ride?

An ice skater is preparing for a jump with turns and has his arms extended. His moment of inertia is 1.8 kg · m 2 1.8kg·m2 while his arms are extended, and he is spinning at 0.5 rev/s. If he launches himself into the air at 9.0 m/s at an angle of 45 ° 45° with respect to the ice, how many revolutions can he execute while airborne if his moment of inertia in the air is 0.5 kg · m 2 0.5kg·m2 ?

A space station consists of a giant rotating hollow cylinder of mass 10 6 kg 106kg including people on the station and a radius of 100.00 m. It is rotating in space at 3.30 rev/min in order to produce artificial gravity. If 100 people of an average mass of 65.00 kg spacewalk to an awaiting spaceship, what is the new rotation rate when all the people are off the station?

Neptune has a mass of 1.0 × 10 26 kg 1.0×1026kg and is 4.5 × 10 9 km 4.5×109km from the Sun with an orbital period of 165 years. Planetesimals in the outer primordial solar system 4.5 billion years ago coalesced into Neptune over hundreds of millions of years. If the primordial disk that evolved into our present day solar system had a radius of 10 11 1011 km and if the matter that made up these planetesimals that later became Neptune was spread out evenly on the edges of it, what was the orbital period of the outer edges of the primordial disk?

#### 11.4 Precession of a Gyroscope

A gyroscope has a 0.5-kg disk that spins at 40 rev/s. The center of mass of the disk is 10 cm from a pivot which is also the radius of the disk. What is the precession angular velocity?

The precession angular velocity of a gyroscope is 1.0 rad/s. If the mass of the rotating disk is 0.4 kg and its radius is 30 cm, as well as the distance from the center of mass to the pivot, what is the rotation rate in rev/s of the disk?

The axis of Earth makes a 23.5 ° 23.5° angle with a direction perpendicular to the plane of Earth’s orbit. As shown below, this axis precesses, making one complete rotation in 25,780 y.

(a) Calculate the change in angular momentum in half this time.

(b) What is the average torque producing this change in angular momentum?

(c) If this torque were created by a pair of forces acting at the most effective point on the equator, what would the magnitude of each force be?

### Additional Problems

A marble is rolling across the floor at a speed of 7.0 m/s when it starts up a plane inclined at 30 ° 30° to the horizontal. (a) How far along the plane does the marble travel before coming to a rest? (b) How much time elapses while the marble moves up the plane?

Repeat the preceding problem replacing the marble with a hollow sphere. Explain the new results.

The mass of a hoop of radius 1.0 m is 6.0 kg. It rolls across a horizontal surface with a speed of 10.0 m/s. (a) How much work is required to stop the hoop? (b) If the hoop starts up a surface at 30 ° 30° to the horizontal with a speed of 10.0 m/s, how far along the incline will it travel before stopping and rolling back down?

Repeat the preceding problem for a hollow sphere of the same radius and mass and initial speed. Explain the differences in the results.

A particle has mass 0.5 kg and is traveling along the line x = 5.0 m x=5.0m at 2.0 m/s in the positive *y*-direction. What is the particle’s angular momentum about the origin?

A 4.0-kg particle moves in a circle of radius 2.0 m. The angular momentum of the particle varies in time according to l = 5.0 t 2 . l=5.0t2. (a) What is the torque on the particle about the center of the circle at t = 3.4 s t=3.4s ? (b) What is the angular velocity of the particle at t = 3.4 s t=3.4s ?

A proton is accelerated in a cyclotron to 5.0 × 10 6 m / s 5.0×106m/s in 0.01 s. The proton follows a circular path. If the radius of the cyclotron is 0.5 km, (a) What is the angular momentum of the proton about the center at its maximum speed? (b) What is the torque on the proton about the center as it accelerates to maximum speed?

(a) What is the angular momentum of the Moon in its orbit around Earth? (b) How does this angular momentum compare with the angular momentum of the Moon on its axis? Remember that the Moon keeps one side toward Earth at all times.

A DVD is rotating at 500 rpm. What is the angular momentum of the DVD if has a radius of 6.0 cm and mass 20.0 g?

A potter’s disk spins from rest up to 10 rev/s in 15 s. The disk has a mass 3.0 kg and radius 30.0 cm. What is the angular momentum of the disk at t = 5 s, t = 1 0 s t=5s,t=10 s ?

Suppose you start an antique car by exerting a force of 300 N on its crank for 0.250 s. What is the angular momentum given to the engine if the handle of the crank is 0.300 m from the pivot and the force is exerted to create maximum torque the entire time?

A solid cylinder of mass 2.0 kg and radius 20 cm is rotating counterclockwise around a vertical axis through its center at 600 rev/min. A second solid cylinder of the same mass is rotating clockwise around the same vertical axis at 900 rev/min. If the cylinders couple so that they rotate about the same vertical axis, what is the angular velocity of the combination?

A boy stands at the center of a platform that is rotating without friction at 1.0 rev/s. The boy holds weights as far from his body as possible. At this position the total moment of inertia of the boy, platform, and weights is 5.0 kg · m 2 . 5.0kg·m2. The boy draws the weights in close to his body, thereby decreasing the total moment of inertia to 1.5 kg · m 2 . 1.5kg·m2. (a) What is the final angular velocity of the platform? (b) By how much does the rotational kinetic energy increase?

Eight children, each of mass 40 kg, climb on a small merry-go-round. They position themselves evenly on the outer edge and join hands. The merry-go-round has a radius of 4.0 m and a moment of inertia 1000.0 kg · m 2 1000.0kg·m2 . After the merry-go-round is given an angular velocity of 6.0 rev/min, the children walk inward and stop when they are 0.75 m from the axis of rotation. What is the new angular velocity of the merry-go-round? Assume there is negligible frictional torque on the structure.

A thin meter stick of mass 150 g rotates around an axis perpendicular to the stick’s long axis at an angular velocity of 240 rev/min. What is the angular momentum of the stick if the rotation axis (a) passes through the center of the stick? (b) Passes through one end of the stick?

A satellite in the shape of a sphere of mass 20,000 kg and radius 5.0 m is spinning about an axis through its center of mass. It has a rotation rate of 8.0 rev/s. Two antennas deploy in the plane of rotation extending from the center of mass of the satellite. Each antenna can be approximated as a rod has mass 200.0 kg and length 7.0 m. What is the new rotation rate of the satellite?

A top has moment of inertia 3.2 × 10 −4 kg · m 2 3.2×10−4kg·m2 and radius 4.0 cm from the center of mass to the pivot point. If it spins at 20.0 rev/s and is precessing, how many revolutions does it precess in 10.0 s?

### Challenge Problems

The truck shown below is initially at rest with solid cylindrical roll of paper sitting on its bed. If the truck moves forward with a uniform acceleration *a*, what distance *s* does it move before the paper rolls off its back end? (*Hint*: If the roll accelerates forward with a ′ a′ , then is accelerates backward relative to the truck with an acceleration a − a ′ a−a′ . Also, R α = a − a ′ Rα=a−a′ .)

A bowling ball of radius 8.5 cm is tossed onto a bowling lane with speed 9.0 m/s. The direction of the toss is to the left, as viewed by the observer, so the bowling ball starts to rotate counterclockwise when in contact with the floor. The coefficient of kinetic friction on the lane is 0.3. (a) What is the time required for the ball to come to the point where it is not slipping? What is the distance d to the point where the ball is rolling without slipping?

A small ball of mass 0.50 kg is attached by a massless string to a vertical rod that is spinning as shown below. When the rod has an angular velocity of 6.0 rad/s, the string makes an angle of 30 ° 30° with respect to the vertical. (a) If the angular velocity is increased to 10.0 rad/s, what is the new angle of the string? (b) Calculate the initial and final angular momenta of the ball. (c) Can the rod spin fast enough so that the ball is horizontal?

A bug flying horizontally at 1.0 m/s collides and sticks to the end of a uniform stick hanging vertically. After the impact, the stick swings out to a maximum angle of 5.0 ° 5.0° from the vertical before rotating back. If the mass of the stick is 10 times that of the bug, calculate the length of the stick.

## Contents

### Orbital angular momentum in two dimensions Edit

Angular momentum is a vector quantity (more precisely, a pseudovector) that represents the product of a body's rotational inertia and rotational velocity (in radians/sec) about a particular axis. However, if the particle's trajectory lies in a single plane, it is sufficient to discard the vector nature of angular momentum, and treat it as a scalar (more precisely, a pseudoscalar). [2] Angular momentum can be considered a rotational analog of linear momentum. Thus, where linear momentum p is proportional to mass m and linear speed v ,

angular momentum L is proportional to moment of inertia I and angular speed ω measured in radians per second. [3]

Unlike mass, which depends only on amount of matter, moment of inertia is also dependent on the position of the axis of rotation and the shape of the matter. Unlike linear velocity, which does not depend upon the choice of origin, orbital angular velocity is always measured with respect to a fixed origin. Therefore, strictly speaking, L should be referred to as the angular momentum *relative to that center*. [4]

This simple analysis can also apply to non-circular motion if only the component of the motion which is perpendicular to the radius vector is considered. In that case,

where r ⊥ = r sin ( θ ) *moment arm*, a line dropped perpendicularly from the origin onto the path of the particle. It is this definition, (length of moment arm)×(linear momentum) to which the term *moment of momentum* refers. [5]

### Scalar—angular momentum from Lagrangian mechanics Edit

Another approach is to define angular momentum as the conjugate momentum (also called **canonical momentum**) of the angular coordinate ϕ

And the potential energy is

The *generalized momentum* "canonically conjugate to" the coordinate ϕ

### Orbital angular momentum in three dimensions Edit

To completely define orbital angular momentum in three dimensions, it is required to know the rate at which the position vector sweeps out angle, the direction perpendicular to the instantaneous plane of angular displacement, and the mass involved, as well as how this mass is distributed in space. [6] By retaining this vector nature of angular momentum, the general nature of the equations is also retained, and can describe any sort of three-dimensional motion about the center of rotation – circular, linear, or otherwise. In vector notation, the orbital angular momentum of a point particle in motion about the origin can be expressed as:

This can be expanded, reduced, and by the rules of vector algebra, rearranged:

The two-dimensional scalar equations of the previous section can thus be given direction:

In the spherical coordinate system the angular momentum vector expresses as

### Orbital angular momentum in four or more dimensions Edit

Angular momentum in higher dimensions can be defined by application of Noether's theorem to rotation groups of higher order. [* citation needed *] Generalization beyond three dimensions is best treated using differential forms. [* citation needed *]

Angular momentum can be described as the rotational analog of linear momentum. Like linear momentum it involves elements of mass and displacement. Unlike linear momentum it also involves elements of position and shape.

Many problems in physics involve matter in motion about some certain point in space, be it in actual rotation about it, or simply moving past it, where it is desired to know what effect the moving matter has on the point—can it exert energy upon it or perform work about it? Energy, the ability to do work, can be stored in matter by setting it in motion—a combination of its inertia and its displacement. Inertia is measured by its mass, and displacement by its velocity. Their product,

is the matter's momentum. [7] Referring this momentum to a central point introduces a complication: the momentum is not applied to the point directly. For instance, a particle of matter at the outer edge of a wheel is, in effect, at the end of a lever of the same length as the wheel's radius, its momentum turning the lever about the center point. This imaginary lever is known as the *moment arm*. It has the effect of multiplying the momentum's effort in proportion to its length, an effect known as a *moment*. Hence, the particle's momentum referred to a particular point,

Because moment of inertia is a crucial part of the spin angular momentum, the latter necessarily includes all of the complications of the former, which is calculated by multiplying elementary bits of the mass by the squares of their distances from the center of rotation. [9] Therefore, the total moment of inertia, and the angular momentum, is a complex function of the configuration of the matter about the center of rotation and the orientation of the rotation for the various bits.

For a rigid body, for instance a wheel or an asteroid, the orientation of rotation is simply the position of the rotation axis versus the matter of the body. It may or may not pass through the center of mass, or it may lie completely outside of the body. For the same body, angular momentum may take a different value for every possible axis about which rotation may take place. [10] It reaches a minimum when the axis passes through the center of mass. [11]

For a collection of objects revolving about a center, for instance all of the bodies of the Solar System, the orientations may be somewhat organized, as is the Solar System, with most of the bodies' axes lying close to the system's axis. Their orientations may also be completely random.

In brief, the more mass and the farther it is from the center of rotation (the longer the moment arm), the greater the moment of inertia, and therefore the greater the angular momentum for a given angular velocity. In many cases the moment of inertia, and hence the angular momentum, can be simplified by, [12]

I = r 2 m

and for any collection of particles m i

Angular momentum's dependence on position and shape is reflected in its units versus linear momentum: kg⋅m 2 /s, N⋅m⋅s, or J⋅s for angular momentum versus kg⋅m/s or N⋅s for linear momentum. When calculating angular momentum as the product of the moment of inertia times the angular velocity, the angular velocity must be expressed in radians per second, where the radian assumes the dimensionless value of unity. (When performing dimensional analysis, it may be productive to use orientational analysis which treats radians as a base unit, but this is outside the scope of the International system of units). Angular momentum's units can be interpreted as torque⋅time or as energy⋅time per angle. An object with angular momentum of *L* N⋅m⋅s can be reduced to zero rotation (all of the rotational energy can be transferred out of it) by an angular impulse of *L* N⋅m⋅s [13] or equivalently, by torque or work of *L* N⋅m for one second, or energy of *L* J for one second. [14]

The plane perpendicular to the axis of angular momentum and passing through the center of mass [15] is sometimes called the *invariable plane*, because the direction of the axis remains fixed if only the interactions of the bodies within the system, free from outside influences, are considered. [16] One such plane is the invariable plane of the Solar System.

### Angular momentum and torque Edit

Newton's second law of motion can be expressed mathematically,

or force = mass × acceleration. The rotational equivalent for point particles may be derived as follows:

which means that the torque (i.e. the time derivative of the angular momentum) is

This is the rotational analog of Newton's Second Law. Note that the torque is not necessarily proportional or parallel to the angular acceleration (as one might expect). The reason for this is that the moment of inertia of a particle can change with time, something that cannot occur for ordinary mass.

### General considerations Edit

A rotational analog of Newton's third law of motion might be written, "In a closed system, no torque can be exerted on any matter without the exertion on some other matter of an equal and opposite torque." [17] Hence, *angular momentum can be exchanged between objects in a closed system, but total angular momentum before and after an exchange remains constant (is conserved).* [18]

Seen another way, a rotational analogue of Newton's first law of motion might be written, "A rigid body continues in a state of uniform rotation unless acted by an external influence." [17] Thus *with no external influence to act upon it, the original angular momentum of the system remains constant*. [19]

The conservation of angular momentum is used in analyzing *central force motion*. If the net force on some body is directed always toward some point, the *center*, then there is no torque on the body with respect to the center, as all of the force is directed along the radius vector, and none is perpendicular to the radius. Mathematically, torque τ = r × F = 0 ,

For a planet, angular momentum is distributed between the spin of the planet and its revolution in its orbit, and these are often exchanged by various mechanisms. The conservation of angular momentum in the Earth–Moon system results in the transfer of angular momentum from Earth to Moon, due to tidal torque the Moon exerts on the Earth. This in turn results in the slowing down of the rotation rate of Earth, at about 65.7 nanoseconds per day, [20] and in gradual increase of the radius of Moon's orbit, at about 3.82 centimeters per year. [21]

The conservation of angular momentum explains the angular acceleration of an ice skater as she brings her arms and legs close to the vertical axis of rotation. By bringing part of the mass of her body closer to the axis, she decreases her body's moment of inertia. Because angular momentum is the product of moment of inertia and angular velocity, if the angular momentum remains constant (is conserved), then the angular velocity (rotational speed) of the skater must increase.

The same phenomenon results in extremely fast spin of compact stars (like white dwarfs, neutron stars and black holes) when they are formed out of much larger and slower rotating stars. Decrease in the size of an object *n* times results in increase of its angular velocity by the factor of *n* 2 .

Conservation is not always a full explanation for the dynamics of a system but is a key constraint. For example, a spinning top is subject to gravitational torque making it lean over and change the angular momentum about the nutation axis, but neglecting friction at the point of spinning contact, it has a conserved angular momentum about its spinning axis, and another about its precession axis. Also, in any planetary system, the planets, star(s), comets, and asteroids can all move in numerous complicated ways, but only so that the angular momentum of the system is conserved.

Noether's theorem states that every conservation law is associated with a symmetry (invariant) of the underlying physics. The symmetry associated with conservation of angular momentum is rotational invariance. The fact that the physics of a system is unchanged if it is rotated by any angle about an axis implies that angular momentum is conserved. [22]

### Relation to Newton's second law of motion Edit

While angular momentum total conservation can be understood separately from Newton's laws of motion as stemming from Noether's theorem in systems symmetric under rotations, it can also be understood simply as an efficient method of calculation of results that can also be otherwise arrived at directly from Newton's second law, together with laws governing the forces of nature (such as Newton's third law, Maxwell's equations and Lorentz force). Indeed, given initial conditions of position and velocity for every point, and the forces at such a condition, one may use Newton's second law to calculate the second derivative of position, and solving for this gives full information on the development of the physical system with time. [23] Note, however, that this is no longer true in quantum mechanics, due to the existence of particle spin, which is angular momentum that cannot be described by the cumulative effect of point-like motions in space.

As an example, consider decreasing of the moment of inertia, e.g. when a figure skater is pulling in her/his hands, speeding up the circular motion. In terms of angular momentum conservation, we have, for angular momentum *L*, moment of inertia *I* and angular velocity *ω*:

0 = d L = d ( I ⋅ ω ) = d I ⋅ ω + I ⋅ d ω

Using this, we see that the change requires an energy of:

so that a decrease in the moment of inertia requires investing energy.

This can be compared to the work done as calculated using Newton's laws. Each point in the rotating body is accelerating, at each point of time, with radial acceleration of:

Let us observe a point of mass *m*, whose position vector relative to the center of motion is parallel to the z-axis at a given point of time, and is at a distance *z*. The centripetal force on this point, keeping the circular motion, is:

Thus the work required for moving this point to a distance *dz* farther from the center of motion is:

For a non-pointlike body one must integrate over this, with *m* replaced by the mass density per unit *z*. This gives:

which is exactly the energy required for keeping the angular momentum conserved.

Note, that the above calculation can also be performed per mass, using kinematics only. Thus the phenomena of figure skater accelerating tangential velocity while pulling her/his hands in, can be understood as follows in layman's language: The skater's palms are not moving in a straight line, so they are constantly accelerating inwards, but do not gain additional speed because the accelerating is always done when their motion inwards is zero. However, this is different when pulling the palms closer to the body: The acceleration due to rotation now increases the speed but because of the rotation, the increase in speed does not translate to a significant speed inwards, but to an increase of the rotation speed.

### In Lagrangian formalism Edit

In Lagrangian mechanics, angular momentum for rotation around a given axis, is the conjugate momentum of the generalized coordinate of the angle around the same axis. For example, L z

where the subscript i stands for the i-th body, and *m*, *v*_{T} and *ω*_{z} stand for mass, tangential velocity around the z-axis and angular velocity around that axis, respectively.

For a body that is not point-like, with density *ρ*, we have instead:

where *I*_{z} is the moment of inertia around the z-axis.

Thus, assuming the potential energy does not depend on *ω*_{z} (this assumption may fail for electromagnetic systems), we have the angular momentum of the i-th object:

We have thus far rotated each object by a separate angle we may also define an overall angle *θ*_{z} by which we rotate the whole system, thus rotating also each object around the z-axis, and have the overall angular momentum:

From Euler-Lagrange equations it then follows that:

Since the lagrangian is dependent upon the angles of the object only through the potential, we have:

which is the torque on the i-th object.

Suppose the system is invariant to rotations, so that the potential is independent of an overall rotation by the angle *θ*_{z} (thus it may depend on the angles of objects only through their differences, in the form V ( θ z i , θ z j ) = V ( θ z i − θ z j ) *,< heta _ >_*

*-< heta _>_*)> ). We therefore get for the total angular momentum:

* *

*And thus the angular momentum around the z-axis is conserved.*

This analysis can be repeated separately for each axis, giving conversation of the angular momentum vector. However, the angles around the three axes cannot be treated simultaneously as generalized coordinates, since they are not independent in particular, two angles per point suffice to determine its position. While it is true that in the case of a rigid body, fully describing it requires, in addition to three translational degrees of freedom, also specification of three rotational degrees of freedom however these cannot be defined as rotations around the Cartesian axes (see Euler angles). This caveat is reflected in quantum mechanics in the non-trivial commutation relations of the different components of the angular momentum operator.

### In Hamiltonian formalism Edit

Equivalently, in Hamiltonian mechanics the Hamiltonian can be described as a function of the angular momentum. As before, the part of the kinetic energy related to rotation around the z-axis for the i-th object is:

which is analogous to the energy dependence upon momentum along the z-axis, p z i 2 2 m i *>^<2>><<2m>_ >>> .*

* *

*Hamilton's equations relate the angle around the z-axis to its conjugate momentum, the angular momentum around the same axis:*

And so we get the same results as in the Lagrangian formalism.

Note, that for combining all axes together, we write the kinetic energy as:

where *p*_{r} is the momentum in the radial direction, and the moment of inertia is a 3-dimensional matrix bold letters stand for 3-dimensional vectors.

For point-like bodies we have:

This form of the kinetic energy part of the Hamiltonian is useful in analyzing central potential problems, and is easily transformed to a quantum mechanical work frame (e.g. in the hydrogen atom problem).

While in classical mechanics the language of angular momentum can be replaced by Newton's laws of motion, it is particularly useful for motion in central potential such as planetary motion in the solar system. Thus, the orbit of a planet in the solar system is defined by its energy, angular momentum and angles of the orbit major axis relative to a coordinate frame.

In astrodynamics and celestial mechanics, a *massless* (or *per unit mass*) angular momentum is defined [24]

called *specific angular momentum*. Note that L = m h .

Angular momentum is also an extremely useful concept for describing rotating rigid bodies such as a gyroscope or a rocky planet. For a continuous mass distribution with density function *ρ*(**r**), a differential volume element *dV* with position vector **r** within the mass has a mass element *dm* = *ρ*(**r**)*dV*. Therefore, the infinitesimal angular momentum of this element is:

and integrating this differential over the volume of the entire mass gives its total angular momentum:

In the derivation which follows, integrals similar to this can replace the sums for the case of continuous mass.

### Collection of particles Edit

For a collection of particles in motion about an arbitrary origin, it is informative to develop the equation of angular momentum by resolving their motion into components about their own center of mass and about the origin. Given,

The total mass of the particles is simply their sum,

The position vector of the center of mass is defined by, [25]

The total angular momentum of the collection of particles is the sum of the angular momentum of each particle,

It can be shown that (see sidebar),

therefore the second and third terms vanish,

The first term can be rearranged,

and total angular momentum for the collection of particles is finally, [26]

The first term is the angular momentum of the center of mass relative to the origin. Similar to **Single particle**, below, it is the angular momentum of one particle of mass *M* at the center of mass moving with velocity **V**. The second term is the angular momentum of the particles moving relative to the center of mass, similar to **Fixed center of mass**, below. The result is general—the motion of the particles is not restricted to rotation or revolution about the origin or center of mass. The particles need not be individual masses, but can be elements of a continuous distribution, such as a solid body.

Rearranging equation (**2**) by vector identities, multiplying both terms by "one", and grouping appropriately,

gives the total angular momentum of the system of particles in terms of moment of inertia I

#### Single particle case Edit

In the case of a single particle moving about the arbitrary origin,

#### Case of a fixed center of mass Edit

For the case of the center of mass fixed in space with respect to the origin,

In modern (20th century) theoretical physics, angular momentum (not including any intrinsic angular momentum – see below) is described using a different formalism, instead of a classical pseudovector. In this formalism, angular momentum is the 2-form Noether charge associated with rotational invariance. As a result, angular momentum is not conserved for general curved spacetimes, unless it happens to be asymptotically rotationally invariant. [* citation needed *]

In classical mechanics, the angular momentum of a particle can be reinterpreted as a plane element:

in which the exterior product ∧ replaces the cross product × (these products have similar characteristics but are nonequivalent). This has the advantage of a clearer geometric interpretation as a plane element, defined from the **x** and **p** vectors, and the expression is true in any number of dimensions (two or higher). In Cartesian coordinates:

or more compactly in index notation:

The angular velocity can also be defined as an antisymmetric second order tensor, with components *ω _{ij}*. The relation between the two antisymmetric tensors is given by the moment of inertia which must now be a fourth order tensor: [27]

Again, this equation in **L** and **ω** as tensors is true in any number of dimensions. This equation also appears in the geometric algebra formalism, in which **L** and **ω** are bivectors, and the moment of inertia is a mapping between them.

in the language of four-vectors, namely the four position *X* and the four momentum *P*, and absorbs the above **L** together with the motion of the centre of mass of the particle.

In each of the above cases, for a system of particles, the total angular momentum is just the sum of the individual particle angular momenta, and the centre of mass is for the system.

Angular momentum in quantum mechanics differs in many profound respects from angular momentum in classical mechanics. In relativistic quantum mechanics, it differs even more, in which the above relativistic definition becomes a tensorial operator.

### Spin, orbital, and total angular momentum Edit

**Left:**"spin" angular momentum**S**is really orbital angular momentum of the object at every point.**Right:**extrinsic orbital angular momentum**L**about an axis.**Top:**the moment of inertia tensor**I**and angular velocity**ω**(**L**is not always parallel to**ω**). [28]**Bottom:**momentum**p**and its radial position**r**from the axis. The total angular momentum (spin plus orbital) is**J**. For a*quantum*particle the interpretations are different particle spin does*not*have the above interpretation.

The classical definition of angular momentum as L = r × p

> can be carried over to quantum mechanics, by reinterpreting **r** as the quantum position operator and **p** as the quantum momentum operator. **L** is then an operator, specifically called the *orbital angular momentum operator*. The components of the angular momentum operator satisfy the commutation relations of the Lie algebra so(3). Indeed, these operators are precisely the infinitesimal action of the rotation group on the quantum Hilbert space. [29] (See also the discussion below of the angular momentum operators as the generators of rotations.)

However, in quantum physics, there is another type of angular momentum, called *spin angular momentum*, represented by the spin operator **S**. Almost all elementary particles have nonzero spin. [30] Spin is often depicted as a particle literally spinning around an axis, but this is a misleading and inaccurate picture: spin is an intrinsic property of a particle, unrelated to any sort of motion in space and fundamentally different from orbital angular momentum. All elementary particles have a characteristic spin (possibly zero), [31] for example electrons have "spin 1/2" (this actually means "spin ħ/2"), photons have "spin 1" (this actually means "spin ħ"), and pi-mesons have spin 0. [32]

Finally, there is total angular momentum **J**, which combines both the spin and orbital angular momentum of all particles and fields. (For one particle, **J** = **L** + **S** .) Conservation of angular momentum applies to **J**, but not to **L** or **S** for example, the spin–orbit interaction allows angular momentum to transfer back and forth between **L** and **S**, with the total remaining constant. Electrons and photons need not have integer-based values for total angular momentum, but can also have fractional values. [33]

In molecules the total angular momentum **F** is the sum of the rovibronic (orbital) angular momentum **N**, the electron spin angular momentum **S**, and the nuclear spin angular momentum **I**. For electronic singlet states the rovibronic angular momentum is denoted **J** rather than **N**. As explained by Van Vleck, [34] the components of the molecular rovibronic angular momentum referred to molecule-fixed axes have different commutation relations from those for the components about space-fixed axes.

### Quantization Edit

In quantum mechanics, angular momentum is quantized – that is, it cannot vary continuously, but only in "quantum leaps" between certain allowed values. For any system, the following restrictions on measurement results apply, where ℏ

If you measure. | The result can be. |

L n ^ | … , − 2 ℏ , − ℏ , 0 , ℏ , 2 ℏ , … |

S n ^ | … , − 3 2 ℏ , − ℏ , − 1 2 ℏ , 0 , 1 2 ℏ , ℏ , 3 2 ℏ , … |

L 2 = L x 2 + L y 2 + L z 2 | [ ℏ 2 n ( n + 1 ) ] |

S 2 | [ ℏ 2 n ( n + 1 ) ] |

(There are additional restrictions as well, see angular momentum operator for details.)

The reduced Planck constant ℏ

Quantization of angular momentum was first postulated by Niels Bohr in his Bohr model of the atom and was later predicted by Erwin Schrödinger in his Schrödinger equation.

### Uncertainty Edit

In the definition L = r × p

> , six operators are involved: The position operators r x

The uncertainty is closely related to the fact that different components of an angular momentum operator do not commute, for example L x L y ≠ L y L x

### Total angular momentum as generator of rotations Edit

is the rotation operator that takes any system and rotates it by angle ϕ

The relationship between the angular momentum operator and the rotation operators is the same as the relationship between Lie algebras and Lie groups in mathematics. The close relationship between angular momentum and rotations is reflected in Noether's theorem that proves that angular momentum is conserved whenever the laws of physics are rotationally invariant.

When describing the motion of a charged particle in an electromagnetic field, the canonical momentum **P** (derived from the Lagrangian for this system) is not gauge invariant. As a consequence, the canonical angular momentum **L** = **r** × **P** is not gauge invariant either. Instead, the momentum that is physical, the so-called *kinetic momentum* (used throughout this article), is (in SI units)

where *e* is the electric charge of the particle and **A** the magnetic vector potential of the electromagnetic field. The gauge-invariant angular momentum, that is *kinetic angular momentum*, is given by

The interplay with quantum mechanics is discussed further in the article on canonical commutation relations.

In *classical Maxwell electrodynamics* the Poynting vector is a linear momentum density of electromagnetic field. [36]

The above identities are valid *locally*, i.e. in each space point r

Newton, in the *Principia*, hinted at angular momentum in his examples of the First Law of Motion,

*A top, whose parts by their cohesion are perpetually drawn aside from rectilinear motions, does not cease its rotation, otherwise than as it is retarded by the air. The greater bodies of the planets and comets, meeting with less resistance in more free spaces, preserve their motions both progressive and circular for a much longer time.* [38]

He did not further investigate angular momentum directly in the *Principia*,

*From such kind of reflexions also sometimes arise the circular motions of bodies about their own centres. But these are cases which I do not consider in what follows and it would be too tedious to demonstrate every particular that relates to this subject.* [39]

However, his geometric proof of the law of areas is an outstanding example of Newton's genius, and indirectly proves angular momentum conservation in the case of a central force.

### The Law of Areas Edit

#### Newton's derivation Edit

As a planet orbits the Sun, the line between the Sun and the planet sweeps out equal areas in equal intervals of time. This had been known since Kepler expounded his second law of planetary motion. Newton derived a unique geometric proof, and went on to show that the attractive force of the Sun's gravity was the cause of all of Kepler's laws.

During the first interval of time, an object is in motion from point **A** to point **B**. Undisturbed, it would continue to point **c** during the second interval. When the object arrives at **B**, it receives an impulse directed toward point **S**. The impulse gives it a small added velocity toward **S**, such that if this were its only velocity, it would move from **B** to **V** during the second interval. By the rules of velocity composition, these two velocities add, and point **C** is found by construction of parallelogram **BcCV**. Thus the object's path is deflected by the impulse so that it arrives at point **C** at the end of the second interval. Because the triangles **SBc** and **SBC** have the same base **SB** and the same height **Bc** or **VC**, they have the same area. By symmetry, triangle **SBc** also has the same area as triangle **SAB**, therefore the object has swept out equal areas **SAB** and **SBC** in equal times.

At point **C**, the object receives another impulse toward **S**, again deflecting its path during the third interval from **d** to **D**. Thus it continues to **E** and beyond, the triangles **SAB**, **SBc**, **SBC**, **SCd**, **SCD**, **SDe**, **SDE** all having the same area. Allowing the time intervals to become ever smaller, the path **ABCDE** approaches indefinitely close to a continuous curve.

Note that because this derivation is geometric, and no specific force is applied, it proves a more general law than Kepler's second law of planetary motion. It shows that the Law of Areas applies to any central force, attractive or repulsive, continuous or non-continuous, or zero.

#### Conservation of angular momentum in the Law of Areas Edit

The proportionality of angular momentum to the area swept out by a moving object can be understood by realizing that the bases of the triangles, that is, the lines from **S** to the object, are equivalent to the radius r , and that the heights of the triangles are proportional to the perpendicular component of velocity v _{⊥} . Hence, if the area swept per unit time is constant, then by the triangular area formula 1 / 2 (base)(height) , the product (base)(height) and therefore the product rv _{⊥} are constant: if r and the base length are decreased, v _{⊥} and height must increase proportionally. Mass is constant, therefore angular momentum rmv _{⊥} is conserved by this exchange of distance and velocity.

### After Newton Edit

Leonhard Euler, Daniel Bernoulli, and Patrick d'Arcy all understood angular momentum in terms of conservation of areal velocity, a result of their analysis of Kepler's second law of planetary motion. It is unlikely that they realized the implications for ordinary rotating matter. [40]

In 1736 Euler, like Newton, touched on some of the equations of angular momentum in his *Mechanica* without further developing them. [41]

Bernoulli wrote in a 1744 letter of a "moment of rotational motion", possibly the first conception of angular momentum as we now understand it. [42]

In 1799, Pierre-Simon Laplace first realized that a fixed plane was associated with rotation—his *invariable plane*.

Louis Poinsot in 1803 began representing rotations as a line segment perpendicular to the rotation, and elaborated on the "conservation of moments".

In 1852 Léon Foucault used a gyroscope in an experiment to display the Earth's rotation.

William J. M. Rankine's 1858 *Manual of Applied Mechanics* defined angular momentum in the modern sense for the first time:

*. a line whose length is proportional to the magnitude of the angular momentum, and whose direction is perpendicular to the plane of motion of the body and of the fixed point, and such, that when the motion of the body is viewed from the extremity of the line, the radius-vector of the body seems to have right-handed rotation.*

## MATERIALS AND METHODS

### Experimental procedures

Kinetic and kinematic walking data were collected at the Gait Laboratory of Spaulding Rehabilitation Hospital, Harvard Medical School, in a study approved by the Spaulding committee on the Use of Humans as Experimental Subjects. Ten healthy adult participants, five male and five female, with an age range from 20 to 38 years, volunteered for the study. The participants walked at a self-selected speed across a 10 m walkway in the Motion Analysis Laboratory. Participants were timed between two fixed points to ensure that the same walking speed was used between experimental trials. Walking speeds within a±5% interval from the self-selected speed were accepted. For each study participant, a total of seven walking trials were collected.

The data collection procedures were based on standard techniques(Kadaba et al., 1989 Winter, 1990 Kadaba et al., 1990 Kerrigan et al., 2000 Kerrigan et al., 2001). An infrared camera system (eight cameras, VICON 512 motion analysis system,Oxford Metrics, Oxford, UK) was used to measure the three-dimensional locations of reflective markers at 120 frames s –1 . A total of 33 markers were placed on various parts of a participant's body: 16 lower-body markers, five trunk markers, eight upper-limb markers and four head markers. The markers were attached to the following bony landmarks: bilateral anterior superior iliac spines, posterior superior iliac spines, lateral femoral condyles, lateral malleoli, forefeet and heels. Additional markers were rigidly attached to wands over the mid-femur and mid-shaft of the tibia. The kinematics of the upper body were also collected with markers placed on the following locations: sternum, clavicle, C7 vertebra, T10 vertebra, head, and bilaterally on the shoulder, elbow and wrist. The VICON 512 system was able to detect marker position with a precision of ∼1 mm.

During the walking trials, ground reaction forces were measured synchronously with the kinematic data at a sampling rate of 1080 Hz using two staggered force platforms (model no. 2222 or OR6-5-1, Advanced Mechanical Technology Inc., Watertown, MA, USA) embedded in the walkway. The platforms measured ground reaction force and CP location at a precision of ∼0.1 N and ∼2 mm, respectively.

### Human model

A human model was constructed in order to calculate physical quantities such as CM position and angular momentum. The model and coordinate system used in the study are shown in Fig. 1. The model comprises 16 rigid body segments: feet, tibias,femurs, hands, forearms, arms, pelvis-abdomen, chest, neck and head. The feet and hands were modeled as rectangular boxes. The tibia segments, femur segments, forearm segments and arm segments were modeled as truncated cones. The pelvis-abdomen and chest segments were modeled as elliptical slabs[ellipses in the horizontal (*x–y*) plane and extruded in the vertical (*z*) direction]. The neck was modeled as a cylinder, and the head was modeled as a sphere. The following 28 anthropometric measurements were taken for each study participant to accurately construct a representative model: (1) body weight, height, and total leg length measured from the medial malleolus to the anterior superior iliac spine (2) lengths, widths and thicknesses of foot and hand segments (3) segment lengths and proximal/distal base radii of tibia, femur, forearm and arm (4) heights, widths and thicknesses of chest and pelvis-abdomen segments and (5) radius of the head. The neck radius was set equal to half the head radius. The human model had a total of 38 degrees of freedom, or 32 internal degrees of freedom (12 for the legs, 14 for the arms, and six for the head, neck and trunk) and six external degrees of freedom.

For acceptance of the human model, we required that each segment's relative mass and density were in reasonable agreement with human morphological data from the literature (Winter,1990). Relative mass was defined as segment mass divided by total body mass, and density as segment mass divided by segment volume. We accepted a segment design if both its relative mass and density fell within one standard deviation of the segment's mean experimental values from the literature. When the relative mass of each model segment was set equal to each segment's mean experimental value from the literature, model segment density often became abnormal, falling beyond two standard deviations from the experimental mean. In distinction, when the density of each model segment was set equal to each segment's mean experimental value from the literature, model relative mass then became abnormal. As a resolution to this difficulty, we performed an optimization where model relative mass was varied until the error between model and experimental density values were minimized. We then confirmed that each segment's relative mass and density fell within one standard deviation of their experimental means reported in Winter(Winter, 1990).

## 1.3 Ready to study?

Study comment In order to study this module you will need to be familiar with the following terms: *angle angular measure* (*degree*, *radian*, the relationship *s* = *r&theta* between arc length *s*, radius *r* and angle swept out *&theta*), *areas* and *volumes* of *solid_of_revolution regular solids*, *Cartesian coordinate system*, *density*, *energy*, *force*, *kinetic energy*, *mass*, *Newton&rsquos laws of motion*, *SI units* (distance, force and energy), *translational equilibrium*, *uniform acceleration equations*, *uniform circular motion* (*angular speed*, *speed* and the relationship between these), *vector vector notation* ( *magnitude_of_a_vector_or_vector_quantity magnitude*, *components_of_a_vector vector component*, *component vector*, *vector_addition addition*, *vector_difference subtraction*), *weight*, *work*, *algebraic_expression algebraic* and *trigonometric_identities trigonometrical equations* and manipulation of these and the *calculus notation of calculus*, including *differention*, and *integration* of simple *polynomial functions*. If you are uncertain about any of these terms then you can review them now by referring to the *Glossary*, which will indicate where in *FLAP* they are introduced. The following questions will allow you to establish whether you need to review some of the topics before embarking on the module.

What is the angular speed of a bicycle wheel of radius 34&thinspcm when the bicycle is travelling forwards, without skidding, at a constant speed of 30&thinspkm per hour?

The speed of a point on the rim of the wheel relative to the hub is *&upsilon* = *&omegar*, where *&omega* is the *angular speed* and *r* the radius of the wheel. Because the wheel is rolling and not skidding, this is also the speed of the bicycle. Thus

Consult *angular speed* in the *Glossary* for further information.

**Figure 1** See Question R2.

Find the components_of_a_vector components of the force * F* of magnitude_of_a_vector_or_vector_quantity magnitude 12&thinspN along both the

*x-*and

*z*–axes in Figure 1. If this force

*were applied to a mass of 3.0&thinspkg, placed at A, write down the*

**F***x-*and

*z*–components of the acceleration of the mass.

The components_of_a_vector component of a force along a given direction is given by the product of the *magnitude_of_a_vector_or_vector_quantity magnitude* *of the force* and the cosine of the angle between the force and the direction concerned. From Figure 1, the force is at an angle of 30° to the positive *x*–direction and 60° to the negative *z*–direction.

The *x*–component of * F* is

The *z*–component of * F* is

*Newton&rsquos second law of motion*, * F* =

*m*, leads to

**a**Consult the relevant terms in the *Glossary* for further information.

A mains water pipe section is of length 24&thinspm, outer diameter 0.80&thinspm, inner diameter 0.74&thinspm and density 2.4 × 10 3 &thinspkg&thinspm &minus3 . Calculate its mass and the magnitude of its weight (taking the magnitude of the acceleration due to gravity to be 10&thinspm&thinsps &minus2 ).

We solve this problem by subtracting a smaller solid cylinder with the inner radius from a larger solid cylinder with the outer radius.

Since mass = density × volume, it is necessary first to calculate the volumes of these solids. *Do not forget to divide the diameter by two in order to calculate the radius*.

The volume of a cylinder = cross–sectional area × height = *&pir* 2 *h*

The volume of the larger cylinder is, therefore, *&pi*_{&thinsp}(0.40&thinspm) 2 × 24&thinspm

The volume of the smaller cylinder is *&pi*_{&thinsp}(0.37&thinspm) 2 × 24&thinspm

The mass of the hollow pipe is, therefore,

and the magnitude of its weight is 4.2 × 10 4 &thinspN.

Consult the relevant terms in the *Glossary* for further information.

## 11.3 Conservation of Angular Momentum

So far, we have looked at the angular momentum of systems consisting of point particles and rigid bodies. We have also analyzed the torques involved, using the expression that relates the external net torque to the change in angular momentum, Equation 11.8. Examples of systems that obey this equation include a freely spinning bicycle tire that slows over time due to torque arising from friction, or the slowing of Earth’s rotation over millions of years due to frictional forces exerted on tidal deformations.

However, suppose there is no net external torque on the system, ∑ τ → = 0 . ∑ τ → = 0 . In this case, Equation 11.8 becomes the law of conservation of angular momentum .

### Law of Conservation of Angular Momentum

The angular momentum of a system of particles around a point in a fixed inertial reference frame is conserved if there is no net external torque around that point:

As an example of conservation of angular momentum, Figure 11.14 shows an ice skater executing a spin. The net torque on her is very close to zero because there is relatively little friction between her skates and the ice. Also, the friction is exerted very close to the pivot point. Both | F → | and | r → | | F → | and | r → | are small, so | τ → | | τ → | is negligible. Consequently, she can spin for quite some time. She can also increase her rate of spin by pulling her arms and legs in. Why does pulling her arms and legs in increase her rate of spin? The answer is that her angular momentum is constant, so that

where the primed quantities refer to conditions after she has pulled in her arms and reduced her moment of inertia. Because I ′ I ′ is smaller, the angular velocity ω ′ ω ′ must increase to keep the angular momentum constant.

It is interesting to see how the rotational kinetic energy of the skater changes when she pulls her arms in. Her initial rotational energy is

whereas her final rotational energy is

The solar system is another example of how conservation of angular momentum works in our universe. Our solar system was born from a huge cloud of gas and dust that initially had rotational energy. Gravitational forces caused the cloud to contract, and the rotation rate increased as a result of conservation of angular momentum (Figure 11.15).

We continue our discussion with an example that has applications to engineering.

### Example 11.7

#### Coupled Flywheels

#### Strategy

#### Solution

b. Before contact, only one flywheel is rotating. The rotational kinetic energy of this flywheel is the initial rotational kinetic energy of the system, 1 2 I 0 ω 0 2 1 2 I 0 ω 0 2 . The final kinetic energy is 1 2 ( 4 I 0 ) ω 2 = 1 2 ( 4 I 0 ) ( ω 0 4 ) 2 = 1 8 I 0 ω 0 2 . 1 2 ( 4 I 0 ) ω 2 = 1 2 ( 4 I 0 ) ( ω 0 4 ) 2 = 1 8 I 0 ω 0 2 .

Therefore, the ratio of the final kinetic energy to the initial kinetic energy is

Thus, 3/4 of the initial kinetic energy is lost to the coupling of the two flywheels.

#### Significance

A merry-go-round at a playground is rotating at 4.0 rev/min. Three children jump on and increase the moment of inertia of the merry-go-round/children rotating system by 25 % 25 % . What is the new rotation rate?

### Example 11.8

#### Dismount from a High Bar

#### Strategy

#### Solution

The moment of inertia in the tuck is I f = 1 12 m L f 2 = 1 12 80.0 kg ( 0 . 9 m ) 2 = 5.4 kg · m 2 I f = 1 12 m L f 2 = 1 12 80.0 kg ( 0 . 9 m ) 2 = 5.4 kg · m 2 .

Conservation of angular momentum: I f ω f = I 0 ω 0 ⇒ ω f = I 0 ω 0 I f = 21.6 kg · m 2 ( 1.0 rev / s ) 5.4 kg · m 2 = 4.0 rev / s I f ω f = I 0 ω 0 ⇒ ω f = I 0 ω 0 I f = 21.6 kg · m 2 ( 1.0 rev / s ) 5.4 kg · m 2 = 4.0 rev / s .

Time interval in the tuck: t = 2 h g = 2 ( 3.0 − 1.8 ) m 9.8 m / s = 0.5 s t = 2 h g = 2 ( 3.0 − 1.8 ) m 9.8 m / s = 0.5 s .

In 0.5 s, he will be able to execute two revolutions at 4.0 rev/s.

#### Significance

### Example 11.9

#### Conservation of Angular Momentum of a Collision

#### Strategy

#### Solution

The moment of inertia of the system with the bullet embedded in the disk is

The final angular momentum of the system is

Thus, by conservation of angular momentum, L i = L f L i = L f and

#### Significance

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## DISCUSSION

Here, we have considered beams close to the paraxial limit and have shown how the independence of spin and orbital angular momenta allows for a new definition of total angular momentum. However, it has recently been shown that independent spin and orbital angular momenta can be defined beyond this limit (

*19*,*34*,*35*). This immediately provides the nonparaxial generalization of the total angular momentum component*J*_{γ}, on replacing the paraxial forms for spin and orbital angular momenta, assumed here, with the nonparaxial ones. The nonparaxial spin and orbital angular momenta are independently conserved hence, this would be a conserved quantity as well. They are modified forms of the rotation operators, specifically the transverse parts of those that rotate the field vectors (spin) and image (orbital) around the specified axis. The corresponding total angular momentum*J*_{γ}generates these modified rotations simultaneously, in a fixed ratio γ.The new form of total angular momentum we have identified gives an alternative representation of the state space in terms of beams with nonuniform polarization, leading to a new understanding of the effects of optical angular momentum. Its half-integer spectrum shows that for light, as is already known for electronic systems, reduced dimensionality allows for new forms of quantization. The half-integer quantization, which we demonstrate through noise measurements, implies fermionic exchange statistics, and an important extension of our work will be to identify the measurable consequences of such photonic fermionization.

**1.**A. Ashkin, J. M. Dziedzic, J. E. Bjorkholm, and S. Chu, “Observation of Single-Beam Gradient Force Optical Trap for Dielectric Particles,” Opt. Lett.**11**, 288–290 (1986). [CrossRef] [PubMed]**2.**S. M. Block, D. F. Blair, and H. C. Berg, “Compliance of bacterial flagella measured with optical tweezers,” Nature**338**, 514–518 (1989). [CrossRef] [PubMed]**3.**A. Ashkin, K. Schutze, J. M. Dziedzic, U. Euteneuer, and M. Schliwa, “Force generation of organelle transport measured in vivo by an infrared laser trap,” Nature**348**, 346–348 (1990). [CrossRef] [PubMed]**4.**K. Svoboda, C. F. Schmidt, B. J. Schnapp, and S. M. Block, “Direct observation of kinesin stepping by optical trapping interferometry,” Nature**365**, 721–727 (1993). [CrossRef] [PubMed]**5.**J. T. Finer, R. M. Simmons, and J. A. Spudich, “Single myosin molecule mechanics: piconewton forces and nanometre steps,” Nature**368**, 113–119 (1994). [CrossRef] [PubMed]**6.**H. Yin, M. D. Wang, K. Svoboda, R. Landick, S. M. Block, and J. Gelles, “Transcription Against an Applied Force,” Science**270**, 1653–1657 (1995). [CrossRef] [PubMed]**7.**M. S. Z. Kellermayer, S. B. Smith, H. L. Granzier, and C. Bustamante, “Folding-Unfolding Transitions in Single Titin Molecules Characterized with Laser Tweezers,” Science**276**, 1112–1116 (1997). [CrossRef] [PubMed]**8.**L. Tskhovrebova, J. Trinick, J. A. Sleep, and R. M. Simmons, “Elasticity and unfolding of single molecules of the giant muscle protein titin,” Nature**387**, 308–312 (1997). [CrossRef] [PubMed]**9.**D. E. Smith, S. J. Tans, S. B. Smith, S. Grimes, D. L. Anderson, and C. Bustamante, “The bacteriophage theta 29 portal motor can package DNA against a larger internal force,” Nature**413**, 748–752 (2001). [CrossRef] [PubMed]**10.**H. He, M. E. J. Friese, N. R. Heckenberg, and H. Rubinsztein-Dunlop, “Direct Observation of Transfer of Angular Momentum to Absorptive Particles from a Laser Beam with a Phase Singularity,” Phys. Rev. Lett.**75**, 826–829 (1995). [CrossRef] [PubMed]**11.**M. E. J. Friese, J. Enger, H. Rubinsztein-Dunlop, and N. R. Heckenberg, “Optical angular-momentum transfer to trapped absorbing particles,” Phys. Rev. A**54**, 1593–1596 (1996). [CrossRef] [PubMed]**12.**M. E. J. Friese, T. A. Nieminen, N. R. Heckenberg, and H. Rubinsztein-Dunlop, “Optical alignment and spinning of laser-trapped microscopic particles,” Nature**394**, 348–.350 (1998). [CrossRef]**13.**T. A. Nieminen, N. R. Heckenberg, and H. Rubinsztein-Dunlop, “Optical measurement of microscopic torques,” J. Mod. Opt.**48**, 405–413 (2001).**14.**A. I. Bishop, T. A. Nieminen, N. R. Heckenberg, and H. Rubinsztein-Dunlop, “Optical Microrheology Using Rotating Laser-Trapped Particles,” Phys. Rev. Lett.**92**, 198104 (2004). [CrossRef] [PubMed]**15.**L. Allen, M. W. Beijersbergen, R. J. C. Spreeuw, and J. P. Woerdman, “Orbital angular momentum and the transformation of Laguerre-Gaussian laser modes,” Phys. Rev. A**45**, 8185–8189 (1992). [CrossRef] [PubMed]**16.**K. D. Bonin and B. Kourmanov, “Light torque nanocontrol, nanomotors and nanorockers,” Opt. Express**10**, 984–989 (2002). [PubMed]**17.**A. I. Bishop, T. A. Nieminen, N. R. Heckenberg, and H. Rubinsztein-Dunlop, “Optical application and measurement of torque on microparticles of isotropic nonabsorbing material,” Phys. Rev. A**68**, 033802 (2003). [CrossRef]**18.**P. Galajda and P. Ormos, “Complex micromachines produced and driven by light,” Appl. Phys. Lett.**78**, 249–251 (2001). [CrossRef]**19.**A. Mair, A. Vaziri, G. Weihs, and A. Zeilinger, “Entanglement of the orbital angular momentum states of photons,” Nature**412**, 313–316 (2001). [CrossRef] [PubMed]**20.**S. J. Parkin, T. A. Nieminen, N. R. Heckenberg, and H. Rubinsztein-Dunlop, “Optical measurement of torque exerted on an elongated object by a noncircular laser beam,” Phys. Rev. A**70**, 023816 (2004). [CrossRef]**21.**J. Courtial, D. A. Robertson, K. Dholakia, L. Allen, and M. J. Padgett, “Rotational Frequency Shift of a Light Beam,” Phys. Rev. Lett.**81**, 4828–4830 (1998). [CrossRef]**22.**I. V. Basistiy, V. V. Slyusar, M. S. Soskin, and M. V. Vasnetsov, “Manifestation of the rotational Doppler effect by use of an off axis optical vortex beam,” Opt. Lett.**28**, 1185–1187 (2003). [CrossRef] [PubMed]**23.**J. Leach, J. Courtial, K. Skeldon, S. M. Barnett, S. Franke-Arnold, and M. J. Padgett, “Interferometric Methods to Measure Orbital and Spin, or the Total Angular Momentum of a Single Photon,” Phys. Rev. Lett.**92**, 013601 (2004). [CrossRef] [PubMed]**24.**G. Knüner, S. Parkin, N. R. Heckenberg, and H. Rubinsztein-Dunlop, “Characterization of optically driven fluid stress fields with optical tweezers,” Phys. Rev. E**72**, 031507 (2005). [CrossRef]**25.**W. S. Ryu, R. M. Berry, and H. C. Berg, “Torque-generating units of the flagellar motor of Escherichia coli have a high duty ratio,” Nature**403**, 444–447 (2000). [CrossRef] [PubMed]### References

- View by:
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- |
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- |
- Publication

- A. Ashkin, J. M. Dziedzic, J. E. Bjorkholm, and S. Chu, “Observation of Single-Beam Gradient Force Optical Trap for Dielectric Particles,” Opt. Lett. 11, 288–290 (1986).

[Crossref][PubMed] - S. M. Block, D. F. Blair, and H. C. Berg, “Compliance of bacterial flagella measured with optical tweezers,” Nature 338, 514–518 (1989).

[Crossref][PubMed] - A. Ashkin, K. Schutze, J. M. Dziedzic, U. Euteneuer, and M. Schliwa, “Force generation of organelle transport measured in vivo by an infrared laser trap,” Nature 348, 346–348 (1990).

[Crossref][PubMed] - K. Svoboda, C. F. Schmidt, B. J. Schnapp, and S. M. Block, “Direct observation of kinesin stepping by optical trapping interferometry,” Nature 365, 721–727 (1993).

[Crossref][PubMed] - J. T. Finer, R. M. Simmons, and J. A. Spudich, “Single myosin molecule mechanics: piconewton forces and nanometre steps,” Nature 368, 113–119 (1994).

[Crossref][PubMed] - H. Yin, M. D. Wang, K. Svoboda, R. Landick, S. M. Block, and J. Gelles, “Transcription Against an Applied Force,” Science 270, 1653–1657 (1995).

[Crossref][PubMed] - M. S. Z. Kellermayer, S. B. Smith, H. L. Granzier, and C. Bustamante, “Folding-Unfolding Transitions in Single Titin Molecules Characterized with Laser Tweezers,” Science 276, 1112–1116 (1997).

[Crossref][PubMed] - L. Tskhovrebova, J. Trinick, J. A. Sleep, and R. M. Simmons, “Elasticity and unfolding of single molecules of the giant muscle protein titin,” Nature 387, 308–312 (1997).

[Crossref][PubMed] - D. E. Smith, S. J. Tans, S. B. Smith, S. Grimes, D. L. Anderson, and C. Bustamante, “The bacteriophage theta 29 portal motor can package DNA against a larger internal force,” Nature 413, 748–752 (2001).

[Crossref][PubMed] - H. He, M. E. J. Friese, N. R. Heckenberg, and H. Rubinsztein-Dunlop, “Direct Observation of Transfer of Angular Momentum to Absorptive Particles from a Laser Beam with a Phase Singularity,” Phys. Rev. Lett. 75, 826–829 (1995).

[Crossref][PubMed] - M. E. J. Friese, J. Enger, H. Rubinsztein-Dunlop, and N. R. Heckenberg, “Optical angular-momentum transfer to trapped absorbing particles,” Phys. Rev. A 54, 1593–1596 (1996).

[Crossref][PubMed] - M. E. J. Friese, T. A. Nieminen, N. R. Heckenberg, and H. Rubinsztein-Dunlop, “Optical alignment and spinning of laser-trapped microscopic particles,” Nature 394, 348–.350 (1998).

[Crossref] - T. A. Nieminen, N. R. Heckenberg, and H. Rubinsztein-Dunlop, “Optical measurement of microscopic torques,” J. Mod. Opt. 48, 405–413 (2001).
- A. I. Bishop, T. A. Nieminen, N. R. Heckenberg, and H. Rubinsztein-Dunlop, “Optical Microrheology Using Rotating Laser-Trapped Particles,” Phys. Rev. Lett. 92, 198104 (2004).

[Crossref][PubMed] - L. Allen, M. W. Beijersbergen, R. J. C. Spreeuw, and J. P. Woerdman, “Orbital angular momentum and the transformation of Laguerre-Gaussian laser modes,” Phys. Rev. A 45, 8185–8189 (1992).

[Crossref][PubMed] - K. D. Bonin and B. Kourmanov, “Light torque nanocontrol, nanomotors and nanorockers,” Opt. Express 10, 984–989 (2002).

[PubMed] - A. I. Bishop, T. A. Nieminen, N. R. Heckenberg, and H. Rubinsztein-Dunlop, “Optical application and measurement of torque on microparticles of isotropic nonabsorbing material,” Phys. Rev. A 68, 033802 (2003).

[Crossref] - P. Galajda and P. Ormos, “Complex micromachines produced and driven by light,” Appl. Phys. Lett. 78, 249–251 (2001).

[Crossref] - A. Mair, A. Vaziri, G. Weihs, and A. Zeilinger, “Entanglement of the orbital angular momentum states of photons,” Nature 412, 313–316 (2001).

[Crossref][PubMed] - S. J. Parkin, T. A. Nieminen, N. R. Heckenberg, and H. Rubinsztein-Dunlop, “Optical measurement of torque exerted on an elongated object by a noncircular laser beam,” Phys. Rev. A 70, 023816 (2004).

[Crossref] - J. Courtial, D. A. Robertson, K. Dholakia, L. Allen, and M. J. Padgett, “Rotational Frequency Shift of a Light Beam,” Phys. Rev. Lett. 81, 4828–4830 (1998).

[Crossref] - I. V. Basistiy, V. V. Slyusar, M. S. Soskin, and M. V. Vasnetsov, “Manifestation of the rotational Doppler effect by use of an off axis optical vortex beam,” Opt. Lett. 28, 1185–1187 (2003).

[Crossref][PubMed] - J. Leach, J. Courtial, K. Skeldon, S. M. Barnett, S. Franke-Arnold, and M. J. Padgett, “Interferometric Methods to Measure Orbital and Spin, or the Total Angular Momentum of a Single Photon,” Phys. Rev. Lett. 92, 013601 (2004).

[Crossref][PubMed] - G. Knüner, S. Parkin, N. R. Heckenberg, and H. Rubinsztein-Dunlop, “Characterization of optically driven fluid stress fields with optical tweezers,” Phys. Rev. E 72, 031507 (2005).

[Crossref] - W. S. Ryu, R. M. Berry, and H. C. Berg, “Torque-generating units of the flagellar motor of Escherichia coli have a high duty ratio,” Nature 403, 444–447 (2000).

[Crossref][PubMed]

#### 2005 (1)

G. Knüner, S. Parkin, N. R. Heckenberg, and H. Rubinsztein-Dunlop, “Characterization of optically driven fluid stress fields with optical tweezers,” Phys. Rev. E 72, 031507 (2005).

[Crossref]#### 2004 (3)

J. Leach, J. Courtial, K. Skeldon, S. M. Barnett, S. Franke-Arnold, and M. J. Padgett, “Interferometric Methods to Measure Orbital and Spin, or the Total Angular Momentum of a Single Photon,” Phys. Rev. Lett. 92, 013601 (2004).

[Crossref] [PubMed]S. J. Parkin, T. A. Nieminen, N. R. Heckenberg, and H. Rubinsztein-Dunlop, “Optical measurement of torque exerted on an elongated object by a noncircular laser beam,” Phys. Rev. A 70, 023816 (2004).

[Crossref]A. I. Bishop, T. A. Nieminen, N. R. Heckenberg, and H. Rubinsztein-Dunlop, “Optical Microrheology Using Rotating Laser-Trapped Particles,” Phys. Rev. Lett. 92, 198104 (2004).

[Crossref] [PubMed]#### 2003 (2)

A. I. Bishop, T. A. Nieminen, N. R. Heckenberg, and H. Rubinsztein-Dunlop, “Optical application and measurement of torque on microparticles of isotropic nonabsorbing material,” Phys. Rev. A 68, 033802 (2003).

[Crossref]#### 2002 (1)

#### 2001 (4)

T. A. Nieminen, N. R. Heckenberg, and H. Rubinsztein-Dunlop, “Optical measurement of microscopic torques,” J. Mod. Opt. 48, 405–413 (2001).

D. E. Smith, S. J. Tans, S. B. Smith, S. Grimes, D. L. Anderson, and C. Bustamante, “The bacteriophage theta 29 portal motor can package DNA against a larger internal force,” Nature 413, 748–752 (2001).

[Crossref] [PubMed]P. Galajda and P. Ormos, “Complex micromachines produced and driven by light,” Appl. Phys. Lett. 78, 249–251 (2001).

[Crossref]A. Mair, A. Vaziri, G. Weihs, and A. Zeilinger, “Entanglement of the orbital angular momentum states of photons,” Nature 412, 313–316 (2001).

[Crossref] [PubMed]#### 2000 (1)

W. S. Ryu, R. M. Berry, and H. C. Berg, “Torque-generating units of the flagellar motor of Escherichia coli have a high duty ratio,” Nature 403, 444–447 (2000).

[Crossref] [PubMed]#### 1998 (2)

J. Courtial, D. A. Robertson, K. Dholakia, L. Allen, and M. J. Padgett, “Rotational Frequency Shift of a Light Beam,” Phys. Rev. Lett. 81, 4828–4830 (1998).

[Crossref]M. E. J. Friese, T. A. Nieminen, N. R. Heckenberg, and H. Rubinsztein-Dunlop, “Optical alignment and spinning of laser-trapped microscopic particles,” Nature 394, 348–.350 (1998).

[Crossref]#### 1997 (2)

M. S. Z. Kellermayer, S. B. Smith, H. L. Granzier, and C. Bustamante, “Folding-Unfolding Transitions in Single Titin Molecules Characterized with Laser Tweezers,” Science 276, 1112–1116 (1997).

[Crossref] [PubMed]L. Tskhovrebova, J. Trinick, J. A. Sleep, and R. M. Simmons, “Elasticity and unfolding of single molecules of the giant muscle protein titin,” Nature 387, 308–312 (1997).

[Crossref] [PubMed]#### 1996 (1)

M. E. J. Friese, J. Enger, H. Rubinsztein-Dunlop, and N. R. Heckenberg, “Optical angular-momentum transfer to trapped absorbing particles,” Phys. Rev. A 54, 1593–1596 (1996).

[Crossref] [PubMed]#### 1995 (2)

H. He, M. E. J. Friese, N. R. Heckenberg, and H. Rubinsztein-Dunlop, “Direct Observation of Transfer of Angular Momentum to Absorptive Particles from a Laser Beam with a Phase Singularity,” Phys. Rev. Lett. 75, 826–829 (1995).

[Crossref] [PubMed]H. Yin, M. D. Wang, K. Svoboda, R. Landick, S. M. Block, and J. Gelles, “Transcription Against an Applied Force,” Science 270, 1653–1657 (1995).

[Crossref] [PubMed]#### 1994 (1)

J. T. Finer, R. M. Simmons, and J. A. Spudich, “Single myosin molecule mechanics: piconewton forces and nanometre steps,” Nature 368, 113–119 (1994).

[Crossref] [PubMed]#### 1993 (1)

K. Svoboda, C. F. Schmidt, B. J. Schnapp, and S. M. Block, “Direct observation of kinesin stepping by optical trapping interferometry,” Nature 365, 721–727 (1993).

[Crossref] [PubMed]#### 1992 (1)

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L. Allen, M. W. Beijersbergen, R. J. C. Spreeuw, and J. P. Woerdman, “Orbital angular momentum and the transformation of Laguerre-Gaussian laser modes,” Phys. Rev. A 45, 8185–8189 (1992).

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S. J. Parkin, T. A. Nieminen, N. R. Heckenberg, and H. Rubinsztein-Dunlop, “Optical measurement of torque exerted on an elongated object by a noncircular laser beam,” Phys. Rev. A 70, 023816 (2004).

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## Total rotational angular momentum estimates for Jupiter? - Astronomy

The typical red giant star rotates slowly. This characteristic is expected from the conservation of angular momentum as these stars expand during their evolution. Nevertheless, a small percentage of red giant stars are rapidly rotating. One possible source of these stars' excess angular momenta is the orbital angular momentum of a planetary companion. The transfer of orbital angular momentum to the stellar envelope decays the planet's orbit, ultimately leading to the rapid in-spiral of the planet into the star. Using the known sample of exoplanets around main sequence host stars, I simulated both the future evolution of these stars and the expected interactions with their planets and found that Jupiter-mass planets residing at inner solar system distances---relatively common in exoplanetary systems---can contribute enough angular momentum to cause rapid rotation in their host stars during the red giant phase. Gas giant planets are also massive enough to alter the chemical composition of their host stars' envelopes when they are accreted. The central experiment of this thesis is to search for abundance anomalies in the rapid rotators that could be indicative of planet accretion. Hypothetical anomalies include the replenishment of light elements that are diluted by giant stars during first dredge-up (such as the stellar surface abundance of lithium), changes in isotopic abundance ratios that were altered by nucleosynthesis (such as increasing the stellar surface 12C/13C), and the preferential enhancement of refractory elements (indicative of the accretion of chemically fractionated material such as a planet). To increase the total number of known rapid rotators, I measured rotational velocities in a large database of spectra collected for the Grid Giant Star Survey developed for NASA's Space Interferometry Mission's astrometric grid. The 28 new rapid rotators discovered in this sample were combined with rapid rotators from the literature and a control sample of slow rotators to form a new sample for the abundance experiment. This thesis presents evidence that the accretion of planets of a few Jupiter masses with "normal" planetary compositions can reproduce both the observed rotational velocities and abundances of red giant rapid rotators.

## 1 Answer 1

The first point to realize is that the bike is not a rigid body. You can decompose it in three different parts (the two wheels and the bicycle frame), evaluate separately the angular momentum of each and sum.

For each wheel, the total angular momentum is given by the angular momentum with respect to the wheel's center of mass, plus the angular momentum of the center of mass. This gives

for the two wheels, where $I_

$ is their momentum of inertia with respect to their center of mass, $R_w$ their radius and $m_w$ their mass (I supposed that the two wheels are identical). Note that $-V/R_w = omega$ is the angular velocity of each wheel. For the bicycle frame we get

because it just translate horizontally. Here $y_

$ is the vertical position of the frame's center of mass, and $M$ the frame's mass. Adding together all the pieces we get $vec

= -2 I_ frac <> >hat - 2 m_ V R_ hat - M V y_ hat $ but this can be rewritten as

is the total mass of the bicycle and

is the horizontal position of the bicycle's center of mass. So

The final result is the sum of the angular momentum of the rotating wheels with respect of their center of mass, plus the angular momentum of the center of mass of the bicycle.

## Watch the video: Διατήρηση στροφορμής και ορμής στην κρούση σφαίρας και ράβδου Ν. Χρόνης (February 2023).