# 6.2: Four Questions

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## 6.2.1 What is angular momentum?

In Chapter 5, we showed that the linear momentum $$\mathbf{P}$$ of a particle with mass $$m$$ and velocity $$\mathbf{V}$$ is the product of the particle mass and particle velocity, e.g. $$\mathbf{P}=\mathrm{mV}$$. The angular momentum of a particle with respect to point $$\mathbf{O}$$ is the vector (cross) product of the position vector $$\mathbf{r}$$ of the particle with respect to point $$O$$ and the linear momentum of the particle $$\mathbf{P}$$ (See Figure $$\PageIndex{1}$$): $\mathbf{L}_{0} = \mathbf{r} \times \mathbf{P}=\mathbf{r} \times(m \mathbf{V}) \nonumber$

Figure $$\PageIndex{1}$$: Calculating linear momentum of a particle.

It is important to recognize that angular momentum, like linear momentum, is a vector—it has both magnitude and direction. The dimensions of angular momentum are $$[\mathrm{L}]^{2} [\mathrm{M}] [\mathrm{T}]^{-1}$$. Typical units in the SI system are $$\mathrm{kg} \cdot \mathrm{m}^{2} / \mathrm{s}$$ or $$\mathrm{N} \cdot \mathrm{m} \cdot \mathrm{s}$$ and in the USCS system are $$\mathrm{lbm} \cdot \mathrm{ft}^{2} / \mathrm{s}$$ or $$\mathrm{lbf} \cdot \mathrm{ft} \cdot \mathrm{s}$$.

The specific angular momentum for the particle can be calculated as the angular momentum per unit mass: $\mathbf{l}_{0}=\frac{\mathbf{L}_{0}}{m}=\mathbf{r} \times \mathbf{V} \nonumber$ The dimensions of specific angular momentum are $$[\mathrm{L}]^{2} [\mathrm{~T}]^{-1} .$$

## 6.2.2 How can angular momentum be stored in a system?

For a system of particles, the angular momentum of the system with respect to the point $$O$$ is the sum of the angular momentum of each particle in the system: \begin{align} \mathbf{L}_{O, \text{ sys}} &= \sum_{j=1}^{n} \mathbf{L}_{O, \ j} = \sum_{j=1}^{n}\left(\mathbf{l}_{o} m\right)_{j} \nonumber \\ &= \sum_{j=1}^{n} \left(\mathbf{r}_{j} \times \mathbf{V}_{j}\right) m_{j} \end{align} \nonumber

For a continuous system, the summation is replaced by an integral over the system volume: $\mathbf{L}_{O, \text{ sys}} = \int\limits_{M_{\text{sys}}} \mathbf{I}_{o} \ dm = \int\limits_{V\kern-0.5em\raise0.3ex-_{\text{sys}}}(\mathbf{r} \times \mathbf{V}) \rho \ d V\kern-0.8em\raise0.3ex- \nonumber$ Because of the kinematics describing the motion of a system, the evaluation of this integral can be very difficult. In this chapter, we will only consider two types of motion for which this integral can be easily evaluated — translation and rotation about a fixed, centroidal axis. A more detailed investigation of angular momentum for general motion will be delayed until a later course, i.e. ES 204–Analysis of Mechanical Systems.

### Plane motion of a plane system

In our applications of angular momentum to a system, we will restrict ourselves to plane motion of a plane system. A plane system is a system that has a plane of symmetry that contains the center of mass of the system. Plane motion as discussed in Section 5.1.2 requires that the plane of symmetry of the system and the $$x \text{-} y$$ plane remain parallel at all times.

Figure $$\PageIndex{2}$$: Plane system with translation.

#### A plane system with translation

When a system is translating every point in the system has the same velocity, $$\mathbf{V}=\mathbf{V}_{\mathbf{x}} + \mathbf{V}_{\mathbf{y}}$$. To calculate the angular momentum of the system about point $$\mathrm{O}$$, we must evaluate Eq. $$\PageIndex{4}$$ as follows:

$\begin{array}{ll} \mathbf{L}_{0, \text{ sys}} &= \int\limits_{V\kern-0.5em\raise0.3ex-_{\text{sys}}} (\mathbf{r} \times \mathbf{V}) \rho \ d V\kern-0.8em\raise0.3ex- & \left| \begin{array}{l} \mathbf{V} \text{ comes outside the} \\ \text{integral because} \\ \text{it is a constant.} \end{array} \right. \\[4pt] &= \left( \int\limits_{\ V\kern-0.5em\raise0.3ex-_{\text{sys}}} \mathbf{r} \rho \ d V\kern-0.8em\raise0.3ex- \right) \times \underbrace{ \mathbf{V} }_{\begin{array}{c} \mathbf{V}_G = \mathbf{V} \\ \text{since} \\ \text{translation} \end{array}} & \left| \begin{array}{l} \text{The integral within () can be} \\ \text{rewritten in terms of the} \\ \text{center of mass (Section } 5.1.2 \text{)} \end{array} \right. \\ &= \left( m_{\text{sys }} \mathbf{r}_G \right) \times \mathbf{V}_G & \left| \begin{array}{l} \mathbf{r}_G \text{ is the position vector of the} \\ \text{center of mass with respect to} \\ \text{point } O. \end{array} \right. \\[4pt] &= m_{\text{sys}} \left( \mathbf{r}_G \times \mathbf{V}_G \right) & \end{array} \nonumber$

Thus the angular momentum of a translating plane system equals the product of the system mass and the cross product of $$\mathbf{r}_{G}$$ and $$\mathbf{V}_{G}$$, the position and velocity vector, respectively, of the system center of mass.

(a) Starting with the result in Eq. $$\PageIndex{5}$$, prove that the following expression is correct for an open system: $\frac{d \mathbf{L}_{0, \text{ sys}}}{dt} = \frac{d}{d t}\left[ m_{\text{sys}} \left(\mathbf{r}_{G} \times \mathbf{V}_{G}\right) \right] = \left(\mathbf{r}_{G} \times \mathbf{V}_{G}\right) \frac{d m_{\text{sys}}}{d t} + m_{\text{sys}} \left(\mathbf{r}_{G} \times \frac{d \mathbf{V}_{G}}{d t}\right) \nonumber$ What happened to $$m_{\text{sys}} \left(\dfrac{d \mathbf{r}_{G}}{d t} \times \mathbf{V}_{G}\right)$$?

(b) The block shown in the figure has a mass of $$50 \mathrm{~kg}$$ and is translating to the right with a velocity of $$10 \mathrm{~m} / \mathrm{s}$$. Determine the magnitude and direction of (a) the linear momentum of the system; (b) the angular momentum of the system with respect to point $$A$$, (c) the angular momentum of the system with respect to point $$B$$, and (d) the angular momentum of the system with respect to point $$C$$.

If a plane system is undergoing linear translation, i.e. it is traveling in a straight line, can it have angular momentum?

Figure $$\PageIndex{3}$$: Block traveling along a horizontal surface, with three vertically aligned reference points.

#### A plane, rigid system rotating about a fixed, centroidal axis

Many systems exhibit this type of motion: the rotation of a ceiling fan, the rotation of the rotor in a motor, and the rotation of a merry-go-round. A centroidal axis is an axis of rotation that intersects the center of mass of the system. The rotation of your windshield wiper blade and arm on a car does not fit this category because it is not rotating about a centroidal axis of the wiper blade-arm combination. The same would be true of the rotation of a simple pendulum about a non-centroidal pivot point.

For a rotating plane system we will make use, without proof, of a kinematic relation for the velocity of any point in a rigid system rotating in the $$x \text{-} y$$ plane about a fixed axis: $\begin{array}{l} \mathbf{V} &= \vec{\omega} \times \mathbf{r} \\ &=\left(\omega_{z} \mathbf{k}\right) \times (x \mathbf{i} + y \mathbf{j}) = \left| \begin{array}{lll} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 0 & \omega_{z} \\ x & y & 0 \end{array} \right| \\ &= -y \omega_{z} \mathbf{i} + x \omega_{z} \mathbf{j} \\[4pt] &= r \omega_{z} \left( \dfrac{-y \mathbf{i} + x \mathbf{j}}{r}\right) = \left(r \omega_{z}\right) [-(\sin \theta) \mathbf{i} + (\cos \theta) \mathbf{j}] \end{array} \nonumber$

Note that the angular velocity vector has a magnitude $$\omega_{z}$$ and points in the direction of the positive $$z$$-axis.

When you think about using the right-hand rule to cross the vector $$\mathbf{\omega} =\omega_{z} \mathbf{k}$$ with the $$\mathbf{r}$$ vector, align your fingers with the positive $$z$$-axis and curl them towards the radial vector $$\mathbf{r}$$. Your thumb should now point in a direction normal to the $$\mathbf{r}$$ vector.

When we further restrict ourselves to plane systems which are rotating around a centroidal axis, an axis that passes through the center of mass, we calculate the angular momentum as follows:

$\begin{array}{ll} L_{G, \text{ sys}} &= \displaystyle \int\limits_{V\kern-0.5em\raise0.3ex-_{\text{sys}}} (\mathbf{r} \times \mathbf{V}) \rho \ d V\kern-0.8em\raise0.3ex- & \left| \begin{array}{l} \text{where } \mathbf{V} = \omega_z \mathbf{k} \times \mathbf{r} \text{ since only} \\ \text{rotation about z-axis is possible} \end{array} \right. \\ &= \displaystyle \int\limits_{V\kern-0.5em\raise0.3ex-_{\text{sys}}} \left[ \mathbf{r} \times \left( \omega_z \mathbf{k} \times \mathbf{r} \right) \right] \rho \ d V\kern-0.8em\raise0.3ex- & \left| \begin{array}{l} \text{The rotational velocity } \omega_z \\ \text{is a constant.} \end{array} \right. \\ &=\omega_z \displaystyle \int\limits_{V\kern-0.5em\raise0.3ex-_{\text{sys}}} \underbrace{ \left[ \mathbf{r} \times \left( \mathbf{k} \times \mathbf{r} \right) \right] }_{= | \mathbf{r} \cdot \mathbf{r} | \mathbf{k} = r^2 \mathbf{k}} \rho \ d V\kern-0.8em\raise0.3ex- & \left| \begin{array}{l} \text{The double cross product can be} \\ \text{simplified using cylindrical coordinates} \\ \text{where } \mathbf{r} = r \ \mathbf{e}_{\mathbf{r}}, \ \mathbf{k} \times \mathbf{e}_{\mathbf{r}} = \mathbf{e}_{\theta}, \text{ and} \\ \mathbf{e}_{\mathbf{r}} \times \mathbf{e}_{\theta} = \mathbf{k}. \text{ (See Section 5.5)} \end{array} \right. \\[4pt] &= \omega_z \displaystyle \int\limits_{V\kern-0.5em\raise0.3ex-_{\text{sys}}} \left[ r^2 \mathbf{k} \right] \rho \ d V\kern-0.8em\raise0.3ex- \\ &= \omega_z \mathbf{k} \underbrace{ \left[ \displaystyle \int\limits_{V\kern-0.5em\raise0.3ex-_{\text{sys}}} r^2 \rho \ d V\kern-0.8em\raise0.3ex- \right] }_{= \text{Mass moment of inertia } I_G } & \left| \begin{array}{l} \text{The quantity in the brackets is} \\ \text{called the mass moment of inertia} \\ \text{about the } z \text{-axis that contains} \\ \text{the center of mass } G. \end{array} \right. \\ &= \omega_z I_G \mathbf{k} \end{array} \nonumber$

Thus the magnitude of the angular momentum of a plane, rigid system rotating about a fixed, centroidal axis equals the product of the rotational velocity $$\omega_{z}$$ about the $$z$$-axis and the mass moment of inertia $$I_{G}$$ of the system about the $$z$$-axis. As shown in Figure $$\PageIndex{4}$$, a mass can have a mass moment of inertia about any axis—$$x$$, $$y$$, or $$z$$.

Figure $$\PageIndex{4}$$: Moment of inertia of a mass.

The mass moment of inertia about the $$z$$-axis would be evaluated using the integral

$I_{G, \ z} = \int\limits_{V\kern-0.5em\raise0.3ex-_{\text{sys}}} r^2 \rho \ d V\kern-1.0em\raise0.3ex- = \begin{cases} \displaystyle \int\limits_{V\kern-0.5em\raise0.3ex-_{\text{sys}}} \left( x^2 + y^2 \right) \rho \ dx \ dy & \text{where } d V\kern-1.0em\raise0.3ex- = dx \ dy \\[4pt] \quad\quad\quad \text{or} \\[4pt] \displaystyle \int\limits_{V\kern-0.5em\raise0.3ex-_{\text{sys}}} r^2 \rho \ r \ d \theta \ dr & \text{where } d V\kern-1.0em\raise0.3ex- = r \ d \theta \ dr \end{cases} \nonumber$

Depending upon the shape of the object, the $$\mathrm{x} \text{-} \mathrm{y}$$ or $$\mathrm{r} \text{-} \theta$$ formulation may be preferable. The dimensions of the mass moment of inertia are $$[\mathrm{M}][\mathrm{L}]^{2}$$. Typical units are $$\mathrm{kg} \cdot \mathrm{m}^{2}$$ in SI and $$\mathrm{lbm} \cdot \mathrm{ft}^{2}$$ in USCS. Values for a rectangular solid, a solid cylinder, and a solid sphere can be found in Figure $$\PageIndex{5}$$.

Figure $$\PageIndex{5}$$: Centroidal mass moment of inertia for some common shapes.

## 6.2.3 How is angular momentum transported across a system boundary?

Angular momentum about point $$O$$ is transported across the boundary of a system by two mechanisms: angular momentum transport with external forces and angular momentum transport with mass flow.

### Angular momentum transport with forces

When an external force is applied to a system as discussed earlier it produces a moment with respect to point $$O$$. This moment represents a transport rate of angular momentum. More specifically, the net transport rate of angular momentum about point $$O$$ into a system with due to external force is written mathematically as the sum of the individual moments: $\dot{\mathbf{L}}_{O, \text { forces}} = \sum_{j} \mathbf{M}_{O, \ j} = \sum_{j}\left(\mathbf{r}_{j} \times \mathbf{F}_{j}\right) = \left[ \begin{array}{c} \text { Transport rate of } \\ \text { angular momentum } \\ \text { with external forces } \end{array} \right] \nonumber$

The moment about point $$O$$ due to the distributed gravitational force or weight can be shown to equal the moment produced by the force of the weight acting at the center of mass (or center of gravity) of the system. Force couples acting on the system produce a moment and also transport angular momentum across the system boundary. Thinking of the moment of a force as a transport rate of linear momentum, the dimensions of a moment are $$[ \text{Force} ][\mathrm{L}] = \left\{ [\mathrm{M}][\mathrm{L}]^{2} / [\mathrm{T}] \right\} / [\mathrm{T}] =[\mathrm{M}] [\mathrm{L}]^{2} [\mathrm{T}]^{-2} .$$

### Angular momentum transport with mass flow

As was shown earlier, every lump of mass with a velocity has angular momentum about a point $$O$$. When mass is allowed to flow across the boundary of an open system, each lump of mass carries with it linear momentum. Thus the angular momentum of an open system can also be changed by mass flow carrying angular momentum across the system boundary.

The rate at which angular momentum about point $$O$$ is transported across the boundary can be represented by the product of the mass flow rate and the local velocity at the boundary assuming that the velocity is uniform: $\dot{\mathbf{L}}_{O, \text { mass}} = \dot{m} (\mathbf{r} \times \mathbf{V}) = \left[\begin{array}{c} \text { Transport rate of } \\ \text { angular momentum } \\ \text { with mass flow } \end{array}\right] \nonumber$ where $$\dot{m}$$ is the mass flow rate at the flow boundary, $$\mathbf{r}$$ is the position of the flow boundary with respect to point $$O$$, and $$\mathbf{V}$$ is the local velocity of the flow. Now combining this for all the flow boundaries, the net transport rate into the system of angular momentum about point $$O$$ can be obtained by summing up the transports at all flow boundaries: $\mathbf{L}_{O, \text { mass, net}} = \sum_{in} \left( \mathbf{r}_{i} \times \mathbf{V}_{i}\right) \dot{m}_{i} - \sum_{out} \left(\mathbf{r}_{e} \times \mathbf{V}_{e}\right) \dot{m}_{e} \nonumber$

## 6.2.4 How can angular momentum be generated or consumed in a system?

Experience has shown that the angular momentum of a system cannot be created or destroyed; thus angular momentum is conserved.

## 6.2.5 Putting It All Together

Using the accounting framework, we can develop the following statement for conservation of angular momentum about point $$O$$: $\left[ \begin{array}{c} \text { Rate of accumulation } \\ \text { of } \\ \text { angular momentum } \\ \text { inside a system } \\ \text { at time } t \end{array}\right] = \left[\begin{array}{c} \text { Net transport rate of } \\ \text { angular momentum } \\ \text { into the system } \\ \text { by external forces } \\ \text { at time } t \end{array}\right] + \left[ \begin{array}{c} \text { Net transport rate of } \\ \text { angular momentum } \\ \text { into the system } \\ \text { by mass flow } \\ \text { at time } t \end{array}\right] \nonumber$

In symbols the rate-form of the conservation of angular momentum about point $$O$$ becomes $\frac{d \mathbf{L}_{O, \text { sys}}}{d t} = \sum_{j} \mathbf{M}_{O, \ j} + \sum_{in} \left(\mathbf{r}_{i} \times \mathbf{V}_{i}\right) \dot{m}_{i} - \sum_{out} \left(\mathbf{r}_{e} \times \mathbf{V}_{e}\right) \dot{m}_{e} \nonumber$ Again, as with linear momentum and mass, the judicious use of modeling assumptions will often simplify this equation as we model systems.

This page titled 6.2: Four Questions is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Donald E. Richards (Rose-Hulman Scholar) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.