Lesson 2 · 35 min

Angular Impulse and Momentum

Newton's second law says forces change linear momentum. Take moments of both sides and you get its rotational twin: moments change angular momentum. When the moments vanish, as they do for any force aimed at a fixed center, angular momentum about that center is conserved.

Learning objectives

Derive \(\sum\Mvec_O = \dot{\Hvec}_O\) for a particle from Newton's second law.
Apply the angular impulse–momentum principle \(\Hvec_{O1} + \sum\int\Mvec_O\,dt = \Hvec_{O2}\).
Recognise when angular momentum is conserved, and use it for central-force motion.
Explain why kinetic energy can change while angular momentum is conserved.

Moments change angular momentum

Differentiate \(\Hvec_O = \rvec \times m\vvec\) with respect to time, with \(O\) a fixed point:

\[ \dot{\Hvec}_O = \dot\rvec \times m\vvec + \rvec \times m\dot\vvec = \underbrace{\vvec \times m\vvec}_{=\,\mathbf{0}} + \rvec \times m\mathbf{a} = \rvec \times \sum\Fvec \]

Moment equation for a particle (\(O\) fixed)

\[ \sum\Mvec_O = \dot{\Hvec}_O \]

The resultant moment about a fixed point equals the rate of change of angular momentum about the same point. It is \(\sum\Fvec = \dot{\Lvec}\) with moments taken on both sides.

Angular impulse

Integrate over a time interval. The time integral of a moment is the angular impulse, in N·m·s:

Angular impulse–momentum principle

\[ \Hvec_{O1} + \sum\int_{t_1}^{t_2}\Mvec_O\,dt = \Hvec_{O2} \]

In a plane, with counterclockwise positive: \((H_O)_1 + \sum\int M_O\,dt = (H_O)_2\).

Example 2.1 — Spinning up a sphere on a light arm

A small \(2\ \text{kg}\) sphere is fixed to the end of a light rod \(0.8\ \text{m}\) long that turns in a horizontal plane about a vertical axis at its other end. Starting from rest, a motor applies a moment \(M = 3t\ \text{N·m}\) (\(t\) in seconds). Find the sphere's speed after \(4\ \text{s}\).

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The only moment about the axis is the motor's (gravity and the axis reaction have none about it). The angular momentum is \(H = m r v\).

\[ 0 + \int_0^4 3t\,dt = m r v_2 \quad\Rightarrow\quad 24 = 2(0.8)\,v_2 \quad\Rightarrow\quad v_2 = 15.0\ \text{m/s} \]

The arm then turns at \(\omega = v_2/r = 18.75\ \text{rad/s}\).

Conservation of angular momentum

If the forces on a particle have no net moment about a fixed point \(O\) over an interval, the angular impulse is zero and

Conservation of angular momentum

\[ \Hvec_{O1} = \Hvec_{O2} \qquad \text{when } \sum\int\Mvec_O\,dt = \mathbf{0} \]

The most important case is a central force, one that always points toward (or away from) the same fixed point: the tension in a cord through a hole, gravity from a planet, the spring in a centrifugal governor. Its line of action passes through \(O\), so its moment about \(O\) is zero at every instant, even though the force itself may be large and changing. With \(H_O = m r v_\theta\):

\[ r_1\,(v_\theta)_1 = r_2\,(v_\theta)_2 \]

Figure 2.1 Top view of a \(0.4\ \text{kg}\) puck on a smooth table, tied to a cord that runs down through a hole \(O\). Pull the cord in at a steady rate: the cord's tension always points at \(O\), so \(H_O = m r v_\theta\) stays constant while the puck spirals in faster and faster. Its kinetic energy grows, because the pull does work.

Example 2.2 — Pulling in the cord

The puck of Figure 2.1 moves in a circle of radius \(0.6\ \text{m}\) at \(2\ \text{m/s}\). The cord is slowly pulled until the radius is \(0.2\ \text{m}\). Find the puck's new speed, the work done by the pull, and the cord tension at the end.

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The tension passes through \(O\), so \(H_O\) is conserved (and a slow pull adds almost no radial speed):

\[ m r_1 v_1 = m r_2 v_2 \quad\Rightarrow\quad v_2 = \frac{0.6}{0.2}(2) = 6.00\ \text{m/s} \] \[ T_1 = \tfrac12(0.4)(2)^2 = 0.80\ \text{J}, \quad T_2 = \tfrac12(0.4)(6)^2 = 7.20\ \text{J}, \quad U_\text{pull} = T_2 - T_1 = 6.40\ \text{J} \]

At the end the tension supplies the centripetal force: \(F_2 = m v_2^2/r_2 = 0.4(36)/0.2 = 72.0\ \text{N}\), against \(F_1 = 0.4(4)/0.6 = 2.67\ \text{N}\) at the start: 27 times larger, because \(F = H_O^2/(m r^3)\) at constant \(H_O\).

Orbits

Gravity on a satellite always points at the planet's center, so \(H_O\) about that center is constant. At perigee and apogee the velocity is perpendicular to \(\rvec\), and \(r_p v_p = r_a v_a\). In between, the area swept per unit time is \(\tfrac12 r v_\theta = H_O/2m\), the same all the way round: Kepler's second law.

Figure 2.2 An Earth orbit with a semi-major axis of \(12\,000\ \text{km}\). The satellite sweeps the shaded sectors in equal times, one twelfth of a period each: thin and long near apogee, short and wide near perigee, all of the same area. Raise the eccentricity and compare the speeds at perigee and apogee: their ratio is \(r_a/r_p\).

Example 2.3 — Speed at apogee

A satellite's orbit has its perigee \(7000\ \text{km}\) and its apogee \(12\,000\ \text{km}\) from the Earth's center. The energy equation gives its perigee speed as \(8.481\ \text{km/s}\). Find its speed at apogee.

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\[ r_p v_p = r_a v_a \quad\Rightarrow\quad v_a = \frac{7000}{12\,000}(8.481) = 4.947\ \text{km/s} \]

The mass of the satellite cancels, and no information about gravity is needed: conservation of angular momentum alone links the two points.

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Key takeaways

\(\sum\Mvec_O = \dot{\Hvec}_O\) about a fixed point; integrated, \(\Hvec_{O1} + \sum\int\Mvec_O\,dt = \Hvec_{O2}\).
A central force has no moment about its center, so \(H_O\) is conserved: \(r_1 v_{\theta 1} = r_2 v_{\theta 2}\).
Conservation of \(H\) does not mean conservation of kinetic energy.
Orbits sweep equal areas in equal times because \(H_O\) is constant.
Next: Lesson 3 extends the moment equation to systems of particles.