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Chapter 11

Rolling, Torque, and Angular Momentum

10.9 Rolling Motion of a Rigid Object (Serway)

Angular Momentum

11.1 The Vector Product and Torque

11.3 Angular Momentum of a Rotating Rigid Object

The Vector Product

There are instances where the product of two vectors is another vector.

The vector product of two vectors is called the cross product.

Section 11.

The Vector Product and Torque

The torque vector lies in a direction

perpendicular to the plane formed by

the position vector and the force vector.

The torque is the vector (or cross)

product of the position vector and the

force vector.

Section 11.

"⃗ = $⃗ ×

More About the Vector Product The quantity AB sin q is equal to the

area of the parallelogram formed by

. The direction of is perpendicular to the plane formed by. The best way to determine this direction is to use the right-hand rule.

Section 11. We now give a formal definition of the vector product. Given any two vectors

A

S and B S , the vector product A S

3 B

S is defined as a third vector C S , which has a magnitude of AB sin u, where u is the angle between A S and B S

. That is, if C

S is given by

C

S

5 A

S

3 B

S

its magnitude is C 5 AB sin u (11.3) The quantity AB sin u is equal to the area of the parallelogram formed by A S and

B

S as shown in Figure 11.2. The direction of C S is perpendicular to the plane formed by A S and B S , and the best way to determine this direction is to use the right-hand rule illustrated in Figure 11.2. The four fingers of the right hand are pointed along

A

S and then “wrapped” in the direction that would rotate A S into B S through the angle u. The direction of the upright thumb is the direction of A S

3 B

S

5 C

S

Because of the notation, A S

3 B

S is often read “ A S cross B S ,” so the vector product is also called the cross product. Some properties of the vector product that follow from its definition are as follows:

1. Unlike the scalar product, the vector product is not commutative. Instead,

the order in which the two vectors are multiplied in a vector product is important:

A

S

3 B

S

5 2 B

S

3 A

S

Therefore, if you change the order of the vectors in a vector product, you must change the sign. You can easily verify this relationship with the right- hand rule.

2. If A

S is parallel to B S (u 5 0 or 180 8 ), then A S

3 B

S 5 0 ; therefore, it follows that A S

3 A

S

3. If A

S is perpendicular to B S , then 0 A S

3 B

S

0 5 AB.

duct A

S

3 B

S

is defined as a third vector C

S

, which has a where u is the angle between A

S

and B

S

. That is, if C

S

is C

S

5 A

S

3 B

S

(11.2) C 5 AB sin u (11.3) equal to the area of the parallelogram formed by A

S

and

1. The direction of C

S

is perpendicular to the plane formed st way to determine this direction is to use the right-hand 11.2. The four fingers of the right hand are pointed along in the direction that would rotate A

S

into B

S

through the of the upright thumb is the direction of A

S

3 B

S

5 C

S

. A

S

3 B

S

is often read “ A

S

cross B

S

,” so the vector product is uct. he vector product that follow from its definition are as roduct, the vector product is not commutative. Instead, the two vectors are multiplied in a vector product is A

S

3 B

S

5 2 B

S

3 A

S

(11.4) Chapter 11 Angular Momentum We now give a formal definition of the vector product. Given any two vecto

A

S and B S , the vector product A S

3 B

S is defined as a third vector C S , which has magnitude of AB sin u, where u is the angle between A S and B S

. That is, if C

S given by

C

S

5 A

S

3 B

S

its magnitude is C 5 AB sin u (11. The quantity AB sin u is equal to the area of the parallelogram formed by A S an

B

S as shown in Figure 11.2. The direction of C S is perpendicular to the plane forme by A S and B S , and the best way to determine this direction is to use the right-han rule illustrated in Figure 11.2. The four fingers of the right hand are pointed alon

A

S and then “wrapped” in the direction that would rotate A S into B S through th angle u. The direction of the upright thumb is the direction of A S

3 B

S

5 C

S Because of the notation, A S

3 B

S is often read “ A S cross B S ,” so the vector product also called the cross product. Some properties of the vector product that follow from its definition are follows:

1. Unlike the scalar product, the vector product is not commutative. Instead,

the order in which the two vectors are multiplied in a vector product is important:

A

S

3 B

S

5 2 B

S

3 A

S

Therefore, if you change the order of the vectors in a vector product, you must change the sign. You can easily verify this relationship with the right- hand rule. S S S S Properties of the  vector product 11.1 The torque vector n a direction perpendicular lane formed by the posi- ctor r S and the applied force F S

. In the situation shown,

F S lie in the xy plane, so the is along the z axis. O P y z f r S r S F S F S t!! S S direction of C is perpendicular

Properties of the vector product rule illustrated in Figure 11.2. The four fingers of the right hand are pointed along

A

S and then “wrapped” in the direction that would rotate A S into B S through the angle u. The direction of the upright thumb is the direction of A S

3 B

S

5 C

S

Because of the notation, A S

3 B

S is often read “ A S cross B S ,” so the vector product is also called the cross product. Some properties of the vector product that follow from its definition are as follows:

1. Unlike the scalar product, the vector product is not commutative. Instead,

the order in which the two vectors are multiplied in a vector product is important:

A

S

3 B

S

5 2 B

S

3 A

S

Therefore, if you change the order of the vectors in a vector product, you must change the sign. You can easily verify this relationship with the right- hand rule.

2. If A

S is parallel to B S (u 5 0 or 180 8 ), then A S

3 B

S 5 0 ; therefore, it follows that A S

3 A

S

3. If A

S is perpendicular to B S , then 0 A S

3 B

S

0 5 AB.

4. The vector product obeys the distributive law:

A

S

3 1 B

S

1 C

S

2 5 A

S

3 B

S

1 A

S

3 C

S

5. The derivative of the vector product with respect to some variable such as t is

d dt

1 A

S

3 B

S

d A S dt

3 B

S

1 A

S

d B S dt

where it is important to preserve the multiplicative order of the terms on the right side in view of Equation 11.4. It is left as an exercise (Problem 4) to show from Equations 11.3 and 11.4 and from the definition of unit vectors that the cross products of the unit vectors i

^

, j

^

,

and k

^

obey the following rules:  r lar

rce , he F S S icular B , d by ct

Signs in Cross Products

Signs are interchangeable in cross products

§ and

Section 11.

cross products of the unit vectors i , j ,

k

^

3 k

^

5 0 (11.7a)

i

^

5 k

^

(11.7b)

3 j

^

5 i

^

(11.7c)

k

^

5 j

^

(11.7d)

For example, A

S

3 12 B

S

2 5 2 A

S

3 B

S

and B

S

can be expressed in the follow-

`

A

y

A

z

B

y

B

z

` i

^

1 `

A

z

A

x

B

z

B

x

` j

^

1 `

A

x

A

y

B

x

B

y

` k

^

the right side in view of Equation 11.4.

It is left as an exercise (Problem 4) to show from Equations 11.3 and 11.4 and

from the definition of unit vectors that the cross products of the unit vectors i

^ , j

^ ,

and k

^ obey the following rules:

i

^ 3 i

^ 5 j

^ 3 j

^ 5 k

^ 3 k

^ 5 0 (11.7a)

i

^ 3 j

^ 5 2 j

^ 3 i

^ 5 k

^ (11.7b)

j

^ 3 k

^ 5 2 k

^ 3 j

^ 5 i

^ (11.7c)

k

^ 3 i

^ 5 2 i

^ 3 k

^ 5 j

^ (11.7d)

Signs are interchangeable in cross products. For example, A

S

3 12 B

S

(^2) 5 2 A

S

3 B

S

and i

^ 3

1 2 j

^ 2 5 2 i

^ 3 j

^ .

The cross product of any two vectors A

S

and B

S

can be expressed in the follow-

ing determinant form:

A

S

3 B

S

5 †

i

^ j

^ k

^

A x

A y

A z

B x

B y

B z

† (^5) `

A y

A z

B y

B z

` i

^ (^1) `

A z

A x

B z

B x

` j

^ (^1) `

A x

A y

B x

B y

` k

^

s of 

tors

tor

f tak-

n two

ation

de of

A

S

roduct

having

to the

own.

Using Determinants

The cross product can be expressed as

Expanding the determinants gives

Section 11.

j

^

3 k

^

5 2 k

^

3 j

^

5 i

^

(11.7c)

k

^ 3 i

^ 5 2 i

^ 3 k

^ 5 j

^ (11.7d)

terchangeable in cross products. For example, A

S

3 12 B

S

(^2) 5 2 A

S

3 B

S

j

^ (^2) 5 2 i

^ 3 j

^ .

s product of any two vectors A

S

and B

S

can be expressed in the follow-

inant form:

A

S

3 B

S

5 †

i

^

j

^

k

^

A x

A y

A z

B x

B y

B z

† 5 `

A y

A z

B y

B z

` i

^

1 `

A z

A x

B z

B x

` j

^

1 `

A x

A y

B x

B y

` k

^

Torque Vector Example

Section 11.

304 CHAPTE R 11 ROLLI NG, TORQU E, AN D ANG U L AR M OM E NTU M

the position vector a

tween the directions

Sample Problem 11.02 Torque on a particle due to a force

In Fig. 11-11 a , three forces, each of magnitude 2.0 N, act on a

particle. The particle is in the xz plane at point A given by

position vector , where r " 3.0 m and u " 30 !. What is the

torque, about the origin O , due to each force?

KEY IDEA

Because the three force vectors do not lie in a plane, we

must use cross products, with magnitudes given by Eq. 11-

(t " rF sin f) and directions given by the right-hand rule.

Calculations: Because we want the torques with respect to

the origin O , the vector required for each cross product

is the given position vector. To determine the angle f be-

tween and each force, we shift the force vectors of Fig. 11-

11 a , each in turn, so that their tails are at the origin. Figures

11-11 b , c , and d , which are direct views of the xz plane, show

the shifted force vectors and , respectively. (Note

how much easier the angles between the force vectors and

F

:

3

F

:

2

F ,

:

1

,

r

:

r

:

r

:

means F is directed

:

3

would use ".)

Now, applying Eq

t 1

" rF 1

sin f 1

"

t 2

" rF 2

sin f 2

"

and t 3

" rF 3

sin f

" 6.0 N #m

Next, we use the

the right hand so as t

the two angles betwe

the direction of the to

Fig. 11-11 b ; t

:

2

is dire

t

:

3

is directed as show

tors are shown in Fig.

304 CHAPTE R 11 ROLLI NG, TORQU E, AN D ANG U L AR M OM E NTU M

the position vector are to see.) In Fig. 11

tween the directions of and F is 90!

:

3

r

:

Sample Problem 11.02 Torque on a particle due to a force

In Fig. 11-11 a , three forces, each of magnitude 2.0 N, act on a

particle. The particle is in the xz plane at point A given by

position vector , where r " 3.0 m and u " 30 !. What is the

torque, about the origin O , due to each force?

KEY IDEA

Because the three force vectors do not lie in a plane, we

must use cross products, with magnitudes given by Eq. 11-

(t " rF sin f) and directions given by the right-hand rule.

Calculations: Because we want the torques with respect to

the origin O , the vector required for each cross product

is the given position vector. To determine the angle f be-

tween and each force, we shift the force vectors of Fig. 11-

11 a , each in turn, so that their tails are at the origin. Figures

11-11 b , c , and d , which are direct views of the xz plane, show

the shifted force vectors and , respectively. (Note

how much easier the angles between the force vectors and

F

:

3

F

:

2

F ,

:

1

r

:

r

:

r

:

means F is directed into the page. (For

:

3

would use ".)

Now, applying Eq. 11-15, we find

t 1

" rF 1

sin f 1

" (3.0 m)(2.0 N)(sin 15

t 2

" rF 2

sin f 2

" (3.0 m)(2.0 N)(sin 12

and t 3

" rF 3

sin f 3

" (3.0 m)(2.0 N)(s

" 6.0 N #m.

Next, we use the right-hand rule, pla

the right hand so as to rotate into thr

the two angles between their directions. T

the direction of the torque.Thus t

:

1

is direc

Fig. 11-11 b ; t

:

2

is directed out of the page

t

:

3

is directed as shown in Fig. 11-11 d. Al

tors are shown in Fig. 11-11 e.

F

:

r

:

(t " rF sin f) and directions given by the right-hand rule.

Calculations: Because we want the torques with respect to

the origin O , the vector required for each cross product

is the given position vector. To determine the angle f be-

tween and each force, we shift the force vectors of Fig. 11-

11 a , each in turn, so that their tails are at the origin. Figures

11-11 b , c , and d , which are direct views of the xz plane, show

the shifted force vectors and , respectively. (Note

how much easier the angles between the force vectors and

F

:

3

F

:

2

F ,

:

1

r

:

r

:

A

z

x

y

( a )

F 3

F 1

F 2

θ (^) O

r

x

z

θ = 30°

( b )

φ 1

= 150°

O

F 1

r

x

z

F 1

r

x

r

Cross into F 1

. Torque 1

is into the figure (negative y).

r τ

x

z

θ = 30°

φ 2

= 120°

O

( c )

F 2

r

x

z

F 2

r

F 2

FFF 2

FFFFFF 2

FF

Cross into

Torque 2

is out of

the figure (positive y).

F 2

r.

τ

x

r

z z z

z

and t 3

" rF 3

sin f 3

" (3.0 m)(2.0 N)(sin 90!)

" 6.0 N #m.

Next, we use the right-hand rule, placing t

the right hand so as to rotate into through t

the two angles between their directions. The thu

the direction of the torque.Thus t

:

1

is directed int

Fig. 11-11 b ; t

:

2

is directed out of the page in Fig

t

:

3

is directed as shown in Fig. 11-11 d. All three

tors are shown in Fig. 11-11 e.

F

:

r

:

Because the three force vectors do not lie in a

plane, we must use cross products, with

magnitudes given by ( rF sin !)

a ) A particle at point A is acted on by three forces, each parallel to a coordinate axis.The angle f (used in finding torque) is shown d ( c ) for. ( d ) Torque t : 3 is perpendicular to both and (force F is directed into the plane of the figure). ( e ) The torques. : 3 F : 3 r : F : 2 dditional examples, video, and practice available at WileyPLUS A x y F 3 F 1 F 2 θ (^) O r x θ = 30° ( b ) φ 1 = 150° O F 1 x F 1 x F 1 Cross into F 1

. Torque

1 is into the figure (negative y). r τ x z θ = 30° φ 2 = 120° O ( c ) F 2 r x z F 2 r F 2 FFF 2 FFFFFF 2 FF Cross into Torque 2 is out of the figure (positive y). F 2 r. τ x z F 2 r z θ = 30° τ 3 θ F 3 r x z τ 3 r 3 r 3 r 3 x z τ 3 r 3 rr 333 r y z x 1 2 O τ 3 θ ( e ) τ τ The three torques. Cross into Torque 3 is in the xz plane. F 3 r. τ , TORQU E, AN D ANG U L AR M OM E NTU M the position vector are to see.) In Fig. 11-11 d , the angle be- tween the directions of and F is 90! and the symbol! : 3 r : rque on a particle due to a force agnitude 2.0 N, act on a ne at point A given by d u " 30 !. What is the h force? not lie in a plane, we udes given by Eq. 11- y the right-hand rule. torques with respect to for each cross product rmine the angle f be- force vectors of Fig. 11- e at the origin. Figures s of the xz plane, show F , respectively. (Note : 3 means F is directed into the page. (For out of the page, we : 3 would use ".) Now, applying Eq. 11-15, we find t 1 " rF 1 sin f 1 " (3.0 m)(2.0 N)(sin 150!) " 3.0 N #m, t 2 " rF 2 sin f 2 " (3.0 m)(2.0 N)(sin 120!) " 5.2 N #m, and t 3 " rF 3 sin f 3 " (3.0 m)(2.0 N)(sin 90!) " 6.0 N #m. (Answer) Next, we use the right-hand rule, placing the fingers of the right hand so as to rotate into through the smaller of the two angles between their directions. The thumb points in the direction of the torque.Thus t : 1 is directed into the page in Fig. 11-11 b ; t : 2 is directed out of the page in Fig. 11-11 c ; and t : 3 is directed as shown in Fig. 11-11 d. All three torque vec- F : r :

at 65.0 8 both start from the origin and form two sides

of a parallelogram. Both angles are measured coun-

terclockwise from the x axis. (a) Find the area of

the parallelogram. (b) Find the length of its longer

diagonal.

3. Two vectors are given by A

S

5 i

^ 1 2 j

^

and B

S

5 2 2 i

^ 1 3 j

^ .

Find (a) A

S

3 B

S

and (b) the angle between A

S

and B

S

.

4. Use the definition of the vector product and the defini-

tions of the unit vectors i

^ , j

^

, and k

^

to prove Equations

11.7. You may assume the x axis points to the right, the

y axis up, and the z axis horizontally toward you (not

away from you). This choice is said to make the coordi-

nate system a right-handed system.

5. Calculate the net torque (magnitude and direction) on

the beam in Figure P11.5 about (a) an axis through O

perpendicular to the page and (b) an axis through C

perpendicular to the page.

C

4.0 m

2.0 m

10 N

25 N

O

M

S

(b) What If? Will

applied not at B bu

A

F

1

S

10. A student claims t

that 12 i

^ 2 3 j

^ 1 4 k

^ 2

believe this claim? (

Section 11.2 Analysis M

(Angular Momentum)

11. A light, rigid rod o

ticles, with masses m

ends. The combina

pivot through the c

mine the angular m

origin when the spe

Q/C

M

Angular Momentum Recall that the concept of linear momentum p : of linear momentum are extremely powerful tools. They allow us to predict the outcome of, say, a collision of two cars without knowing the details of the collision.

ance between the fixed point and

given by the right-hand rule: Position t the fingers are in the direction of. nd the palm to be in the direction of. mb gives the direction of!.

:

p

:

r

: A φ φ z x y p (= r × p ) p ( redrawn, with tail at origin) z ( a ) O φ r p ! !

A particle of mass m with linear momentum

p : (= m v ) as it passes through point A in an xy plane. The angular momentum ⃗ " of this particle with respect to the origin O is a vector quantity defined as

! r ! p! r ! mv , ! rp ! ! rmv ! !! rmv sin f your right hand so that Then rotate them arou Your outstretched thum ngular Momentum ecall that the concept of linear momentum and the principle of conservation f linear momentum are extremely powerful tools. They allow us to predict e outcome of, say, a collision of two cars without knowing the details of the col- sion. Here we begin a discussion of the angular counterpart of , winding up in odule 11-8 with the angular counterpart of the conservation principle, which an lead to beautiful (almost magical) feats in ballet, fancy diving, ice skating, and any other activities. Figure 11-12 shows a particle of mass m with linear momentum as passes through point A in an xy plane. The angular momentum of this parti- le with respect to the origin O is a vector quantity defined as (angular momentum defined), (11-18) here is the position vector of the particle with respect to O. As the particle oves relative to O in the direction of its momentum p , position vector : (! mv : ) r : ! : ! r : " p : ! m ( r : " v : ) ! : p : (! mv : ) p : p : where # ⃗ : is the position vector of the particle with respect to O.

Direction.

To find the direction of the angular momentum vector ∶

⃗

in the figure

that the fingers are in the direction of.

around the palm to be in the direction of.

thumb gives the direction of!.

:

p

:

r

Figure 11-12 Defining angular momentum.A

A

φ

φ

z

x

y

p

(= r × p )

p ( redrawn, with

tail at origin)

A

z

x

y

Extension of p

φ

( a )

( b )

O

O

φ

r r

r

p

p

!

!

we slide the vector p :

until its tail is at the

origin O. Then we use

the right-hand rule for

vector products,

sweeping the fingers

from : r into p.

The outstretched

thumb then shows that

the direction of

⃗

is in

the positive direction

of the z axis