Exercises — Angular momentum L = Iω (fixed axis), L = r × p (general)

• Single particle: $\vec{L}=\vec{r}\times \vec{p}$, magnitude $L=rp\sin \theta =p{r}_{\perp }$.
• Rigid body, fixed/symmetry axis: $L=I\omega$, with $I=\sum m_i{r}_i^2$.
• No external torque ⇒ $L$ conserved: $I_1{\omega }_1=I_2{\omega }_2$.

WHAT we ask: $L=rp\sin \theta$. WHY: only the "going-around" part of the motion counts. The velocity points along the line joining the ball to $O$, so $\vec{r}$ and $\vec{p}$ are parallel: $\theta =0^{\circ }$, hence $\sin \theta =0$. $L=rp\sin 0^{\circ }=0.$ Meaning: motion straight toward (or away from) the origin has zero lever arm $r_{\perp }=r\sin \theta =0$ — nothing swirls around $O$.

WHY $L=I\omega$? Rigid body, fixed symmetry axis — the exact case where the sum $\sum m_i{r}_i^2\omega$ collapses to $I\omega$. $L=I\omega =(0.5)(10)=5\mathrm{kg}{\mathrm{m}}^2/s.$

Step 1 — momentum: $\vec{p}=m\vec{v}=(0,1,0)\mathrm{kg}m/s$. Step 2 — cross product (see Cross Product). For $\vec{r}=(r_x,r_y,0)$ and $\vec{p}=(p_x,p_y,0)$ only the $z$-component survives: $L_z=r_x{p}_y-r_y{p}_x=(3)(1)-(4)(0)=3.$ So $\vec{L}=(0,0,3)\mathrm{kg}{\mathrm{m}}^2/s$, magnitude $|\vec{L}|=3\mathrm{kg}{\mathrm{m}}^2/s$, pointing out of the page (+$z$) by the right-hand rule. Check with lever arm: $r=\sqrt{3^2+4^2}=5$, $p=1$, and $\sin \theta$ works out so that $r_{\perp }p=3$. Consistent. ✓

Step 1 — $I$ about the chosen axis (see Moment of Inertia): $I=\frac{1}{3}M{\ell }^2=\frac{1}{3}(3)(2^2)=4\mathrm{kg}{\mathrm{m}}^2.$ Step 2 — apply $L=I\omega$: $L=I\omega =(4)(4)=16\mathrm{kg}{\mathrm{m}}^2/s.$ Why the end axis matters: the same rod about its centre has $I=\frac{1}{12}M{\ell }^2=1\mathrm{kg}{\mathrm{m}}^2$ — four times smaller. The axis choice changes $I$, hence $L$.

(a) Conserve $L$ (no external torque — internal muscular pull only, see Conservation of Angular Momentum): $I_1{\omega }_1=I_2{\omega }_2\Rightarrow {\omega }_2=\frac{(6)(3)}{2}=9rad/s.$ (b) Using $KE=\frac{1}{2}I{\omega }^2$ (see Rotational Kinetic Energy): $K{E}_1=\frac{1}{2}(6)(3^2)=27\mathrm{J},K{E}_2=\frac{1}{2}(2)(9^2)=81\mathrm{J}.$ (c) The extra $54\mathrm{J}$ came from the skater's muscles doing work pulling the mass inward against the outward "centrifugal" tendency. $L$ is conserved; $KE$ is not — energy was added internally.

The tool: ${\vec{\tau }}_{\text{ext}}=d\vec{L}/dt$ (differentiate $\vec{L}=\vec{r}\times \vec{p}$; see Torque). No force ⇒ $\vec{F}=0$ ⇒ torque $\vec{\tau }=\vec{r}\times \vec{F}=0$ ⇒ $d\vec{L}/dt=0$, so $\vec{L}$ is constant. Geometric picture: $L=p{r}_{\perp }$. As the ball glides, $\vec{r}$ swings and lengthens, but the perpendicular distance $r_{\perp }$ from $O$ to the (fixed) line of motion never changes, and $p$ is constant. Two constants multiplied ⇒ constant $L$. The lever arm, not $r$, is what nature holds fixed here.

(a) Gravity is a central force: $\vec{F}$ is antiparallel to $\vec{r}$, so $\vec{\tau }=\vec{r}\times \vec{F}=0$. Hence $d\vec{L}/dt=0$ and $\vec{L}$ is conserved (this is Kepler's Second Law — equal areas in equal times). (b) At perihelion and aphelion the velocity is perpendicular to $\vec{r}$, so $\theta =90^{\circ }$ and $L=mvr$. Conserve $L$ (mass cancels): $m{v}_p{r}_p=m{v}_a{r}_a\Rightarrow v_a=\frac{v_p{r}_p}{r_a}=\frac{(6.0\times 10^4)(1.0\times 10^{11})}{1.5\times 10^{11}}=4.0\times 10^{4}m/s.$ Meaning: farther out ⇒ slower, exactly matching "sweeps equal areas in equal times."

(a) The only forces between them are internal (friction at contact) ⇒ no external torque about the common axis ⇒ $L$ conserved: $I_A{\omega }_A=(I_A+I_B){\omega }_f\Rightarrow {\omega }_f=\frac{(2)(12)}{2+4}=4rad/s.$ (b) $K{E}_i=\frac{1}{2}{I}_A{\omega }_A^2=\frac{1}{2}(2)(144)=144\mathrm{J}.$ $K{E}_f=\frac{1}{2}(6)(4^2)=48\mathrm{J}.$ Fraction lost $=\frac{144-48}{144}=\frac{96}{144}=\frac{2}{3}\approx 0.667$. Meaning: like a perfectly inelastic linear collision — $L$ (the rotational "momentum") is conserved, but sliding friction burns $\frac{2}{3}$ of the energy as heat.

(a) Point mass at radius $R$: $I_{\text{beetle}}=m{R}^2=(0.10)(0.50^2)=0.025\mathrm{kg}{\mathrm{m}}^2$. (b) Conserve $L$ (beetle's legs push the disc but that's an internal torque; nothing external twists the system):

• Initial: beetle at centre contributes $0$, so $L_i=I_d{\omega }_0=(0.20)(6.0)=1.2\mathrm{kg}{\mathrm{m}}^2/s$.
• Final: $L_f=(I_d+m{R}^2){\omega }_1=(0.225){\omega }_1$. ${\omega }_1=\frac{1.2}{0.225}=5.333rad/s\approx 5.33rad/s.$ (c) $K{E}_i=\frac{1}{2}(0.20)(6.0^2)=3.6\mathrm{J}$; $K{E}_f=\frac{1}{2}(0.225)(5.333^2)=3.2\mathrm{J}$. Energy falls by $0.4\mathrm{J}$: the beetle's muscles absorb net negative work (it must brake against the rim's motion as it moves outward), so $KE$ decreases even though nothing external acted.

(a) $\vec{p}=(1,3,0)$; $L_z=r_x{p}_y-r_y{p}_x=(2)(3)-(0)(1)=6$. So $\vec{L}=(0,0,6)\mathrm{kg}{\mathrm{m}}^2/s$. (b) $\vec{r}$ points along $+x$, so the radial velocity is $v_x=1m/s$ (straight out — contributes nothing, $\theta =0$). The tangential velocity is $v_y=3m/s$, perpendicular to $\vec{r}$. Only it swirls: $L=r(m{v}_y)=2\times (1\times 3)=6\mathrm{kg}{\mathrm{m}}^2/s.✓$ The radial $v_x$ moves the particle toward/away from $O$ — zero lever arm, zero $L$. (c) $L=I\omega$ needs a rigid body about a fixed axis with a well-defined single $\omega$. A lone particle drifting on a straight/general path has no fixed rotation axis and no rigid $I$ — you must use the general $\vec{L}=\vec{r}\times \vec{p}$.

• Straight-at-origin motion has $L=0$ ::: because $\theta =0\Rightarrow r_{\perp }=0$
• Central force ⇒ $L$ conserved ::: torque $\vec{r}\times \vec{F}=0$ when $\vec{F}\parallel \vec{r}$
• Sticking/inelastic rotation ::: conserve $L$, expect $KE$ loss
• Person walks outward on turntable ::: $\omega$ drops (larger $I$), $KE$ changes (they do work)
• $L=I\omega$ requires ::: rigid body, fixed/symmetry axis — else use $\vec{r}\times \vec{p}$