17  Momenta and Impulses of Particles

17.1 Linear Momentum and Its Conservation

Recall the linear momentum \({\bf G} = m{\bf v}\).

17.1.1 Linear Impulse and Linear Momentum

The integral form of the balance of linear momentum: \[\begin{align} {\bf G}(t_1)-{\bf G}(t_0) = \int_{t_0}^{t_1}{\bf F}\,dt. \end{align}\]

The time integral of a force is its linear impulse. This form is more general than \({\bf F}=m{\bf a}\) because it does not require \({\bf v}\) to be differentiable.

17.1.2 Conservation of Linear Momentum

TipThink!

Question: Under what conditions is the linear momentum of a system conserved in a direction \({\bf c}\)?

NoteAnswer

\[\begin{align} \frac{d}{dt}({\bf G}\cdot{\bf c}) = {\bf F}\cdot{\bf c}+{\bf G}\cdot\dot{\bf c} = 0. \end{align}\] For a constant \({\bf c}\), \({\bf G}\cdot{\bf c}\) is conserved if and only if \({\bf F}\cdot{\bf c} = 0\) — i.e., no force in that direction.

17.1.3 Example

WarningExample

Question: Is the linear momentum of a projectile conserved? In a certain direction?

NoteAnswer

With \({\bf W} = -mg{\bf E}_y\), linear momentum is conserved in the \({\bf E}_x\) direction (no force there) but not in \({\bf E}_y\).

17.2 The Moment of a Force

TipThink!

Question: How do you calculate the moment of a force about a point?

NoteAnswer

The moment of force \({\bf F}\) applied at \(A\) about point \(P\) is: \[\begin{align} {\bf M}^P = \lp{\bf r}_A-{\bf r}_P\rp\times {\bf F}. \end{align}\]

WarningExample

Question: Given \[\begin{align*} {\bf F} &= F_x{\bf E}_x+F_y{\bf E}_y+F_z{\bf E}_z,\\ {\bf r}_P &= -2{\bf E}_x+3{\bf E}_y+{\bf E}_z,\\ {\bf r}_A &= -{\bf E}_x+2{\bf E}_y-{\bf E}_z, \end{align*}\] calculate \({\bf M}^O\) and \({\bf M}^P\).

NoteAnswer

…

17.3 Angular Momentum and Its Conservation

Let \({\bf r}\) be the position vector of a particle relative to a fixed point \(O\), and \({\bf v}\) its absolute velocity. The angular momentum relative to \(O\): \[\begin{align} {\bf H}^O = {\bf r}\times m{\bf v} = {\bf r}\times{\bf G}. \end{align}\]

17.3.1 Angular Momentum Theorem

From the BoLM: \[\begin{align} \dot{\bf H}^O = \frac{d}{dt}\lp{\bf r}\times m{\bf v}\rp = \underbrace{{\bf v}\times m{\bf v}}_{{\bf 0}}+{\bf r}\times m\dot{\bf v} = {\bf r}\times {\bf F}. \end{align}\] So \(\dot{\bf H}^O = {\bf M}^O\) (the angular momentum theorem).

17.3.2 Conservation of Angular Momentum

TipThink!

Question: Under what conditions is \({\bf H}^O\cdot{\bf c}\) conserved?

NoteAnswer

For a constant \({\bf c}\): if and only if \(\lp{\bf r}\times{\bf F}\rp\cdot{\bf c} = 0\).

17.3.3 Central Force Problems

A central force problem has \({\bf F}\) parallel to \({\bf r}\), so \(\dot{\bf H}^O = {\bf 0}\) and \({\bf H}^O = h{\bf h}\) = constant.

The vectors \({\bf r}\) and \({\bf v}\) remain in a fixed plane through \(O\). Choosing \({\bf E}_z = {\bf h}\): \[\begin{align} {\bf H}^O = h{\bf E}_z = {\bf r}(t_0)\times m{\bf v}(t_0). \end{align}\]

17.3.4 Kepler’s Problem

The gravitational force on a planet of mass \(m\) by the sun of mass \(M\): \[\begin{align} {\bf F} = -\frac{GmM}{\lnorm{\bf r}\rnorm^3}{\bf r}, \qquad U = -\frac{GmM}{\lnorm{\bf r}\rnorm}. \end{align}\]

Two conserved quantities: \[\begin{align} E &= \frac{1}{2}m(\dot{r}^2+r^2\dot{\theta}^2)-\frac{GMm}{r},\\ h &= {\bf H}^O\cdot{\bf E}_z = mr^2\dot{\theta}. \end{align}\]

Watch this video on Kepler’s laws.

17.3.5 Particle on a Smooth Cone

Show that \({\bf H}^O\cdot{\bf E}_z\) is conserved.

17.4 Summary

Linear impulse–momentum: \(\int_{t_A}^{t_B}\mathbf{F}\,dt = \mathbf{G}_B - \mathbf{G}_A\).

Angular momentum about fixed \(O\): \(\mathbf{H}^O = \mathbf{r}\times m\mathbf{v}\).

BoAM: \(\mathbf{M}^O = \dot{\mathbf{H}}^O\).

Conservation of \(\mathbf{G}\) if \(\mathbf{F}=\mathbf{0}\); conservation of \(\mathbf{H}^O\) if \(\mathbf{M}^O=\mathbf{0}\).

17.5 Lecture Videos

17.6 Exercises

The following problems are from Set 12 – Impulse and Momentum.

1. [MKB 03-149] Straightforward application of the linear impulse–momentum equation. (ans. \(v=1.218\) m/s down)

2. [MKB 03-159] The block is subjected to a time-varying force \(P(t)\) shown in the plot; \(P=0\) for \(t>3\) s. (ans. \(t_s=3.69\) s)

3. [MKB 03-161] Use Newton’s third law. Recall \(\int_{x_A}^{x_B}f(x)\,dx = F_{\mathrm{avg}}(x_B-x_A)\). (ans. \(v_f=0.00264\) m/s, \(F_{\mathrm{avg}}=59.5\) N)

4. [03-167] (ans. \(R_x=559\) lb, \(R_y=218\) lb)

5. [03-177] Central force problem – what quantities are conserved? (ans. \(v_P=17\,723\) mi/hr)

6. [03-185] (ans. \(v_B=5.43\) m/s)

7. [03-181] (ans. \(|H|=389\) N·m·s, \(|M|=260\) N·m)

8. [03-192] Label \(r\) as \(R\) and the requested angle \(\theta\) as \(\beta\). The particle moves on a surface of revolution \(z^2+(r-1.15R)^2=R^2\). Use conservation of total energy and \(\mathbf{E}_z\)-component of \(\mathbf{H}^O\). (ans. \(\theta=52.9^\circ\))