Translational Angular Momentum

From Physics Book
Jump to navigation Jump to search

"Orbital" Angular Momentum Motion

Claimed by: Dhanusha Subramani 2015

Claimed by Nicole Harris Fall 2017

The Main Idea

Angular momentum describes the rotational motion of an object; it encompasses the translational and rotational angular momentum. The translational angular momentum deals with motion that can be described as orbital angular motion such as the orbits of the planets around the Sun in a bound system. When compared to the rotational angular momentum, we can see that the translational angular momentum doesn't take into account angular velocity or the inertia of the object(s); it simply deals with the distance(radius) and the momentum.

A Mathematical Model

The magnitude of Translational Angular Momentum is calculated using the magnitudes of the position and momentum vectors and the sin of the angle between the two vectors (θ). This is also technically equal to the cross product of the position and momentum vectors. (The position and momentum of the object are taken at a particular instant, so it is difficult to determine the object's future or past motion or path.)

The direction of the Translational Angular Momentum can be determined using the right-hand rule. The positive z axis (+z) indicates that the object is moving counterclockwise in an xy plane while the negative z axis (-z) indicates that the object is moving counterclockwise. See: Right Hand Rule for how to determine the direction of angular momentum.

The perpendicular component of the r vector represents the product of the position times the sine of the angle (θ). This distance can be determined by drawing a line perpendicular to the momentum vector. The perpendicular distance is then the line that goes from the tail of the r vector to the line that the momentum vector points to.

It is useful to think of the translational angular momentum as relative to a certain location, A, because this will help determine the momentum as specific to the location and instant, making the calculations more accurate and useful for application.

Keep in mind that you are taking the cross product of the two vectors, not multiplying them. Remember the formula for calculating the cross product is on the back of the formula sheet.

A Computational Model

Here is some VPython coding that I screen captured that allows us to visualize angular momentum in its two components (translational and rotational). When we run the code in VIDLE, we can see that there are two separate rods: the barbell which consists of the rod and two small masses and another rod which connects to the barbell. Only the barbell exhibits translational angular momentum because of the change in its displacement as it rotates; the barbell has no rotational angular momentum because it doesn't have its own axis about which it rotates (the second rod does have rotational angular momentum however).



Example 1 At a particular instant the location of an object relative to location A is given by the vector [math]\displaystyle{ { \vec{ r}_{A} = \left \langle {6,6,0} m \right \rangle } }[/math]. At this instant the momentum of the object is [math]\displaystyle{ \vec{p} = \left \langle {-11, 13, 0} \right \rangle kgm/s }[/math]. What is the angular momentum of the object about location A?

Solution 1

[math]\displaystyle{ {\vec{L}_{trans,A} = \vec{ r}_{A} ⨯ \vec{p} = \left \langle {0, 0, {r}_{x}{p}_{y} – {p}_{x}{r}_{y}} \right \rangle } }[/math]

[math]\displaystyle{ {=\left \langle {0, 0, (6)(13)-(-11)(6)} \right \rangle kg·m^2 /s } }[/math]

[math]\displaystyle{ {=\left \langle {0, 0, 144} \right \rangle kg·m^2 /s } }[/math]

Example 2

Calculate the magnitude of Earht’s translational (orbital) angular momentum relative to the Sun when the Earth is at location A and when the Earth is at location B as shown in the figure belwo. The mass of the Earth is 6*10^24 kg and its distance from the sun is 1.5*10^11 m.

Solution 2 The Earth makes one complete orbit around the sun in a time period of 1 year, so its average speed its:

[math]\displaystyle{ { v = \frac{distance}{time} = \frac{2pi(1.5*10^{11}) m }{(365)(24)(60)(60)} = 3.0*10^4\ m/s} }[/math]

At location A,

[math]\displaystyle{ \vec{p} = \left \langle {0,( 6*10^{24} kg)(3.0*10^4 m/s, 0} \right \rangle = \left \langle {0, 1.8*10^{29}, 0} \right \rangle kg·m/s }[/math]

[math]\displaystyle{ |\vec{p}| = 1.8*10^{29}\ kg·m/s }[/math]

[math]\displaystyle{ |\vec{L}_{trans,Sun}| = (1.5*10^{11} m)(1.8*10^{29} kg·m/s) sin(90) = 2.7*10^{40} kg·m^2/s }[/math]

This example uses the main magnitude formula for calculating the translational angular momentum. The velocity is calculated by dividing the circumference of the circular orbit by the time (seconds) then is multiplied by the Earth's mad to determine the momentum vector. The magnitudes of the position and momentum vectors are multiplied by the sin of 90 degrees because that's the angle between the position and momentum vectors. Since the angle and magnitudes of the position and momentum vectors are the same at position A and B, the magnitude of the translational angular momentum remains the same, as well.


Example 1 The Figure shows seven particles, each with the same magnitude of momentum [math]\displaystyle{ |\vec{p}| kgm/s }[/math] but with different directions of momentum and different positions relative to location A. The distances shown in the diagram have these values: 15 m, h = 30 m, and d = 25 m. Calculate the [math]\displaystyle{ z }[/math] component of the translational angular momentum [math]\displaystyle{ L_{Az} }[/math] for each particle (x to the right, y up, z out of the page). Make sure you give the correct sign.

Solution 1

[math]\displaystyle{ L_{Az,1} = |\vec{p}|h(-\hat{z}) = -700\ kg·m^2/s }[/math]

[math]\displaystyle{ L_{Az,2} = |\vec{p}|h(-\hat{z}) = -700\ kg·m^2/s }[/math]

[math]\displaystyle{ L_{Az,3} = |\vec{p}|h(-\hat{z}) = -700\ kg·m^2/s }[/math]

[math]\displaystyle{ L_{Az,4} = |\vec{p}|w(-\hat{z}) = -450\ kg·m^2/s }[/math]

[math]\displaystyle{ L_{Az,5} = |\vec{p}|0 = 0\ kg·m^2/s }[/math]

[math]\displaystyle{ L_{Az,6} = |\vec{p}|0 = 0\ kg·m^2/s }[/math]

[math]\displaystyle{ L_{Az,7} = |\vec{p}|d sin(90)\hat{z} = 675\ kg·m^2/s }[/math]

Particles 1, 2, 3, and 4 all have negative values for the z component via the right hand rule comparing the position and momentum vector directions which shows that the thumb extends into the page (-z direction). Particle 5 has a value of 0 because the r vector magnitude would be 0. Particle 6 has a value of 0 because even though it has a position magnitude, the position vector and momentum vector point in the same direction, meaning that the angle in between the two is 0.

Example 2 Here is also a similar example from Exam 4 following the same methodology and approach. Remember to keep the right-hand rule in mind when determining the direction of the transnational angular momentum.


Example Riders on the Ferris wheel sit in chairs that are on pivots so that they remain level as the wheel turns at a constant rate. The Ferris wheel pictured above has a radius of 24 meters, and it makes one complete revolution around its axle in 20 s. In answering the following questions, consider location A, at the center of the axle, to be the location around which we calculate the angular momentum. At the instance shown in the picture, a child of mass 40 kg, witting at location F, is traveling with velocity [math]\displaystyle{ \left \langle {7.5, 0, 0} \right \rangle m/s }[/math].

a) What is the linear momentum of the child?

b) In the definition [math]\displaystyle{ \vec{L} = \vec{r} × \vec {p} }[/math], what is the vector [math]\displaystyle{ \vec{r} }[/math]?

c) What is the [math]\displaystyle{ \vec{r}_{tangent} }[/math]?

d) What is the magnitude of the angular momentum of the child about location A?

e) What is the plane defined by [math]\displaystyle{ \vec{r} }[/math] and [math]\displaystyle{ \vec{p} }[/math]? (As in, which plane contains both vectors?)

f) Use the right-hand rule to determine the z component of the angular momentum of the child about location A.

g) You used the right-hand rule to determine the z component of the angular momentum, but as a check, calculate in terms of position an momentum: What is [math]\displaystyle{ xp_y }[/math]? What is [math]\displaystyle{ yp_x }[/math]? Therefore, what is the z component of the angular momentum of the child about location A?

h) The Ferris wheel keeps turning with coordinates [math]\displaystyle{ \left \langle {16.971, -16.971, 0} \right \rangle m }[/math] relative to locaton A, moving with velocity [math]\displaystyle{ \left \langle {5.303, 5.303, 0} \right \rangle m/s }[/math]. Now what is the magnitude of the angular momentum of the child about location A?


a) [math]\displaystyle{ \vec{p} = \vec{v} = \left \langle {300, 0, 0} \right \rangle kg m^2 /s }[/math]

b) [math]\displaystyle{ \vec{r} = radius }[/math] (-y direction at position F)

[math]\displaystyle{ \qquad = \left \langle {0, -24, 0} \right \rangle m }[/math]

c) [math]\displaystyle{ \vec{r}_{tangent} = \vec{r} = \left \langle {0, -24, 0} \right \rangle m }[/math]

d) [math]\displaystyle{ |\vec{L}_{trans,A}| = |\vec{p}||\vec{r}| = 7200\ kgm^2/s }[/math]

e) In the xy plane: [math]\displaystyle{ y = \vec{r} }[/math] [math]\displaystyle{ x=\vec{p} }[/math]

f) [math]\displaystyle{ L_{Az} = |\vec{L}_{trans,A}| = 7200\ kgm^2/s }[/math]

in the +z direction (right hand rule)

g) [math]\displaystyle{ xp_{y} = 0\ kgm^2/s }[/math]

[math]\displaystyle{ \qquad yp_{y} = -7200\ kgm^2/s }[/math]

[math]\displaystyle{ \qquad L_{z} = xp_{y} - yp_{y} = 7200\ kgm^2/s }[/math]

h) [math]\displaystyle{ \vec{p} = \left \langle {212.12, 212.12, 0} \right \rangle kgm/s }[/math]

[math]\displaystyle{ \qquad xp_{y} = 2599.89\ kgm^2/s }[/math]

[math]\displaystyle{ \qquad yp_{y} = 2599.89\ kgm^2/s }[/math]

[math]\displaystyle{ \qquad L_{z} = 7200\ kgm^2/s }[/math]


Angular momentum can be useful especially when considering the conservation of angular momentum. It can be applied to areas of research and study in physics that help to understand the Universe since it's a concept that has been noted in small particles, atoms, and various motions of most objects. For me, I found it interesting to connect angular momentum to medicine (my career path interest). Angular momentum is taken into account when dealing with sports injuries such as those involving skis and skates. The translational and rotational components of the angular momentum will determine how much force is placed on knees and can be useful for understanding injuries. Also, certain medical technology and machinery have to account for angular momentum such as the MRI (magnetic Resonance Imaging) which operates by taking advantage of the electron spin angular momentum and taken into account the magnetic field it operates in.


We can contribute the idea of orbits to Kepler who discovered that planets have elliptical orbits around the sun (Kepler's First Law). The idea of angular momentum itself, however, is credited to Newton based on his First Law of Motion which states that all objects remain at rest until a force is exerted upon them. Derivations of the first law come from observations and calculations of orbits by Newton and found that there was conservation of angular momentum in the Law of Areas. Later on, Euler, Bernoulli, and d'Arcy analyzed Kepler's Second Law of Planetary Motion and further understood the conservation; however, they probably didn't fully understand the rotational aspects.

See also

This topic relates to the entirety of the angular momentum section as a whole since it's the basis on which to build upon.

Total Angular Momentum

The Angular Momentum Principle

Rotational Angular Momentum

Right Hand Rule

Further reading

External links


Georgia Tech Physics 2211 Exam 4 Fall 2015

Matter and Interactions by Ruth Chabay and Bruce Sherwood (4th Edition)