# Angular moment (Angular Momentum): quantity, conservation, examples, exercises

We explain the angular moment with quantity, conservation, examples and exercises. The **angular moment** **or angular momentum** is, for the movement of rotation, the momentum is for translational movement. It is a vector quantity that characterizes the rotation of a point particle or an object extended around an axis that passes through a point. angular moment

This means that whenever angular momentum is to be calculated, the axis of rotation must be specified appropriately.

Starting with a material point of mass m, the angular momentum is denoted by **L,** the linear momentum as **p** and the position of the particle with respect to an axis that passes through a certain point O is **r** , then:

**L** = **r** x **p**

The bold letters are reserved for vector quantities and the cross means that the angular momentum is the vector product between the position vector **r** and the momentum **p** of the particle. The vector that results from a vector product is perpendicular to the plane formed by the participating vectors.

This means that the direction and sense of **L** can be found by the right-hand rule for the cross product.

In the International System of Units SI, the units of angular momentum are kg⋅m ^{2} / s, which do not have a special name. And for an extended body, which is composed of many particles, the above definition is conveniently extended.

**Amount of angular movement**

The magnitude of the angular momentum vector is according to the definition of the vector product:

L = r⋅m⋅v⋅sen ϕ = mv (r⋅sen ϕ) = mvℓ

Where ϕ is the angle between the vectors **r** and **v** . Then ℓ = r sin ϕ is the perpendicular distance between the line of **v** and the point O.

For the case of the particle that moves describing the circumference shown in the upper image, this angle is 90º, since the speed is always tangent to the circumference and therefore perpendicular to the radius.

Therefore sin 90º = 1 and the magnitude of **L** is:

L = m⋅r⋅v

**Moment of inertia angular moment**

The moment of inertia of a rigid body describes the body’s inertia versus rotation about a certain axis.

It depends not only on the mass of the body, but also on the distance to the axis of rotation. This is easily understandable when you think that for some objects, it is easier to rotate about some axes than others.

For a system of particles, the moment of inertia, denoted by the letter I, is given by:

*I = ∑ r _{i }^{2} Δm _{i}*

Where *Δm _{i}*

_{ }is a small portion of mass and r

_{i}is its distance from the axis of rotation. An extended body is composed of numerous particles, hence its total moment of inertia is the sum of all the products between mass and distance of the particles that compose it.

If it is an extended body, the sum changes to an integral and *Δm* becomes a mass differential *dm* . The limits of integration depend on the geometry of the object:

*I = ∫ _{M } (r ^{2} ) dm*

The concept of moment of inertia is closely related to the angular momentum of an extended object, as we will see below.

**Angular Moment of a System of Particles**

Consider a system of particles, composed of masses *Δm _{i}* that is rotating following a circumference in the

*xy*plane , each one has a linear speed related to its angular speed, the latter the same for all particles:

*v _{i} = ωr _{i}*

Where r _{i} is the distance to the axis of rotation O. Then the magnitude of the angular momentum is:

*L _{i} *=

*Δm*.

_{i}*r*.

_{i}*(ωr*

_{i}) =**r**_{i }^{2}ω Δm_{i}The angular momentum of the system will be given by the sum:

L = *ω ∑ r _{i }^{2} Δm _{i}*

We quickly identify the moment of inertia, as defined in the previous section, and therefore the magnitude of its angular momentum is as follows:

*L = Iω*

As we have said that the particle system was in the xy plane, it turns out that the angular momentum is directed along the z axis, perpendicular to said plane. The direction is given by that of the rotation: the angular momentum is positive if the rotation is carried out counterclockwise.

An extended body can be divided into slices, each with angular momentum given by *L = Iω* directed along the z axis. If the object’s axis of symmetry coincides with the z axis, there is no problem, since even for points that are not in the xy plane, the components of angular momentum perpendicular to that axis cancel out.

Vectorially:

*L** = I ω*

This equation is valid for three-dimensional objects that rotate around an axis of symmetry.

**When does the angular momentum vary?**

When a net force acts on a particle or a body, its momentum can change, and consequently so will its angular momentum. To know when it varies, we use the derivative, which will give us the rate of change over time, if there is one:

Applying the product rule for the derivative:

The term **v** xm **v** is null, since it is the product of a vector with itself, and in the second term we find the net force **F** = m **a** , therefore:

The vector product **r** x **F** is nothing other than the net torque or torque, sometimes denoted by the Greek letter **τ** or as **M** , always in bold, since it is a vector quantity. So, in analogy with linear momentum, the angular momentum varies as long as there is a net torque or torque:

d **L** / dt = **M**

**Conservation of angular momentum**

From the preceding sections we have seen that:

d **L** / dt = **M**

That is, the angular momentum varies when there is a net torque. If there is no net torque, then:

d **L** / dt = **0 → L** is constant

In other words:

Initial angular momentum = Final angular momentum

This result is still valid even if a body is not rigid, as we will see in the following examples.

**Examples angular moment**

Angular momentum is an important magnitude that is revealed in many situations, which shows how universal it is:

**Figure skating and other sports**

Whenever a rotating body contracts, its rotational speed increases, this is well known to ice skaters.

This is due to the fact that when arms and legs contract, the moment of inertia I decreases, since the distance between their parts decreases, but since the angular momentum is conserved, to keep the product Iω constant, the angular velocity must increase.

This is true not only in skating, but also in sports and activities that require turning, such as divers and trapeze artists in the circus.

**Cats land on their feet**

Cats always manage to land on all fours when they fall. Although they have no initial momentum, they make sure to quickly turn their legs and tail to change their rotational inertia and manage to land on their feet.

Likewise, while they maneuver, their angular momentum is zero, since their rotation is not continuous.

**The movement of a frisbee**

A frisbee must be thrown by giving it a spin so that it flies, otherwise it will fall. In effect, the angular momentum provided by the launcher gives the puck sufficient stability to move further in the air.

**Balls in sports angular moment**

Balls in baseball, soccer, basketball, and other sports have angular momentum. As they are spherical, they have moment of inertia and are rotated during the game. Since the moment of inertia of a sphere is:

I = (2/5) MR ^{2}

Where M is the mass of the ball and R its radius, the moment of inertia about a certain (fixed) axis is:

L = (2/5) MR ^{2} ω

**The departure from the moon**

The Moon is moving away from the Earth, since the speed of the Earth’s rotation decreases due to the friction between large bodies of water and the bottom of the sea.

The Earth-Moon system conserves its angular momentum, therefore, if the Earth decreases its contribution, the Moon increases its contribution, moving away from the Earth.

**The atom angular moment**

The first postulate of Bohr’s atomic model states that an electron only occupies orbits where the angular momentum is an integer multiple of *h / 2π* , where h is Planck’s constant.

**Exercise resolved**

A thin steel rod has a mass of 500 g and a length of 30 cm. It rotates around an axis passing through its center at a rate of 300 revolutions per minute. Determine the modulus of its angular momentum.

**Solution angular moment**

We will need the moment of inertia of the rod referred to an axis that passes through its center. Consulting the tables of moment of inertia it is found that:

I = (1/12) ML ^{2} = (1/12) × 0.5 kg x (30 × 10 ^{-2} m) ^{2} = 3.75 × 10 ^{-3} kg.m ^{2}

Since it is an extended body, of which we know the angular speed, we use:

*L = Iω*

Before we transform the angular speed or angular frequency *ω* to radians / s:

ω = (300 revolutions / minute) × (1 minute / 60 seconds) x (2π radians / revolution) = 10 π rad / s

Substituting:

L = 3.75 x10 ^{-3} kg⋅m ^{2} × 10 π rad / s = 0.118 kg⋅m ^{2} / s