Example

Angular Momentum and Torque on a Meteor

A meteor enters Earth’s atmosphere ((Figure)) and is observed by someone on the ground before it burns up in the atmosphere. The vector $$ \overset{\to }{r}=25\,\text{km}\hat{i}+25\,\text{km}\hat{j} $$ gives the position of the meteor with respect to the observer. At the instant the observer sees the meteor, it has linear momentum $$ \overset{\to }{p}=15.0\,\text{kg}(-2.0\text{km}\text{/}\text{s}\hat{j})$$, and it is accelerating at a constant $$ 2.0\,\text{m}\text{/}{\text{s}}^{2}(\text{−}\hat{j}) $$ along its path, which for our purposes can be taken as a straight line. (a) What is the angular momentum of the meteor about the origin, which is at the location of the observer? (b) What is the torque on the meteor about the origin?

Figure 11.10 An observer on the ground sees a meteor at position $$ \overset{\to }{r} $$ with linear momentum $$ \overset{\to }{p}$$.

Strategy

We resolve the acceleration into x– and y-components and use the kinematic equations to express the velocity as a function of acceleration and time. We insert these expressions into the linear momentum and then calculate the angular momentum using the cross-product. Since the position and momentum vectors are in the xy-plane, we expect the angular momentum vector to be along the z-axis. To find the torque, we take the time derivative of the angular momentum.

Solution

The meteor is entering Earth’s atmosphere at an angle of $$ 90.0\text{°} $$ below the horizontal, so the components of the acceleration in the x– and y-directions are

We write the velocities using the kinematic equations.

1. The angular momentum is
   $$\begin{array}{cc}\hfill \overset{\to }{l}& =\overset{\to }{r}\,×\,\overset{\to }{p}=(25.0\,\text{km}\hat{i}+25.0\,\text{km}\hat{j})\,×\,15.0\,\text{kg}(0\hat{i}+{v}_{y}\hat{j})\hfill \\ & =15.0\,\text{kg}[25.0\,\text{km}({v}_{y})\hat{k}]\hfill \\ & =15.0\,\text{kg[}2.50\,×\,{10}^{4}\,\text{m}(-2.0\,×\,{10}^{3}\,\text{m}\text{/}\text{s}-(2.0\,\text{m}\text{/}{\text{s}}^{2})t)\hat{k}].\hfill \end{array}$$

   At $$ t=0$$, the angular momentum of the meteor about the origin is

   $${\overset{\to }{l}}_{0}=15.0\,\text{kg}[2.50\,×\,{10}^{4}\,\text{m}(-2.0\,×\,{10}^{3}\,\text{m}\text{/}\text{s})\hat{k}]=7.50\,×\,{10}^{8}\,\text{kg}·{\text{m}}^{2}\text{/}\text{s}(\text{−}\hat{k}).$$

   This is the instant that the observer sees the meteor.

2. To find the torque, we take the time derivative of the angular momentum. Taking the time derivative of $$ \overset{\to }{l} $$ as a function of time, which is the second equation immediately above, we have
   $$\frac{d\overset{\to }{l}}{dt}=-15.0\,\text{kg}(2.50\,×\,{10}^{4}\,\text{m})(2.0\,\text{m}\text{/}{\text{s}}^{2})\hat{k}.$$

   Then, since $$ \frac{d\overset{\to }{l}}{dt}=\sum \overset{\to }{\tau }$$, we have

   $$\sum \overset{\to }{\tau }=-7.5\,×\,{10}^{5}\text{N}·\text{m}\hat{k}.$$

   The units of torque are given as newton-meters, not to be confused with joules. As a check, we note that the lever arm is the x-component of the vector $$ \overset{\to }{\text{r}} $$ in (Figure) since it is perpendicular to the force acting on the meteor, which is along its path. By Newton’s second law, this force is

   $$\overset{\to }{F}=ma(\text{−}\hat{j})=15.0\,\text{kg}(2.0\,\text{m}\text{/}{\text{s}}^{2})(\text{−}\hat{j})=30.0\,\text{kg}·\text{m}\text{/}{\text{s}}^{2}(\text{−}\hat{j}).$$

   The lever arm is

   $${\overset{\to }{r}}_{\perp }=2.5\,×\,{10}^{4}\,\text{m}\,\hat{i}.$$

   Thus, the torque is

   $$\begin{array}{cc}\hfill \sum \overset{\to }{\tau }={\overset{\to }{r}}_{\perp }\,×\,\overset{\to }{F}& =(2.5\,×\,{10}^{4}\,\text{m}\,\hat{i})\,×\,(-30.0\,\text{kg}·\text{m}\text{/}{\text{s}}^{2}\hat{j}),\hfill \\ & =7.5\,×\,{10}^{5}\,\text{N}·\text{m}(\text{−}\hat{k}).\hfill \end{array}$$

Significance

Since the meteor is accelerating downward toward Earth, its radius and velocity vector are changing. Therefore, since $$ \overset{\to }{l}=\overset{\to }{r}\,×\,\overset{\to }{p}$$, the angular momentum is changing as a function of time. The torque on the meteor about the origin, however, is constant, because the lever arm $$ {\overset{\to }{r}}_{\perp } $$ and the force on the meteor are constants. This example is important in that it illustrates that the angular momentum depends on the choice of origin about which it is calculated. The methods used in this example are also important in developing angular momentum for a system of particles and for a rigid body.

Example

Angular Momentum of Three Particles

Referring to (Figure)(a), determine the total angular momentum due to the three particles about the origin. (b) What is the rate of change of the angular momentum?

Figure 11.11 Three particles in the xy-plane with different position and momentum vectors.

Strategy

Write down the position and momentum vectors for the three particles. Calculate the individual angular momenta and add them as vectors to find the total angular momentum. Then do the same for the torques.

Solution

1. Particle 1: $$ {\overset{\to }{r}}_{1}=-2.0\,\text{m}\hat{i}+1.0\,\text{m}\hat{j},\enspace{\overset{\to }{p}}_{1}=2.0\,\text{kg}(4.0\,\text{m}\text{/}\text{s}\hat{j})=8.0\,\text{kg}·\text{m}\text{/}\text{s}\hat{j},$$
   $${\overset{\to }{l}}_{1}={\overset{\to }{r}}_{1}\,×\,{\overset{\to }{p}}_{1}=-16.0\,\text{kg}·{\text{m}}^{2}\text{/}\text{s}\hat{k}.$$

   Particle 2: $$ {\overset{\to }{r}}_{2}=4.0\,\text{m}\hat{i}+1.0\,\text{m}\hat{j},\enspace{\overset{\to }{p}}_{2}=4.0\,\text{kg}(5.0\,\text{m}\text{/}\text{s}\hat{i})=20.0\,\text{kg}·\text{m}\text{/}\text{s}\hat{i}$$,

   $${\overset{\to }{l}}_{2}={\overset{\to }{r}}_{2}\,×\,{\overset{\to }{p}}_{2}=-20.0\,\text{kg}·{\text{m}}^{2}\text{/}\text{s}\hat{k}.$$

   Particle 3: $$ {\overset{\to }{r}}_{3}=2.0\,\text{m}\hat{i}-2.0\,\text{m}\hat{j},\enspace{\overset{\to }{p}}_{3}=1.0\,\text{kg}(3.0\,\text{m}\text{/}\text{s}\hat{i})=3.0\,\text{kg}·\text{m}\text{/}\text{s}\hat{i}$$,

   $${\overset{\to }{l}}_{3}={\overset{\to }{r}}_{3}\,×\,{\overset{\to }{p}}_{3}=6.0\,\text{kg}·{\text{m}}^{2}\text{/}\text{s}\hat{k}.$$

   We add the individual angular momenta to find the total about the origin:

   $${\overset{\to }{l}}_{T}={\overset{\to }{l}}_{1}+{\overset{\to }{l}}_{2}+{\overset{\to }{l}}_{3}=-30\,\text{kg}·{\text{m}}^{2}\text{/}\text{s}\hat{k}.$$
2. The individual forces and lever arms are
   $$\begin{array}{c}{\overset{\to }{r}}_{1\perp }=1.0\,\text{m}\hat{j},\enspace{\overset{\to }{F}}_{1}=-6.0\,\text{N}\hat{i},\enspace{\overset{\to }{\tau }}_{1}=6.0\text{N}·\text{m}\hat{k}\hfill \\ {\overset{\to }{r}}_{2\perp }=4.0\,\text{m}\hat{i},\enspace{\overset{\to }{F}}_{2}=10.0\,\text{N}\hat{j},\enspace{\overset{\to }{\tau }}_{2}=40.0\,\text{N}·\text{m}\hat{k}\hfill \\ {\overset{\to }{r}}_{3\perp }=2.0\,\text{m}\hat{i},\enspace{\overset{\to }{F}}_{3}=-8.0\,\text{N}\hat{j},\enspace{\overset{\to }{\tau }}_{3}=-16.0\,\text{N}·\text{m}\hat{k}.\hfill \end{array}$$

   Therefore:

   $$\sum _{i}{\overset{\to }{\tau }}_{i}={\overset{\to }{\tau }}_{1}+{\overset{\to }{\tau }}_{2}+{\overset{\to }{\tau }}_{3}=30\,\text{N}·\text{m}\hat{k}.$$

Significance

This example illustrates the superposition principle for angular momentum and torque of a system of particles. Care must be taken when evaluating the radius vectors $$ {\overset{\to }{r}}_{i} $$ of the particles to calculate the angular momenta, and the lever arms, $$ {\overset{\to }{r}}_{i\perp } $$ to calculate the torques, as they are completely different quantities.

Example

Angular Momentum of a Robot Arm

A robot arm on a Mars rover like Curiosity shown in (Figure) is 1.0 m long and has forceps at the free end to pick up rocks. The mass of the arm is 2.0 kg and the mass of the forceps is 1.0 kg. See (Figure). The robot arm and forceps move from rest to $$ \omega =0.1\pi \,\text{rad}\text{/}\text{s} $$ in 0.1 s. It rotates down and picks up a Mars rock that has mass 1.5 kg. The axis of rotation is the point where the robot arm connects to the rover. (a) What is the angular momentum of the robot arm by itself about the axis of rotation after 0.1 s when the arm has stopped accelerating? (b) What is the angular momentum of the robot arm when it has the Mars rock in its forceps and is rotating upwards? (c) When the arm does not have a rock in the forceps, what is the torque about the point where the arm connects to the rover when it is accelerating from rest to its final angular velocity?

Figure 11.13 A robot arm on a Mars rover swings down and picks up a Mars rock. (credit: modification of work by NASA/JPL-Caltech)

Strategy

We use (Figure) to find angular momentum in the various configurations. When the arm is rotating downward, the right-hand rule gives the angular momentum vector directed out of the page, which we will call the positive z-direction. When the arm is rotating upward, the right-hand rule gives the direction of the angular momentum vector into the page or in the negative z-direction. The moment of inertia is the sum of the individual moments of inertia. The arm can be approximated with a solid rod, and the forceps and Mars rock can be approximated as point masses located at a distance of 1 m from the origin. For part (c), we use Newton’s second law of motion for rotation to find the torque on the robot arm.

Solution

1. Writing down the individual moments of inertia, we haveRobot arm: $$ {I}_{\text{R}}=\frac{1}{3}{m}_{\text{R}}{r}^{2}=\frac{1}{3}(2.00\,\text{kg}){(1.00\,\text{m})}^{2}=\frac{2}{3}\,\text{kg}·{\text{m}}^{2}.$$Forceps: $$ {I}_{\text{F}}={m}_{\text{F}}{r}^{2}=(1.0\,\text{kg}){(1.0\,\text{m})}^{2}=1.0\,\text{kg}·{\text{m}}^{2}.$$Mars rock: $$ {I}_{\text{MR}}={m}_{\text{MR}}{r}^{2}=(1.5\,\text{kg}){(1.0\,\text{m})}^{2}=1.5\,\text{kg}·{\text{m}}^{2}.$$Therefore, without the Mars rock, the total moment of inertia is
   $${I}_{\text{Total}}={I}_{\text{R}}+{I}_{\text{F}}=1.67\,\text{kg}·{\text{m}}^{2}$$

   and the magnitude of the angular momentum is

   $$L=I\omega =1.67\,\text{kg}·{\text{m}}^{2}(0.1\pi \,\text{rad}\text{/}\text{s})=0.17\pi \,\text{kg}·{\text{m}}^{2}\text{/}\text{s}.$$

   The angular momentum vector is directed out of the page in the $$ \hat{k} $$ direction since the robot arm is rotating counterclockwise.

2. We must include the Mars rock in the calculation of the moment of inertia, so we have
   $${I}_{\text{Total}}={I}_{\text{R}}+{I}_{\text{F}}+{I}_{\text{MR}}=3.17\,\text{kg}·{\text{m}}^{2}$$

   and

   $$L=I\omega =3.17\,\text{kg}·{\text{m}}^{2}(0.1\pi \,\text{rad}\text{/}\text{s})=0.32\pi \,\text{kg}·{\text{m}}^{2}\text{/}\text{s}\text{.}$$

   Now the angular momentum vector is directed into the page in the $$ \text{−}\hat{k} $$ direction, by the right-hand rule, since the robot arm is now rotating clockwise.

3. We find the torque when the arm does not have the rock by taking the derivative of the angular momentum using (Figure) $$ \frac{d\overset{\to }{L}}{dt}=\sum \overset{\to }{\tau }. $$ But since $$ L=I\omega $$, and understanding that the direction of the angular momentum and torque vectors are along the axis of rotation, we can suppress the vector notation and find
   $$\frac{dL}{dt}=\frac{d(I\omega )}{dt}=I\frac{d\omega }{dt}=I\alpha =\sum \tau ,$$

   which is Newton’s second law for rotation. Since $$ \alpha =\frac{0.1\pi \,\text{rad}\text{/}\text{s}}{0.1\,\text{s}}=\pi \,\text{rad}\text{/}{\text{s}}^{2}$$, we can calculate the net torque:

   $$\sum \tau =I\alpha =1.67\,\text{kg}·{\text{m}}^{2}(\pi \,\text{rad}\text{/}{\text{s}}^{2})=1.67\pi \,\text{N}·\text{m}.$$

Significance

The angular momentum in (a) is less than that of (b) due to the fact that the moment of inertia in (b) is greater than (a), while the angular velocity is the same.