At This Instant, What Is the Magnitude of Its Angular Momentum Relative to Point O?

Learning Objectives

By the finish of this section, you lot will be able to:

• Describe the vector nature of angular momentum
• Notice the total angular momentum and torque nearly a designated origin of a system of particles
• Calculate the angular momentum of a rigid body rotating virtually a fixed axis
• Calculate the torque on a rigid body rotating near a stock-still centrality
• Use conservation of angular momentum in the analysis of objects that change their rotation rate

Why does Globe keep on spinning? What started information technology spinning to begin with? Why doesn't Earth's gravitational attraction non bring the Moon crashing in toward Earth? And how does an ice skater manage to spin faster and faster only by pulling her artillery in? Why does she not take to exert a torque to spin faster?

The answer is in a new conserved quantity, since all of these scenarios are in closed systems. This new quantity, angular momentum, is analogous to linear momentum. In this chapter, we start ascertain and then explore athwart momentum from a variety of viewpoints. Offset, however, nosotros investigate the angular momentum of a single particle. This allows u.s. to develop angular momentum for a arrangement of particles and for a rigid body that is cylindrically symmetric.

Angular Momentum of a Unmarried Particle

(Figure) shows a particle at a position [latex] \overset{\to }{r} [/latex] with linear momentum [latex] \overset{\to }{p}=m\overset{\to }{v} [/latex] with respect to the origin. Even if the particle is not rotating about the origin, we can still define an angular momentum in terms of the position vector and the linear momentum.

Angular Momentum of a Particle

The angular momentum [latex] \overset{\to }{l} [/latex] of a particle is divers equally the cantankerous-product of [latex] \overset{\to }{r} [/latex] and [latex] \overset{\to }{p} [/latex], and is perpendicular to the plane containing [latex] \overset{\to }{r} [/latex] and [latex] \overset{\to }{p}: [/latex]

[latex] \overset{\to }{l}=\overset{\to }{r}\,×\,\overset{\to }{p}. [/latex]

Figure 11.9 In three-dimensional space, the position vector [latex] \overset{\to }{r} [/latex] locates a particle in the xy-aeroplane with linear momentum [latex] \overset{\to }{p} [/latex]. The angular momentum with respect to the origin is [latex] \overset{\to }{50}=\overset{\to }{r}\,×\,\overset{\to }{p} [/latex], which is in the z-direction. The direction of [latex] \overset{\to }{l} [/latex] is given by the correct-hand dominion, equally shown.

The intent of choosing the direction of the angular momentum to be perpendicular to the aeroplane containing [latex] \overset{\to }{r} [/latex] and [latex] \overset{\to }{p} [/latex] is similar to choosing the direction of torque to be perpendicular to the plane of [latex] \overset{\to }{r}\,\text{and}\,\overset{\to }{F}, [/latex] as discussed in Fixed-Axis Rotation. The magnitude of the angular momentum is found from the definition of the cross-product,

[latex] l=rp\,\text{sin}\,\theta , [/latex]

where [latex] \theta [/latex] is the angle between [latex] \overset{\to }{r} [/latex] and [latex] \overset{\to }{p}. [/latex] The units of angular momentum are [latex] \text{kg}·{\text{m}}^{2}\text{/}\text{south} [/latex].

Every bit with the definition of torque, we tin can define a lever arm [latex] {r}_{\perp } [/latex] that is the perpendicular distance from the momentum vector [latex] \overset{\to }{p} [/latex] to the origin, [latex] {r}_{\perp }=r\,\text{sin}\,\theta . [/latex] With this definition, the magnitude of the athwart momentum becomes

[latex] l={r}_{\perp }p={r}_{\perp }mv. [/latex]

Nosotros run across that if the management of [latex] \overset{\to }{p} [/latex] is such that it passes through the origin, then [latex] \theta =0, [/latex] and the angular momentum is zero because the lever arm is zero. In this respect, the magnitude of the angular momentum depends on the selection of origin.

If we take the time derivative of the angular momentum, we arrive at an expression for the torque on the particle:

[latex] \frac{d\overset{\to }{50}}{dt}=\frac{d\overset{\to }{r}}{dt}\,×\,\overset{\to }{p}+\overset{\to }{r}\,×\,\frac{d\overset{\to }{p}}{dt}=\overset{\to }{v}\,×\,chiliad\overset{\to }{five}+\overset{\to }{r}\,×\,\frac{d\overset{\to }{p}}{dt}=\overset{\to }{r}\,×\,\frac{d\overset{\to }{p}}{dt}. [/latex]

Here nosotros have used the definition of [latex] \overset{\to }{p} [/latex] and the fact that a vector crossed into itself is zero. From Newton's 2d law, [latex] \frac{d\overset{\to }{p}}{dt}=\sum \overset{\to }{F}, [/latex] the net force acting on the particle, and the definition of the net torque, we can write

[latex] \frac{d\overset{\to }{l}}{dt}=\sum \overset{\to }{\tau }. [/latex]

Note the similarity with the linear effect of Newton's second constabulary, [latex] \frac{d\overset{\to }{p}}{dt}=\sum \overset{\to }{F} [/latex]. The following problem-solving strategy can serve as a guideline for computing the angular momentum of a particle.

Problem-Solving Strategy: Angular Momentum of a Particle

1. Choose a coordinate organisation about which the athwart momentum is to be calculated.
2. Write down the radius vector to the point particle in unit vector notation.
3. Write the linear momentum vector of the particle in unit vector notation.
4. Take the cantankerous production [latex] \overset{\to }{50}=\overset{\to }{r}\,×\,\overset{\to }{p} [/latex] and use the right-hand dominion to establish the management of the angular momentum vector.
5. See if there is a fourth dimension dependence in the expression of the angular momentum vector. If there is, then a torque exists about the origin, and apply [latex] \frac{d\overset{\to }{50}}{dt}=\sum \overset{\to }{\tau } [/latex] to calculate the torque. If in that location is no fourth dimension dependence in the expression for the angular momentum, so the internet torque is goose egg.

Example

Athwart Momentum and Torque on a Meteor

A meteor enters Earth's atmosphere ((Effigy)) and is observed by someone on the basis earlier it burns upward in the atmosphere. The vector [latex] \overset{\to }{r}=25\,\text{km}\hat{i}+25\,\text{km}\hat{j} [/latex] gives the position of the meteor with respect to the observer. At the instant the observer sees the shooting star, information technology has linear momentum [latex] \overset{\to }{p}=fifteen.0\,\text{kg}(-2.0\text{km}\text{/}\text{s}\hat{j}) [/latex], and it is accelerating at a constant [latex] 2.0\,\text{m}\text{/}{\text{southward}}^{ii}(\text{−}\lid{j}) [/latex] along its path, which for our purposes can exist taken as a straight line. (a) What is the angular momentum of the falling star about the origin, which is at the location of the observer? (b) What is the torque on the meteor about the origin?

Effigy 11.10 An observer on the ground sees a falling star at position [latex] \overset{\to }{r} [/latex] with linear momentum [latex] \overset{\to }{p} [/latex].

Strategy

We resolve the acceleration into ten– and y-components and utilise the kinematic equations to express the velocity every bit a function of acceleration and fourth dimension. 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-airplane, we expect the angular momentum vector to be forth the z-centrality. To find the torque, we take the fourth dimension derivative of the angular momentum.

Solution

The meteor is entering Globe's atmosphere at an angle of [latex] 90.0\text{°} [/latex] beneath the horizontal, and then the components of the acceleration in the 10– and y-directions are

[latex] {a}_{ten}=0,\enspace{a}_{y}=-2.0\,\text{m}\text{/}{\text{s}}^{2}. [/latex]

Nosotros write the velocities using the kinematic equations.

[latex] {v}_{x}=0,\enspace{v}_{y}=-2.0\,×\,{10}^{three}\,\text{m}\text{/}\text{s}-(2.0\,\text{k}\text{/}{\text{south}}^{two})t. [/latex]

1. The athwart momentum is

   [latex] \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})\,×\,xv.0\,\text{kg}(0\hat{i}+{v}_{y}\hat{j})\hfill \\ & =15.0\,\text{kg}[25.0\,\text{km}({v}_{y})\hat{1000}]\hfill \\ & =15.0\,\text{kg[}2.50\,×\,{10}^{4}\,\text{m}(-2.0\,×\,{x}^{three}\,\text{chiliad}\text{/}\text{s}-(ii.0\,\text{m}\text{/}{\text{southward}}^{2})t)\lid{grand}].\hfill \end{assortment} [/latex]

   At [latex] t=0 [/latex], the athwart momentum of the meteor about the origin is

   [latex] {\overset{\to }{50}}_{0}=15.0\,\text{kg}[2.50\,×\,{ten}^{iv}\,\text{k}(-ii.0\,×\,{10}^{iii}\,\text{thousand}\text{/}\text{due south})\hat{k}]=7.50\,×\,{x}^{eight}\,\text{kg}·{\text{m}}^{2}\text{/}\text{s}(\text{−}\hat{grand}). [/latex]

   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 [latex] \overset{\to }{50} [/latex] equally a part of time, which is the second equation immediately above, we have

   [latex] \frac{d\overset{\to }{l}}{dt}=-fifteen.0\,\text{kg}(2.50\,×\,{10}^{4}\,\text{1000})(ii.0\,\text{g}\text{/}{\text{s}}^{2})\hat{k}. [/latex]

   So, since [latex] \frac{d\overset{\to }{l}}{dt}=\sum \overset{\to }{\tau } [/latex], we take

   [latex] \sum \overset{\to }{\tau }=-7.5\,×\,{10}^{v}\text{N}·\text{m}\lid{thou}. [/latex]

   The units of torque are given every bit newton-meters, not to exist dislocated with joules. As a check, we note that the lever arm is the x-component of the vector [latex] \overset{\to }{\text{r}} [/latex] in (Figure) since it is perpendicular to the forcefulness acting on the meteor, which is along its path. By Newton's second law, this force is

   [latex] \overset{\to }{F}=ma(\text{−}\hat{j})=15.0\,\text{kg}(2.0\,\text{one thousand}\text{/}{\text{s}}^{2})(\text{−}\hat{j})=xxx.0\,\text{kg}·\text{one thousand}\text{/}{\text{southward}}^{two}(\text{−}\hat{j}). [/latex]

   The lever arm is

   [latex] {\overset{\to }{r}}_{\perp }=two.five\,×\,{10}^{4}\,\text{m}\,\lid{i}. [/latex]

   Thus, the torque is

   [latex] \begin{array}{cc}\hfill \sum \overset{\to }{\tau }={\overset{\to }{r}}_{\perp }\,×\,\overset{\to }{F}& =(2.five\,×\,{10}^{4}\,\text{g}\,\hat{i})\,×\,(-xxx.0\,\text{kg}·\text{m}\text{/}{\text{due south}}^{2}\lid{j}),\hfill \\ & =7.5\,×\,{10}^{5}\,\text{North}·\text{k}(\text{−}\hat{k}).\hfill \stop{assortment} [/latex]

Significance

Since the falling star is accelerating down toward Earth, its radius and velocity vector are irresolute. Therefore, since [latex] \overset{\to }{50}=\overset{\to }{r}\,×\,\overset{\to }{p} [/latex], the angular momentum is changing as a function of time. The torque on the shooting star about the origin, nonetheless, is constant, because the lever arm [latex] {\overset{\to }{r}}_{\perp } [/latex] and the force on the falling star are constants. This case is of import 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 organisation of particles and for a rigid body.

Check Your Understanding

A proton spiraling effectually a magnetic field executes round motion in the aeroplane of the paper, every bit shown below. The circular path has a radius of 0.4 m and the proton has velocity [latex] 4.0\,×\,{10}^{6}\,\text{m}\text{/}\text{s} [/latex]. What is the athwart momentum of the proton well-nigh the origin?

Testify Respond

From the effigy, nosotros see that the cross product of the radius vector with the momentum vector gives a vector directed out of the page. Inserting the radius and momentum into the expression for the angular momentum, we have [latex] \overset{\to }{l}=\overset{\to }{r}\,×\,\overset{\to }{p}=(0.4\,\text{one thousand}\hat{i}\text{)}\,×\,(ane.67\,×\,{x}^{-27}\,\text{kg}(4.0\,×\,{10}^{6}\,\text{m}\text{/}\text{s})\chapeau{j})=2.vii\,×\,{10}^{-21}\,\text{kg}·{\text{thou}}^{two}\text{/}\text{southward}\hat{k} [/latex]

Angular Momentum of a Organisation of Particles

The athwart momentum of a system of particles is important in many scientific disciplines, one beingness astronomy. Consider a screw galaxy, a rotating island of stars like our own Milky Style. The private stars can be treated equally point particles, each of which has its ain angular momentum. The vector sum of the individual angular momenta requite the total angular momentum of the galaxy. In this section, we develop the tools with which we tin summate the full angular momentum of a system of particles.

In the preceding section, nosotros introduced the angular momentum of a single particle most a designated origin. The expression for this angular momentum is [latex] \overset{\to }{l}=\overset{\to }{r}\,×\,\overset{\to }{p}, [/latex] where the vector [latex] \overset{\to }{r} [/latex] is from the origin to the particle, and [latex] \overset{\to }{p} [/latex] is the particle's linear momentum. If we have a system of North particles, each with position vector from the origin given by [latex] {\overset{\to }{r}}_{i} [/latex] and each having momentum [latex] {\overset{\to }{p}}_{i}, [/latex] then the full angular momentum of the system of particles about the origin is the vector sum of the individual angular momenta well-nigh the origin. That is,

[latex] \overset{\to }{L}={\overset{\to }{l}}_{i}+{\overset{\to }{l}}_{two}+\cdots +{\overset{\to }{l}}_{Northward}. [/latex]

Similarly, if particle i is bailiwick to a net torque [latex] {\overset{\to }{\tau }}_{i} [/latex] about the origin, so we tin can discover the cyberspace torque about the origin due to the system of particles by differentiating (Figure):

[latex] \frac{d\overset{\to }{L}}{dt}=\sum _{i}\frac{d{\overset{\to }{l}}_{i}}{dt}=\sum _{i}{\overset{\to }{\tau }}_{i}. [/latex]

The sum of the individual torques produces a net external torque on the system, which nosotros designate [latex] \sum \overset{\to }{\tau }. [/latex] Thus,

[latex] \frac{d\overset{\to }{50}}{dt}=\sum \overset{\to }{\tau }. [/latex]

(Figure) states that the charge per unit of modify of the total angular momentum of a system is equal to the internet external torque acting on the system when both quantities are measured with respect to a given origin. (Effigy) tin can be applied to whatsoever arrangement that has net angular momentum, including rigid bodies, every bit discussed in the next section.

Example

Angular Momentum of Three Particles

Referring to (Figure)(a), make up one's mind the total angular momentum due to the three particles well-nigh the origin. (b) What is the rate of change of the angular momentum?

Figure eleven.11 Three particles in the xy-plane with unlike position and momentum vectors.

Strategy

Write down the position and momentum vectors for the iii particles. Summate the individual athwart momenta and add them as vectors to find the total angular momentum. And then exercise the same for the torques.

Solution

1. Particle 1: [latex] {\overset{\to }{r}}_{1}=-2.0\,\text{m}\chapeau{i}+ane.0\,\text{m}\chapeau{j},\enspace{\overset{\to }{p}}_{1}=2.0\,\text{kg}(4.0\,\text{one thousand}\text{/}\text{due south}\hat{j})=8.0\,\text{kg}·\text{m}\text{/}\text{s}\lid{j}, [/latex]

   [latex] {\overset{\to }{fifty}}_{1}={\overset{\to }{r}}_{1}\,×\,{\overset{\to }{p}}_{1}=-sixteen.0\,\text{kg}·{\text{g}}^{2}\text{/}\text{southward}\hat{chiliad}. [/latex]

   Particle two: [latex] {\overset{\to }{r}}_{two}=iv.0\,\text{m}\hat{i}+one.0\,\text{m}\hat{j},\enspace{\overset{\to }{p}}_{2}=4.0\,\text{kg}(v.0\,\text{m}\text{/}\text{due south}\hat{i})=twenty.0\,\text{kg}·\text{m}\text{/}\text{southward}\chapeau{i} [/latex],

   [latex] {\overset{\to }{50}}_{2}={\overset{\to }{r}}_{2}\,×\,{\overset{\to }{p}}_{2}=-20.0\,\text{kg}·{\text{m}}^{2}\text{/}\text{s}\hat{k}. [/latex]

   Particle 3: [latex] {\overset{\to }{r}}_{iii}=two.0\,\text{k}\hat{i}-2.0\,\text{m}\hat{j},\enspace{\overset{\to }{p}}_{3}=1.0\,\text{kg}(3.0\,\text{one thousand}\text{/}\text{s}\hat{i})=3.0\,\text{kg}·\text{chiliad}\text{/}\text{due south}\chapeau{i} [/latex],

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

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

   [latex] {\overset{\to }{l}}_{T}={\overset{\to }{l}}_{1}+{\overset{\to }{l}}_{2}+{\overset{\to }{l}}_{3}=-xxx\,\text{kg}·{\text{m}}^{ii}\text{/}\text{s}\chapeau{k}. [/latex]

2. The individual forces and lever arms are

   [latex] \begin{array}{c}{\overset{\to }{r}}_{i\perp }=1.0\,\text{one thousand}\lid{j},\enspace{\overset{\to }{F}}_{1}=-6.0\,\text{N}\chapeau{i},\enspace{\overset{\to }{\tau }}_{1}=half-dozen.0\text{North}·\text{m}\hat{k}\hfill \\ {\overset{\to }{r}}_{2\perp }=4.0\,\text{m}\hat{i},\enspace{\overset{\to }{F}}_{2}=x.0\,\text{Northward}\hat{j},\enspace{\overset{\to }{\tau }}_{2}=twoscore.0\,\text{N}·\text{1000}\hat{k}\hfill \\ {\overset{\to }{r}}_{3\perp }=2.0\,\text{chiliad}\hat{i},\enspace{\overset{\to }{F}}_{three}=-8.0\,\text{N}\hat{j},\enspace{\overset{\to }{\tau }}_{3}=-xvi.0\,\text{N}·\text{m}\hat{k}.\hfill \end{array} [/latex]

   Therefore:

   [latex] \sum _{i}{\overset{\to }{\tau }}_{i}={\overset{\to }{\tau }}_{one}+{\overset{\to }{\tau }}_{2}+{\overset{\to }{\tau }}_{3}=30\,\text{N}·\text{grand}\chapeau{k}. [/latex]

Significance

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

Angular Momentum of a Rigid Torso

We take investigated the athwart momentum of a single particle, which nosotros generalized to a system of particles. Now we can use the principles discussed in the previous section to develop the concept of the angular momentum of a rigid body. Angelic objects such equally planets accept angular momentum due to their spin and orbits around stars. In engineering, anything that rotates about an centrality carries angular momentum, such as flywheels, propellers, and rotating parts in engines. Cognition of the angular momenta of these objects is crucial to the design of the system in which they are a part.

To develop the angular momentum of a rigid body, we model a rigid body as being made upwardly of small mass segments, [latex] \text{Δ}{m}_{i}. [/latex] In (Figure), a rigid trunk is constrained to rotate about the z-axis with angular velocity [latex] \omega [/latex]. All mass segments that make upwardly the rigid trunk undergo round motion near the z-axis with the aforementioned athwart velocity. Part (a) of the figure shows mass segment [latex] \text{Δ}{grand}_{i} [/latex] with position vector [latex] {\overset{\to }{r}}_{i} [/latex] from the origin and radius [latex] {R}_{i} [/latex] to the z-centrality. The magnitude of its tangential velocity is [latex] {v}_{i}={R}_{i}\omega [/latex]. Because the vectors [latex] {\overset{\to }{v}}_{i}\,\text{and}\,{\overset{\to }{r}}_{i} [/latex] are perpendicular to each other, the magnitude of the athwart momentum of this mass segment is

[latex] {l}_{i}={r}_{i}(\text{Δ}m{v}_{i})\text{sin}\,90\text{°}. [/latex]

Figure 11.12 (a) A rigid torso is constrained to rotate around the z-centrality. The rigid body is symmetrical about the z-axis. A mass segment [latex] \text{Δ}{one thousand}_{i} [/latex] is located at position [latex] {\overset{\to }{r}}_{i}, [/latex] which makes angle [latex] {\theta }_{i} [/latex] with respect to the z-axis. The circular move of an infinitesimal mass segment is shown. (b) [latex] {\overset{\to }{50}}_{i} [/latex] is the angular momentum of the mass segment and has a component along the z-axis [latex] {({\overset{\to }{50}}_{i})}_{z} [/latex].

Using the correct-mitt rule, the angular momentum vector points in the direction shown in function (b). The sum of the angular momenta of all the mass segments contains components both along and perpendicular to the centrality of rotation. Every mass segment has a perpendicular component of the athwart momentum that will be cancelled by the perpendicular component of an identical mass segment on the opposite side of the rigid body. Thus, the component along the axis of rotation is the only component that gives a nonzero value when summed over all the mass segments. From part (b), the component of [latex] {\overset{\to }{l}}_{i} [/latex] forth the axis of rotation is

[latex] \begin{array}{cc}\hfill {({l}_{i})}_{z}& ={l}_{i}\text{sin}\,{\theta }_{i}=({r}_{i}\text{Δ}{yard}_{i}{v}_{i})\text{sin}\,{\theta }_{i},\hfill \\ & =({r}_{i}\text{sin}\,{\theta }_{i})(\text{Δ}{m}_{i}{v}_{i})={R}_{i}\text{Δ}{one thousand}_{i}{v}_{i}.\hfill \end{array} [/latex]

The net athwart momentum of the rigid body along the axis of rotation is

[latex] L=\sum _{i}{({\overset{\to }{50}}_{i})}_{z}=\sum _{i}{R}_{i}\text{Δ}{m}_{i}{v}_{i}=\sum _{i}{R}_{i}\text{Δ}{m}_{i}({R}_{i}\omega )=\omega {\sum _{i}\text{Δ}{grand}_{i}({R}_{i})}^{2}. [/latex]

The summation [latex] {\sum _{i}\text{Δ}{m}_{i}({R}_{i})}^{2} [/latex] is simply the moment of inertia I of the rigid body near the centrality of rotation. For a thin hoop rotating virtually an axis perpendicular to the plane of the hoop, all of the [latex] {R}_{i} [/latex]'s are equal to R so the summation reduces to [latex] {R}^{two}\sum _{i}\text{Δ}{m}_{i}=m{R}^{2}, [/latex] which is the moment of inertia for a thin hoop found in (Effigy). Thus, the magnitude of the athwart momentum along the axis of rotation of a rigid body rotating with angular velocity [latex] \omega [/latex] about the centrality is

[latex] 50=I\omega . [/latex]

This equation is coordinating to the magnitude of the linear momentum [latex] p=mv [/latex]. The direction of the angular momentum vector is directed along the centrality of rotation given by the correct-mitt rule.

Case

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 complimentary finish to option upwards rocks. The mass of the arm is ii.0 kg and the mass of the forceps is 1.0 kg. See (Figure). The robot arm and forceps move from rest to [latex] \omega =0.1\pi \,\text{rad}\text{/}\text{s} [/latex] in 0.1 south. It rotates down and picks upwards a Mars rock that has mass 1.5 kg. The axis of rotation is the bespeak 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 later 0.1 s when the arm has stopped accelerating? (b) What is the athwart momentum of the robot arm when it has the Mars stone in its forceps and is rotating upwards? (c) When the arm does not have a stone in the forceps, what is the torque nearly the point where the arm connects to the rover when it is accelerating from rest to its concluding angular velocity?

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

Strategy

We use (Figure) to find athwart momentum in the diverse configurations. When the arm is rotating downward, the right-manus rule gives the athwart momentum vector directed out of the page, which we volition phone call the positive z-direction. When the arm is rotating upward, the right-mitt rule gives the direction of the athwart momentum vector into the page or in the negative z-management. 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 tin can be approximated as point masses located at a altitude of 1 k from the origin. For office (c), nosotros use Newton's second law of movement for rotation to find the torque on the robot arm.

Solution

1. Writing down the individual moments of inertia, nosotros haveRobot arm: [latex] {I}_{\text{R}}=\frac{one}{3}{m}_{\text{R}}{r}^{2}=\frac{1}{three}(2.00\,\text{kg}){(i.00\,\text{m})}^{2}=\frac{2}{three}\,\text{kg}·{\text{yard}}^{2}. [/latex]Forceps: [latex] {I}_{\text{F}}={chiliad}_{\text{F}}{r}^{2}=(one.0\,\text{kg}){(1.0\,\text{m})}^{2}=1.0\,\text{kg}·{\text{chiliad}}^{ii}. [/latex]Mars rock: [latex] {I}_{\text{MR}}={m}_{\text{MR}}{r}^{ii}=(1.5\,\text{kg}){(1.0\,\text{one thousand})}^{2}=1.5\,\text{kg}·{\text{m}}^{two}. [/latex]Therefore, without the Mars stone, the total moment of inertia is

   [latex] {I}_{\text{Total}}={I}_{\text{R}}+{I}_{\text{F}}=1.67\,\text{kg}·{\text{m}}^{2} [/latex]

   and the magnitude of the angular momentum is

   [latex] 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}. [/latex]

   The angular momentum vector is directed out of the page in the [latex] \hat{k} [/latex] 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 accept

   [latex] {I}_{\text{Total}}={I}_{\text{R}}+{I}_{\text{F}}+{I}_{\text{MR}}=3.17\,\text{kg}·{\text{one thousand}}^{ii} [/latex]

   and

   [latex] L=I\omega =3.17\,\text{kg}·{\text{grand}}^{ii}(0.1\pi \,\text{rad}\text{/}\text{due south})=0.32\pi \,\text{kg}·{\text{m}}^{two}\text{/}\text{south}\text{.} [/latex]

   At present the angular momentum vector is directed into the page in the [latex] \text{−}\chapeau{thousand} [/latex] direction, by the right-manus 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 athwart momentum using (Figure) [latex] \frac{d\overset{\to }{L}}{dt}=\sum \overset{\to }{\tau }. [/latex] Just since [latex] L=I\omega [/latex], and understanding that the direction of the angular momentum and torque vectors are forth the axis of rotation, we tin can suppress the vector note and notice

   [latex] \frac{dL}{dt}=\frac{d(I\omega )}{dt}=I\frac{d\omega }{dt}=I\alpha =\sum \tau , [/latex]

   which is Newton's second law for rotation. Since [latex] \alpha =\frac{0.1\pi \,\text{rad}\text{/}\text{south}}{0.i\,\text{s}}=\pi \,\text{rad}\text{/}{\text{s}}^{2} [/latex], we can summate the cyberspace torque:

   [latex] \sum \tau =I\alpha =one.67\,\text{kg}·{\text{thousand}}^{2}(\pi \,\text{rad}\text{/}{\text{s}}^{2})=1.67\pi \,\text{N}·\text{m}. [/latex]

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 athwart velocity is the aforementioned.

Bank check Your Agreement

Which has greater angular momentum: a solid sphere of mass m rotating at a constant angular frequency [latex] {\omega }_{0} [/latex] about the z-axis, or a solid cylinder of aforementioned mass and rotation charge per unit about the z-axis?

Summary

• The angular momentum [latex] \overset{\to }{l}=\overset{\to }{r}\,×\,\overset{\to }{p} [/latex] of a single particle about a designated origin is the vector production of the position vector in the given coordinate system and the particle's linear momentum.
• The angular momentum [latex] \overset{\to }{l}=\sum _{i}{\overset{\to }{l}}_{i} [/latex] of a system of particles almost a designated origin is the vector sum of the private momenta of the particles that brand upwards the system.
• The cyberspace torque on a system about a given origin is the fourth dimension derivative of the angular momentum well-nigh that origin: [latex] \frac{d\overset{\to }{Fifty}}{dt}=\sum \overset{\to }{\tau } [/latex].
• A rigid rotating torso has angular momentum [latex] 50=I\omega [/latex] directed along the axis of rotation. The time derivative of the angular momentum [latex] \frac{dL}{dt}=\sum \tau [/latex] gives the cyberspace torque on a rigid torso and is directed along the axis of rotation.

Conceptual Questions

Can you assign an angular momentum to a particle without first defining a reference signal?

For a particle traveling in a straight line, are in that location any points about which the athwart momentum is goose egg? Assume the line intersects the origin.

Show Solution

All points on the straight line will give zero athwart momentum, considering a vector crossed into a parallel vector is zero.

Under what conditions does a rigid body have athwart momentum simply not linear momentum?

If a particle is moving with respect to a chosen origin it has linear momentum. What conditions must exist for this particle's angular momentum to be zero about the chosen origin?

Show Solution

The particle must be moving on a straight line that passes through the chosen origin.

If you know the velocity of a particle, can you say anything most the particle's athwart momentum?

Problems

A 0.2-kg particle is travelling along the line [latex] y=2.0\,\text{m} [/latex] with a velocity [latex] 5.0\,\text{m}\text{/}\text{s} [/latex]. What is the athwart momentum of the particle about the origin?

A bird flies overhead from where you stand up at an altitude of 300.0 yard and at a speed horizontal to the ground of xx.0 g/s. The bird has a mass of two.0 kg. The radius vector to the bird makes an angle [latex] \theta [/latex] with respect to the basis. The radius vector to the bird and its momentum vector lie in the xy-airplane. What is the bird'south athwart momentum well-nigh the point where you are standing?

A Formula One race car with mass 750.0 kg is speeding through a course in Monaco and enters a circular turn at 220.0 km/h in the counterclockwise direction about the origin of the circumvolve. At some other part of the course, the car enters a 2nd circular plow at 180 km/h also in the counterclockwise direction. If the radius of curvature of the kickoff turn is 130.0 m and that of the second is 100.0 m, compare the angular momenta of the race car in each turn taken well-nigh the origin of the circular plow.

A particle of mass 5.0 kg has position vector [latex] \overset{\to }{r}=(2.0\hat{i}-3.0\hat{j})\text{m} [/latex] at a detail instant of time when its velocity is [latex] \overset{\to }{v}=(three.0\lid{i})\text{thousand}\text{/}\text{s} [/latex] with respect to the origin. (a) What is the angular momentum of the particle? (b) If a force [latex] \overset{\to }{F}=5.0\hat{j}\,\text{Northward} [/latex] acts on the particle at this instant, what is the torque near the origin?

Use the right-hand dominion to make up one's mind the directions of the angular momenta nearly the origin of the particles as shown beneath. The z-axis is out of the page.

Suppose the particles in the preceding trouble have masses [latex] {m}_{1}=0.10\,\text{kg,}\enspace{k}_{2}=0.20\,\text{kg,}\enspace{m}_{3}=0.30\,\text{kg,} [/latex] [latex] {m}_{iv}=0.40\,\text{kg} [/latex]. The velocities of the particles are [latex] {5}_{1}=2.0\hat{i}\text{chiliad}\text{/}\text{s} [/latex], [latex] {v}_{ii}=(3.0\lid{i}-three.0\lid{j})\text{m}\text{/}\text{s} [/latex], [latex] {v}_{3}=-1.v\hat{j}\text{one thousand}\text{/}\text{s} [/latex], [latex] {v}_{four}=-4.0\hat{i}\text{grand}\text{/}\text{s} [/latex]. (a) Calculate the angular momentum of each particle well-nigh the origin. (b) What is the total angular momentum of the four-particle system near the origin?

Two particles of equal mass travel with the same speed in contrary directions along parallel lines separated by a distance d. Testify that the athwart momentum of this 2-particle system is the same no matter what point is used every bit the reference for calculating the athwart momentum.

An aeroplane of mass [latex] 4.0\,×\,{10}^{4}\,\text{kg} [/latex] flies horizontally at an altitude of 10 km with a constant speed of 250 m/s relative to Globe. (a) What is the magnitude of the airplane'southward angular momentum relative to a ground observer straight below the plane? (b) Does the angular momentum modify as the airplane flies forth its path?

Prove Solution

a. [latex] L=1.0\,×\,{10}^{xi}\,\text{kg}·{\text{yard}}^{ii}\text{/}\text{s} [/latex]; b. No, the angular momentum stays the aforementioned since the cantankerous-product involves only the perpendicular distance from the plane to the ground no matter where it is forth its path.

At a detail instant, a 1.0-kg particle's position is [latex] \overset{\to }{r}=(2.0\hat{i}-4.0\hat{j}+6.0\lid{k})\text{grand} [/latex], its velocity is [latex] \overset{\to }{v}=(-1.0\hat{i}+4.0\hat{j}+1.0\hat{m})\text{m}\text{/}\text{s} [/latex], and the forcefulness on it is [latex] \overset{\to }{F}=(x.0\hat{i}+xv.0\hat{j})\text{North} [/latex]. (a) What is the angular momentum of the particle about the origin? (b) What is the torque on the particle near the origin? (c) What is the time rate of change of the particle's angular momentum at this instant?

A particle of mass k is dropped at the point [latex] (\text{−}d,0) [/latex] and falls vertically in Earth's gravitational field [latex] \text{−}grand\hat{j}. [/latex] (a) What is the expression for the angular momentum of the particle around the z-axis, which points directly out of the folio as shown below? (b) Summate the torque on the particle around the z-axis. (c) Is the torque equal to the time rate of change of the angular momentum?

a. [latex] \overset{\to }{v}=\text{−}gt\hat{j},\enspace{\overset{\to }{r}}_{\perp }=\text{−}d\chapeau{i},\enspace\overset{\to }{l}=mdgt\hat{1000} [/latex];

b. [latex] \overset{\to }{F}=\text{−}mg\hat{j},\enspace\sum \overset{\to }{\tau }=dmg\lid{thousand} [/latex]; c. yes

(a) Summate the angular momentum of Earth in its orbit around the Sun. (b) Compare this angular momentum with the athwart momentum of World virtually its centrality.

A boulder of mass 20 kg and radius xx cm rolls downwards a hill xv m loftier from rest. What is its angular momentum when it is half way down the hill? (b) At the bottom?

A satellite is spinning at 6.0 rev/s. The satellite consists of a main body in the shape of a sphere of radius 2.0 g and mass 10,000 kg, and ii antennas projecting out from the center of mass of the main body that tin can exist approximated with rods of length 3.0 m each and mass 10 kg. The antenna's lie in the plane of rotation. What is the angular momentum of the satellite?

A propeller consists of ii blades each three.0 m in length and mass 120 kg each. The propeller tin be approximated by a single rod rotating well-nigh its heart of mass. The propeller starts from residue and rotates up to 1200 rpm in 30 seconds at a constant rate. (a) What is the athwart momentum of the propeller at [latex] t=10\,\text{s;}\,t=twenty\,\text{southward?} [/latex] (b) What is the torque on the propeller?

A pulsar is a rapidly rotating neutron star. The Crab nebula pulsar in the constellation Taurus has a period of [latex] 33.5\,×\,{10}^{-3}\,\text{s} [/latex], radius 10.0 km, and mass [latex] two.eight\,×\,{10}^{xxx}\,\text{kg}. [/latex] The pulsar'south rotational menses will increase over time due to the release of electromagnetic radiation, which doesn't change its radius merely reduces its rotational energy. (a) What is the angular momentum of the pulsar? (b) Suppose the angular velocity decreases at a rate of [latex] {10}^{-xiv}\,\text{rad}\text{/}{\text{south}}^{two} [/latex]. What is the torque on the pulsar?

The blades of a air current turbine are 30 m in length and rotate at a maximum rotation charge per unit of 20 rev/min. (a) If the blades are 6000 kg each and the rotor associates has three blades, summate the angular momentum of the turbine at this rotation rate. (b) What is the torque require to rotate the blades up to the maximum rotation rate in v minutes?

A roller coaster has mass 3000.0 kg and needs to make information technology safely through a vertical circular loop of radius 50.0 grand. What is the minimum angular momentum of the coaster at the lesser of the loop to make it safely through? Neglect friction on the rails. Take the coaster to be a point particle.

A mountain biker takes a jump in a race and goes airborne. The mountain bike is travelling at 10.0 m/s before it goes airborne. If the mass of the front end wheel on the bicycle is 750 thousand and has radius 35 cm, what is the angular momentum of the spinning wheel in the air the moment the cycle leaves the basis?

Evidence Solution

[latex] \omega =28.half dozen\,\text{rad}\text{/}\text{due south}⇒L=two.6\,\text{kg}·{\text{grand}}^{2}\text{/}\text{southward} [/latex]

Glossary

angular momentum

rotational analog of linear momentum, establish past taking the product of moment of inertia and angular velocity

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Source: https://courses.lumenlearning.com/suny-osuniversityphysics/chapter/11-2-angular-momentum/

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