Torque and Angular MomentumTorque and Angular Momentum

, C C B B A A

r r r r r

++ ==

to show that to show that AC AC sinsinbb

==

AB ABsinsinccor or

cc C C b

b B B

sin sin sin

sin

==

. Proceeding similarly,. Proceeding similarly, we can extend the law of sines to the third side of the triangle and the angle

we can extend the law of sines to the third side of the triangle and the angle opposite it.

opposite it.

A B B

C C

b a

c c

Express the magnitude of the vector Express the magnitude of the vector product of

product of

r r

and : and B B:

r A A

××

B B

==

AB ABsinsin

( (

180180

°° −−

)) ==

AB ABsinsincc

r r r r

Express the magnitude of the vector Express the magnitude of the vector product of

product of

r r

and : and C C :

r A A

××

C C

==

AC AC sinsinbb

r r r r

Form the vector product of Form the vector product of

r r

with with C

r r

to obtain:

( ( ))

B B A A

B B A A A A A A

B B A A A A C C A A

r r r r

r r r r r r r r

r r r r r r r r r r

××

==

××

++

××

==

++

××

==

××

because

because A A

××

A A

==

r r r r

Because :

Because A A C C A A B B :

r r r r r r r

×× == ××

A A C C A A B B

r r r r r r r

×× == ××

and and

cc AB AB b

b AC

AC sinsin

==

sinsin Simplify and rewrite this expression

Simplify and rewrite this expression to obtain:

to obtain: cc

C C b

b B B

sin sin sin

sin

==

Proceed similarly to extend this Proceed similarly to extend this result to the law of sines:

result to the law of sines: cc

C C b

b B B a

a A A

sin sin sin

sin sin

sin

== ==

Torque and Angular Momentum Torque and Angular Momentum

37 • • [SSM][SSM] A 2.0-kg A 2.0-kg particle particle moves moves directly directly eastward eastward at at a a constant constant speedspeed of 4.5 m/s along an e

of 4.5 m/s along an east-west line. (ast-west line. (aa) What is its angular momentum (including) What is its angular momentum (including direction) about a point that lies 6.0 m north of the line? (

direction) about a point that lies 6.0 m north of the line? (bb) What is its angular) What is its angular momentum (including direction) about a point

momentum (including direction) about a point that lies 6.0 m south of the line?that lies 6.0 m south of the line?

((cc) What is its angular momentum (including direction) about a point that lies 6.0) What is its angular momentum (including direction) about a point that lies 6.0 m directly east of the particle?

m directly east of the particle?

Picture the Problem

Picture the Problem The angular momentum of the particle isThe angular momentum of the particle is L L r r ^r^r p p^r^r

== ××

where where r

r ^r^r is the vector locating the particle relative to the reference point and is the vector locating the particle relative to the reference point and p p^r^r is the is the particle’s linear momentum.

particle’s linear momentum.

((aa) The magnitude of the ) The magnitude of the particle’sparticle’s angular momentum is given by:

angular momentum is given by:

( (

φ φ

))

φ φ φ

φ sinsin sinsin sin

sin rmvrmv mvmv r r rp

rp L

== == ==

Substitute numerical values and Substitute numerical values and evaluate

evaluate L L::

(

( ))( ( ))( ( ))

/s /s m m kg kg 54 54

m m 6.0 6.0 m/s m/s 4.5 4.5 kg kg 2.0 2.0

⋅⋅

==

Use a right-hand rule to establish Use a right-hand rule to establish the direction of

the direction of L L

r r

upward upward /s,

/s, m m kg kg 54

⋅⋅

²²

==

L L

((bb) Because the distance to the line) Because the distance to the line along which the particle is moving is along which the particle is moving is the same, only the direction of

the same, only the direction of L

r r

differs:

downward downward /s,

/s, m m kg kg 54

⋅⋅

²²

==

((cc) Because) Because r r ^r^r

××

p p^r^r

==

00for a point onfor a point on the line along which the particle is the line along which the particle is moving:

moving:

==

0 L L

r r

38 •• You You observe observe a a 2.0-kg 2.0-kg particle particle moving moving at at a a constant constant speed speed ofof 3.5 m/s in a clockwise direction

3.5 m/s in a clockwise direction around a circle of radius 4.0 m. around a circle of radius 4.0 m. ((aa) What is its) What is its angular momentum (including direction) about the center of the circle? (

angular momentum (including direction) about the center of the circle? (bb) What) What is its moment of inertia about an a

is its moment of inertia about an axis through the center of the cxis through the center of the circle andircle and perpendicular to the plane of the motion? (

perpendicular to the plane of the motion? (cc) What is the angular velocity of the) What is the angular velocity of the particle?

particle?

Picture the Problem

Picture the ProblemThe angular momentum of the The angular momentum of the particle isparticle is L L r r ^r^r p p^r^r

== ××

where where r

r ^r^ris the vector locating the particle is the vector locating the particle relative to the reference point andrelative to the reference point and p p^r^r is the is the particle’s linear momentum.

particle’s linear momentum.

((aa) The magnitude of the ) The magnitude of the particle’sparticle’s angular momentum is given by:

angular momentum is given by:

( (

φ φ

))

φ φ φ

φ sinsin sinsin sin

sin rmvrmv mvmv r r rp

rp L

== == ==

Substitute numerical values and Substitute numerical values and evaluate the magnitude of

evaluate the magnitude of L L::

(

( ))( ( ))( ( ))

/s /s m m kg kg 28 28

m m 4.0 4.0 m/s m/s 3.5 3.5 kg kg 2.0 2.0

⋅⋅

==

L L

Use a right-hand rule to establish the Use a right-hand rule to establish the direction of

direction of L L

r r

you you from from away away /s,

/s, m m kg kg 8 8 2

⋅⋅

²²

==

((bb) Treat the 2.0-kg particle as a) Treat the 2.0-kg particle as a point particle to obtain:

point particle to obtain:

mr 2

mr I

==

Substitute numerical values and Substitute numerical values and evaluate

evaluate I I ::

(

2.02.0kgkg

))( ( ))

4.04.0mm ²²

==

3232kgkg

⋅⋅

mm²²

==

I I

((cc) Because) Because L L == I I ω ω , , the angularthe angular speed of the particle is the ratio of its speed of the particle is the ratio of its angular momentum and its moment angular momentum and its moment of inertia:

of inertia:

I I L

==

L ω ω

Substitute numerical values and Substitute numerical values and evaluate

evaluate ω ω ::

2 2 2

rad/s rad/s 0.88 m 0.88

m kg kg 32 32

/s /s m m kg kg 28

==

⋅⋅

== ⋅⋅

ω ω

39 ••

39 •• ((aa) A particle moving at constant velocity has zero angular momentum) A particle moving at constant velocity has zero angular momentum about a particular point. Use

about a particular point. Use the definition of angular momentum to show the definition of angular momentum to show thatthat under this condition the particle is moving either directly toward or directly away under this condition the particle is moving either directly toward or directly away from the point. (

from the point. (bb) You are a right-handed b) You are a right-handed batter and let a waist-high fastball goatter and let a waist-high fastball go past you without swinging. What is the direction of its angular momentum past you without swinging. What is the direction of its angular momentum

relative to your navel? (Assume the ball travels in a straight horizontal line as it relative to your navel? (Assume the ball travels in a straight horizontal line as it passes you.)

passes you.)

Picture the Problem Picture the Problem L L

r r

and

and p p^r^r are related according to are related according to L L r r ^r^r p p^r^r..

== ××

IfIf L L

r r

= 0, then

= 0, then examination of the magnitude of

examination of the magnitude of r r ^r^r

××

p p^r^r will allow us to conclude that will allow us to conclude thatsinsinφ φ

==

00 and that the particle is moving either directly toward the point, directly away from and that the particle is moving either directly toward the point, directly away from the point, or through the point.

the point, or through the point.

((aa) Because) Because L L

r r

= 0:

= 0: r r ^r^r

××

p p^r^r

==

r r ^r^r

××

mmvv^r^r

==

mmr r ^r^r

××

vv^r^r

==

00 or

==

××

vv r r ^r^r ^r^r

Express

Express the the magnitude magnitude of of r r ^r^r

××

vv^r^r :: r r ^r^r

××

vv^r^r

==

rvrvsinsinφ φ

==

Because neither

Because neither r r nor nor vv is zero: is zero: sinsinφ φ

==

00 where

where φ φ is the angle between is the angle between r r ^r^rand .and .vv^r^r Solving for

Solving for φ φ yields: yields: φ φ

==

sinsin⁻⁻¹¹

( ( ))

==

°°

or or 180180

°°

((bb) Use the right-hand rule to establish that the ) Use the right-hand rule to establish that the ball’s angular momentum isball’s angular momentum is downward.

downward.

40 ••

40 •• A A particle particle that that has has a a massmass mm is traveling with a constant velocity is traveling with a constant velocity vv^r^r along a straight line that is a distance

along a straight line that is a distance bb from the originfrom the origin OO (Figure 10-44). Let(Figure 10-44). Let dAdA be the area swept out by the position vector from

be the area swept out by the position vector from OO to the particle during a timeto the particle during a time interval

interval dt dt . Show that. Show that dAdA//dt dt is constant and is equal to is constant and is equal to L L 22mm , where, where L L is theis the magnitude of the angular momentum of the particle about the origin.

magnitude of the angular momentum of the particle about the origin.

Picture the Problem

Picture the Problem The area swept out by the position vector (the shaded area The area swept out by the position vector (the shaded area in Figure 10-44) is the difference between the area of a trapezoid and the area of a in Figure 10-44) is the difference between the area of a trapezoid and the area of a triangle. Let

triangle. Let x x11 be the be the x x component ofcomponent of r r ^r^r₁₁ and and x x11 + +

Δ Δ

x be the x be the x x component ofcomponent of r r ^r^r₂₂..

Use the formulas for the areas of a trapezoid and a triangle to express

Δ Δ

A A and then and then take the limit as

take the limit as

Δ Δ

→ →

0 to express 0 to express dAdA//dt.dt. Letting Letting θ θ be be the the angle angle betweenbetween r r ^r^r₁₁ and and the horizontal axis, we can express

the horizontal axis, we can express bb as a function of as a function of r r ^r^r₁₁ andand θ θ ..

The area swept out by the position The area swept out by the position vector (the shaded area in Figure vector (the shaded area in Figure 10-44) is given by:

44) is given by:

[ [ ( ( )) ]] ( ( ))

x x b b

x x x

x b b x

x x x x x b b

A A A

Δ Δ

==

Δ Δ ++

−−

Δ Δ ++

++

Δ Δ

==

−−

==

Δ Δ

2 2 1 1

1 2 1 2 1 1 1

2 1 2 1 1

triangle triangle trap

trap

In the limit as

Δ Δ

→ →

0: 0:

==

₂₂¹¹

==

₂₂¹¹bvbv

==

constantconstant dt

dt dx bdx dt b

dt dA dA

Because

Because bb

==

r r ^r^r₁₁ sinsinθ θ ::

( ( )) ( ( ))

( ( ))

m m L p L

m p m

m m

mv vv mv

bv dt bv

dt dA dA

2 sin 2

1 1

2 2 sin sin sin

sin

1 1

1 1 1

2 1 2 1 1 2

2 1 1

==

θ θ

θ θ θ

r r

r r r

r r

r r r

41 ••

41 •• A A 15-g 15-g coin tcoin that hat has has a diama diameter eter of of 1.5 cm 1.5 cm is is spinning spinning at at 10 10 rev/s rev/s aboutabout a fixed vertical axis. The coin

a fixed vertical axis. The coin is spinning on edge with its center dis spinning on edge with its center directly aboveirectly above the point of contact with the tabletop. As you look down on the tabletop, the coin the point of contact with the tabletop. As you look down on the tabletop, the coin spins clockwise. (

spins clockwise. (aa) What is the angular momentum (including direction) of the) What is the angular momentum (including direction) of the coin about its center of mass? (To

coin about its center of mass? (To find the moment of inertia about find the moment of inertia about the axis, seethe axis, see Table 9-1.) Model the coin as a cylinder of length

Table 9-1.) Model the coin as a cylinder of length L L and take the limit asand take the limit as L L approaches zero. (

approaches zero. (bb) What is the coin’s angular momentum (including direction)) What is the coin’s angular momentum (including direction) about a point on

about a point on the tabletop 10 cm the tabletop 10 cm from the from the axis? axis? ((cc) Now the coin’s center of) Now the coin’s center of mass travels at 5.0 cm/s in a straight line e

mass travels at 5.0 cm/s in a straight line east across the tabletop, while spinningast across the tabletop, while spinning the same way as in Part (

the same way as in Part (aa). What is the angular momentum (including direction)). What is the angular momentum (including direction) of the coin about a point on the line of motion of the center of mass? (

of the coin about a point on the line of motion of the center of mass? (d d ) When it) When it is both spinning and sliding, what

is both spinning and sliding, what is the angular momentum of the coin is the angular momentum of the coin (including(including direction) about a point 10 cm north of the line of motion of the center of mass?

direction) about a point 10 cm north of the line of motion of the center of mass?

Picture the Problem

Picture the Problem We can find the total angular momentum of the coin from We can find the total angular momentum of the coin from the sum of its spin and o

the sum of its spin and orbital angular momenta.rbital angular momenta.

((aa) The spin angular momentum of) The spin angular momentum of the coin is:

the coin is:

spin spin spin

spin I I L

==

From Table 9-1, for

From Table 9-1, for L L negligiblenegligible compared to

compared to R R::

2 2 4 4 1 1 MR MR I

==

Substitute for

Substitute for I I to obtain: to obtain: L L_spin_spin

==

₄₄¹¹ MR MR²²ω ω _spin_spin

Substitute numerical values and evaluate Substitute numerical values and evaluate L Lspinspin::

(

( ))( ( ))

11..3333 1010 kgkg mm /s/s rev

rev rad rad 2 2 ss rev 10rev 10 m m 0.0075 0.0075 kg

kg 0.015

0.015 ²² ⁵⁵ ²²

4 4 1 1 spin

spin

⎟⎟ == ×× ⋅⋅

⎠ ⎠ ⎞

⎜⎜ ⎞

⎝

⎛

⎛ ××

==

^π^π ⁻⁻

L L

Use a right-hand rule to establish the direction of Use a right-hand rule to establish the direction of L L_spin_spin

r r

you.

from from away away /s,

/s, m m kg kg 10 10 3 3 ..

1 ⁵⁵ ²²

spin

== ××

⁻⁻

⋅⋅

L L

((bb)The total angular momentum of)The total angular momentum of the coin is the sum of

the coin is the sum of its orbital andits orbital and spin angular momenta:

spin angular momenta:

spin spin orbital

orbital total

total L L L L L

== ++

Substitute numerical values and Substitute numerical values and evaluate

evaluate L Ltotaltotal::

/s /s m m kg kg 10 10 3 3 ..

1 1 0

0 _spin_spin ⁵⁵ ²²

total

== ++

L L

== ××

⁻⁻

⋅⋅

L L

Use a right-hand rule to establish the direction of Use a right-hand rule to establish the direction of L L_total_total

r r

you you from from away away /s,

/s, m m kg kg 10 10 3 3 ..

1 ⁵⁵ ²²

total

== ××

⁻⁻

⋅⋅

L L ((cc) ) Because Because L L_orbital_orbital

==

00::

you you from from away away /s,

/s, m m kg kg 10 10 3 3 ..

1 ⁵⁵ ²²

total

== ××

⁻⁻

⋅⋅

L L

((d d ) When it is both spinning and) When it is both spinning and sliding, the total angular momentum sliding, the total angular momentum of the coin is:

of the coin is:

spin spin orbital

orbital total

total L L L L L

== ++

The orbital angular momentum of The orbital angular momentum of the coin is:

the coin is:

MvR MvR L

L_orbital_orbital

==

The spin angular momentum of the The spin angular momentum of the coin is:

coin is:

spin spin 2 2 4 4 1 1 spin spin spin spin spin

spin I I ω ω MR MR ω ω L

== ==

Substituting for

Substituting for L L_orbital_orbitalandand L L_spin_spinyields:yields: L L_total_total

==

MvR MvR

++

₄₄¹¹ MR MR²²ω ω _spin_spin

Substitute numerical values and evaluate

Substitute numerical values and evaluate L L_total_total::

(

( ))( ( ))( ( ) ) (

( ))( ( ))

you you toward toward pointing

pointing /s,

/s, m m kg kg 10 10 8 8 ..

8 8

rev rev

rad rad 2 2 ss rev 10 rev 10 m m 0075 0075 ..

0 0 kg kg 015 015 ..

0 0

m m 0.10 0.10 m/s

m/s 0.050 0.050 kg

kg 0.015 0.015

2 2 5

2 2 4

4 1 1 total

total

⋅⋅

××

==

⎠ ⎟⎟

⎠ ⎞

⎜⎜ ⎞

⎝

⎛

⎛ ××

++

==

−−

π π L

42 ••

42 •• ((aa) Two stars of masses) Two stars of masses mm11 and and mm22 are located at are located at r r ^r^r₁₁ and and r r ^r^r₂₂ relative to relative to some origin

some origin O,O, as shown in Figure 10-45. They exert equal and oppositeas shown in Figure 10-45. They exert equal and opposite

attractive gravitational forces on each other. For this two-star system, calculate attractive gravitational forces on each other. For this two-star system, calculate the net torque exerted by these internal forces about the origin

the net torque exerted by these internal forces about the origin OO and show that itand show that it is zero only if both forces lie al

is zero only if both forces lie along the line joining the ong the line joining the particles. (particles. (bb)The fact that)The fact that the Newton’s third-law pair of forces are not only equal and oppositely directed the Newton’s third-law pair of forces are not only equal and oppositely directed but also lie along the line connecting the two objects is sometimes called the but also lie along the line connecting the two objects is sometimes called the

strong form of Newton’s third law. Why is it important to add that last phrase?

Hint: Consider what would happen to these two objects if the forces were offset Hint: Consider what would happen to these two objects if the forces were offset from each other.

from each other.

Picture the Problem

Picture the Problem Both the forces acting on the particles exert torques withBoth the forces acting on the particles exert torques with respect to an axis perpendicular to the page and through point O and the net respect to an axis perpendicular to the page and through point O and the net torque about this axis is their vector sum.

torque about this axis is their vector sum.

((aa) The net torque about an axis) The net torque about an axis perpendicular to the page and perpendicular to the page and through point O is given by:

through point O is given by:

2 2 2 2 1 1 1 1 ii

ii net

net r r F F r r F F

r r r r r r r r r r r

== ∑ ∑ == ×× ++ ××

or, because

or, because F F ₂₂ F F ₁₁

r r r

== −−

( (

₁₁ ₂₂

))

₁₁

net

net r r r r F F

r r r r r r r

== −− ××

Because

Because r r ^r^r₁₁

−−

r r ^r^r₂₂ points along points along F F ₁₁

−−

r :: _net_net

== ( (

r r ₁₁

−−

r r ₂₂

)) ××

F F ₁₁

==

r r r r r r r

((bb) If the forces are not along ) If the forces are not along the same line, there will be a nethe same line, there will be a net torque (but still not torque (but still no net force) acting on the system. This net torque would cause the system to

net force) acting on the system. This net torque would cause the system to accelerate angularly, contrary to observation, and hence makes no sense accelerate angularly, contrary to observation, and hence makes no sense physically.

physically.

43 ••

43 •• A 1.8-kg particle moves in a circle of radius 3.4 m. As you look downA 1.8-kg particle moves in a circle of radius 3.4 m. As you look down on the plane of its orbit, it is initially moving clockwise. If we call the clockwise on the plane of its orbit, it is initially moving clockwise. If we call the clockwise direction positive, its angular momentum relative to the center of the circle varies direction positive, its angular momentum relative to the center of the circle varies with time according to

with time according to L L

( (

t t

) ) ==

1010 N N

⋅⋅

−− ( (

4.04.0 N N

⋅⋅

))

t t . (. (aa) Find the magnitude and) Find the magnitude and direction of the torque acting on the particle. (

direction of the torque acting on the particle. (bb) Find the angular velocity of the) Find the angular velocity of the particle as a function of time.

particle as a function of time.

Picture the Problem

Picture the Problem The angular momentum of the particle changes because aThe angular momentum of the particle changes because a net

net torque acts on it. Because we know how the angular momentum depends on torque acts on it. Because we know how the angular momentum depends on time, we can find the net torque acting on the particle by differentiating its time, we can find the net torque acting on the particle by differentiating its angular momentum. We can use a constant-acceleration equation and Newton’s angular momentum. We can use a constant-acceleration equation and Newton’s second law to relate the angular speed of the particle to its angular acceleration.

second law to relate the angular speed of the particle to its angular acceleration.

((aa) The magnitude of the torque) The magnitude of the torque acting on the particle is the rate at acting on the particle is the rate at which its angular momentum which its angular momentum changes:

changes:

dt dt dL

==

net

τ net

Evaluate

Evaluate dLdL//dt dt to obtain: to obtain:

[ [ ( ( )) ]]

m m N N 0 0 ..

4 4

m m N N 0 0 ..

4 4 ss m m N N 10

net 10

net

⋅⋅

−−

==

⋅⋅

−−

⋅⋅

==

t t

dt dt d τ d

Note that, because

Note that, because L L decreases as the particle rotates clockwise, the angular decreases as the particle rotates clockwise, the angular acceleration and the net torque are both upward.

acceleration and the net torque are both upward.

((bb) The angular speed of the particle) The angular speed of the particle is given by:

is given by: _orbital_orbital

orbital orbital orbital

orbital

I I L

==

L ω

Treating the 1.8-kg particle as a Treating the 1.8-kg particle as a point particle, express its moment of point particle, express its moment of inertia relative to an axis through the inertia relative to an axis through the center of the circle and normal to it:

center of the circle and normal to it:

2 2 orbital

orbital MR MR I

==

Substitute for

Substitute for I I _orbital_orbital and and L L_orbital_orbital toto obtain:

obtain:

( ( ))

2 orbital 2

orbital

m m N N 4.0 4.0 ss

m m N N 10 10

R R

⋅⋅

−−

⋅⋅

== ⋅⋅

ω ω

Substitute numerical values and evaluate

Substitute numerical values and evaluate ω ω orbitalorbital::

( ( ))

(

( ))( ( ))

⁰⁰^..⁴⁸⁴⁸^rad/s^rad/s

( (

^0.19^0.19^rad/s^rad/s

))

m m 3.4 3.4 kg kg 8 8 ..

1 1

m m N N 0 0 ..

4 4 ss m m N N 10

10 ₂₂

2 orbital 2

orbital t t t t

−−

⋅⋅ ==

−−

⋅⋅

== ⋅⋅

ω ω

Note that the direction of the angular velocity is downward.

In document Cap10 (Page 25-36)