Lesson 1 · 30 min

Angular Momentum of a Particle

Linear momentum \(m\vvec\) tells you how hard it is to stop a moving particle. Angular momentum tells you how much "turning" that motion carries about a point: it is the moment of the linear momentum, just as a moment is the turning effect of a force.

Learning objectives

Define the angular momentum of a particle about a point, \(\Hvec_O = \rvec \times m\vvec\), and give its units.
Compute it with the scalar form \(H_O = m v d\) and with components in 2D and 3D.
Use the polar form \(H_O = m r^2\dot\theta\) for motion described by \(r\) and \(\theta\).
Explain why angular momentum depends on the point about which it is taken.

The moment of momentum

A particle of mass \(m\) at position \(\rvec\) from a point \(O\) moves with velocity \(\vvec\). Its linear momentum is \(\Lvec = m\vvec\). Take the moment of that vector about \(O\), exactly as you would take the moment of a force:

Angular momentum of a particle about \(O\)

\[ \Hvec_O = \rvec \times m\vvec \]

Units: \(\text{kg·m}^2/\text{s}\) (equivalently N·m·s). \(\Hvec_O\) is perpendicular to the plane that contains \(\rvec\) and \(\vvec\), in the direction given by the right-hand rule.

The moment analogy runs all the way through the subject: a force \(\Fvec\) has a moment \(\Mvec_O = \rvec \times \Fvec\), and the momentum \(m\vvec\) has a moment \(\Hvec_O = \rvec \times m\vvec\). Lesson 2 shows that the first is the rate of change of the second.

Scalar form

For motion in a plane, \(\Hvec_O\) points straight out of or into the plane, so a single signed number describes it. Its magnitude is the momentum times its moment arm, the perpendicular distance \(d\) from \(O\) to the line along which the particle is moving:

\[ H_O = m v d \qquad \text{(counterclockwise positive)} \]

Components

With \(\rvec = x\ihat + y\jhat + z\khat\) and \(\vvec = v_x\ihat + v_y\jhat + v_z\khat\), the cross product gives

\[ \Hvec_O = m\begin{vmatrix} \ihat & \jhat & \khat \\ x & y & z \\ v_x & v_y & v_z \end{vmatrix} = m\left[(y v_z - z v_y)\,\ihat + (z v_x - x v_z)\,\jhat + (x v_y - y v_x)\,\khat\right] \]

In the \(xy\)-plane only the last term survives: \(H_O = m(x v_y - y v_x)\).

Figure 1.1 A \(2\ \text{kg}\) particle coasts in a straight line at constant velocity. Drag the point \(\colM{O}\): the dashed moment arm \(d\) and the angular momentum \(H_O = m v d\) change with the point, but for any one point they stay constant as the particle moves, even though \(\rvec\) keeps changing. The shaded triangles are swept by \(\rvec\) in equal times; they all have the same area, \(\tfrac12 v d\) per second.

Example 1.1 — Two ways in the plane

A \(2\ \text{kg}\) particle is at \((3,\ 4)\ \text{m}\) and moves with \(\vvec = -2\,\ihat + 5\,\jhat\ \text{m/s}\). Find its angular momentum about the origin.

Show solution

Components.

\[ H_O = m(x v_y - y v_x) = 2\left[3(5) - 4(-2)\right] = 46\ \text{kg·m}^2/\text{s} \]

Positive, so counterclockwise: \(\Hvec_O = 46\,\khat\ \text{kg·m}^2/\text{s}\).

Moment arm. The speed is \(v = \sqrt{29} = 5.385\ \text{m/s}\). The distance from \(O\) to the line of motion is \(d = |x v_y - y v_x|/v = 23/5.385 = 4.271\ \text{m}\), so \(H_O = m v d = 2(5.385)(4.271) = 46\ \text{kg·m}^2/\text{s}\). The component form finds the moment arm for you.

Angular momentum in three dimensions

In general \(\rvec\) and \(\vvec\) are not in a coordinate plane, and \(\Hvec_O\) has all three components. It is always perpendicular to both \(\rvec\) and \(\vvec\).

Figure 1.2 \(\Hvec_O = \rvec \times m\vvec\) in 3D. Steer the velocity with the sliders: the red arrow \(\colH{\Hvec_O}\) stays perpendicular to the shaded plane of \(\rvec\) and \(\colV{m\vvec}\), and its length is proportional to the area of the parallelogram they span. Aim \(\vvec\) along \(\rvec\) and \(\Hvec_O\) vanishes.

Example 1.2 — A particle in space

A \(0.5\ \text{kg}\) particle is at \(\rvec = (1,\ 2,\ -1)\ \text{m}\) with \(\vvec = (3,\ -1,\ 2)\ \text{m/s}\). Find \(\Hvec_O\) and its magnitude.

Show solution

\[ \rvec \times \vvec = \begin{vmatrix} \ihat & \jhat & \khat \\ 1 & 2 & -1 \\ 3 & -1 & 2 \end{vmatrix} = (4 - 1)\,\ihat - (2 + 3)\,\jhat + (-1 - 6)\,\khat = 3\,\ihat - 5\,\jhat - 7\,\khat \] \[ \Hvec_O = 0.5(3\,\ihat - 5\,\jhat - 7\,\khat) = 1.5\,\ihat - 2.5\,\jhat - 3.5\,\khat\ \text{kg·m}^2/\text{s}, \qquad |\Hvec_O| = \sqrt{20.75} = 4.555\ \text{kg·m}^2/\text{s} \]

Check: \(\Hvec_O\cdot\rvec = 1.5 - 5 + 3.5 = 0\), so \(\Hvec_O\) is perpendicular to \(\rvec\), as it must be.

Polar form

When a particle moves in a plane and is described by polar coordinates \((r, \theta)\) about \(O\), its velocity has a radial part \(v_r = \dot r\) and a transverse part \(v_\theta = r\dot\theta\). The radial part points straight through \(O\) and has no moment. Only the transverse part counts, with moment arm \(r\):

Angular momentum in polar coordinates

\[ H_O = m\,r\,v_\theta = m\,r^2\dot\theta \]

Example 1.3 — A bead on a rotating arm

A \(1.5\ \text{kg}\) collar slides outward along a rotating arm. At an instant it is \(0.5\ \text{m}\) from the pivot, moving out along the arm at \(2\ \text{m/s}\), while the arm turns at \(4\ \text{rad/s}\). Find its angular momentum about the pivot.

Show solution

\[ H_O = m r^2\dot\theta = 1.5(0.5)^2(4) = 1.5\ \text{kg·m}^2/\text{s} \]

The outward speed \(\dot r = 2\ \text{m/s}\) does not appear: that part of the velocity points straight away from the pivot.

Check your understanding

Key takeaways

\(\Hvec_O = \rvec \times m\vvec\): the moment of the linear momentum about \(O\), in kg·m²/s.
In a plane: \(H_O = m v d = m(x v_y - y v_x)\), counterclockwise positive.
In polar coordinates: \(H_O = m r^2\dot\theta\); radial motion contributes nothing.
\(\Hvec\) depends on the point. A particle coasting in a straight line has a constant \(\Hvec\) about any fixed point.
Next: Lesson 2 connects \(\Hvec\) to the moments of the forces.