Chapter 12

Rotation of a Rigid Body

Vector (or “cross”) Product

A

B AB sin

Cross Product is a vector perpendicular to the plane of vectors A and B

AP Physics C 3

Cross Product of Vectors

A

B

A

B AB sin

sinPosition Force AB

Right hand rule: Curl your right hand around the center of rotation with the fingers going from the first vector to the second vector and the thumb will be pointing in the torque direction

sinB

sinA

B

A

A×B≠B×A

AP Physics C 4

Cross Product Problem

Find:E =D×C ifC =2NandD=1m

C

D110°

E =CDsinθ 2E = 1 sin 110 =1.88Nm

Csin(110)

1. Clockwise.

2. Counter-clockwise.

3. Not at all.

4. Not sure what will happen.

CoM

Pivot

Which way will it rotate once the support is removed?

Test your Understanding

AP Physics C 6

Torque

If the forces are equal, which will open the heavy door more easily?

Interpretation of torque

Measures tendency of any force to cause rotation

Torque is defined with respect to some origin – must talk about “torque of force about point X”, etc.

Torques can cause clockwise (+) or anticlockwise rotation (-) about pivot point

AP Physics C 8

Torque con’t

AP Physics C 9

Torque con’t

Definition of Torque:

r

F

||r F sin

where is the vector from the reference point (generally either the pivot point or the center of mass) to the point of application of the force . If r and F are not perpendicular then:

r

F

where q is the angle between the vectors and .r

Definition of Torque:

||r F sin

AP Physics C 12

Torque Problem

Adrienne (50 kg) and Bo (90kg) are playing on a 100 kg rigid plank resting on the supports seen below. If Adrienne stands on the left end, can Bo walk all the way to the right end with out the plank tipping over? If not, How far can he get past the support on the right?

AP Physics C 13

Torque Problem con’t

2m 3m 4m

50kg 100kg 90kg

N1 N2

x

Netτ =90x-100( .5)-50( 5)=0

x=3.3

AP Physics C 14

Moments

M1 M2

d1 d2

Suppose we have masses m1 and m2 on the seesaw at distances d1 and d2, respectively, from the fulcrum, when does the seesaw balance?

By Archimedes’ Law of the lever, this occurs whenm1d1 + m2d2 = 0

15

Moments con’t

AP Physics C

M1 M2

x1 x2

If we place a coordinate system so that 0 is at the fulcrum and if we let xi be the coordinate at which is placed then:

m1x1 + m2x2 = m1d1 + m2d2 = 0

16

Moments con’t

AP Physics C

More generally, if we place masses m1, m2, …, mr at points x1, x2, …. , xr, respectively, then the see saw balances with the fulcrum at the origin, if and only if

m1x1 + m2x2 + …+ mrxr = 0

M1M4

x1 x4

M2 M3

x2 x3

M5

x5

17

Moments con’t

AP Physics C

Now, suppose that we place masses m1, m2, … , mr at points x1, x2, … xr, respectively, then where should we place the fulcrum so that the seesaw balances?

The answer is that we place the fulcrum at x-bar where:

m1(x1 - (x-bar) ) + m2(x2 - (x-bar) ) + …+ mr(xr - (x-bar) )= 0

M1M4

x1 x4

M2 M3

x2 x3

M5

x5

18

Moments con’t

AP Physics C

m1(x1 - (x-bar) ) + m2(x2 - (x-bar) ) + …+ mr(xr - (x-bar) )is called the moment about x-bar.

Moment is from the Greek word for movement, not time.

If positive, movement is counter-clockwise, negative it is clockwise.

M1M4

x1 x4

M2 M3

x2 x3

M5

x5x-bar

19

Moments con’t

AP Physics C

M1M4

x1 x4

M2 M3

x2 x3

M5

x5x-bar

1 1 2 2 ... 0r rm x x m x x m x x Suppose that

We want to solve for x

1 1 2 2 1 2... ... 0r r rm x m x m x m x m x m x

1 1 2 2 1 2... ... 0r r rm x m x m x m m m x

1

1

r

i ii

r

ii

m xx

m

20

Center of Mass

AP Physics C

M1M4

x1 x4

M2 M3

x2 x3

M5

x5x-bar

01

1

r

i ii

rtot

ii

m xM

xm

m

AP Physics C 21

Center of Mass con’t

M1

M4

M2

M3

Suppose m1, m2, … , mr are masses located at points (x1, y1), (x2, y2), … , (xr, yr).

The moment about the y-axis is:1

r

y i ii

M m x

The moment about the x-axis is:

1

r

x i ii

M m y

1

1

r

i iyi

rtot

ii

m yM

xm

m

1

1

r

i ii x

rtot

ii

m xM

ym

m

x

y

Center of Mass is ,y x

tot tot

M M

m m

AP Physics C 22

Center of Mass con’t

Now lets find the center of mass of a thin plate with uniform density, ρ.

First we need the mass of the plate: ( ) ( )b

aM Area f x g x dx

23

Center of Mass con’t

AP Physics C

Next we need the moments of the region:

f(x)

g(x)

∆x

1 1 12 2 2( ) ( ) ( ) ( )xM y m f x f x x g x g x x

2 212 ( ) ( )

b

x aM f x g x dx

( ) ( )b

y aM x f x g x dx

( ) ( )yM x m x f x g x x

2 2 2 212 2 2

( ) ( ) ( ) ( ) 1( ) ( )

2( ) ( ) 2 ( ) ( )

b b

ba ax

b b atot

a a

f x g x dx f x g x dxMy f x g x dx

m Af x g x dx f x g x dx

1( ) ( )

by

a

Mx x f x g x dx

M A

To find the center of mass we divide by mass:

AP Physics C 24

Center of Mass Problem

Determine the center of mass of the region bounded by y = 2 sin (2x) and y = 0 on the interval, [0, π/2]

Given the symmetry of the curve it is obvious that x-bar is at π/4.

First find the area. 2

200

2sin(2 ) (2 ) | 2A x dx cos x

2

22 2 2

00

1 1 1 1( ) ( ) sin (2 ) sin 4

2 4 4 2 4

b

a

xy f x g x dx x dx x

A

Using the table of integrals:

AP Physics C 25

Moment of Inertia

MCR

r

F

a rF maT Fr mra

2T mr r mr T IF ma

AP Physics C 26

Moment of Inertia con’t

2b

aI y w y dy

( )M w y dy

AP Physics C 27

Calculating Moment of Inertia

AP Physics C 28

Calculating Moment of Inertia

dm dA

M A

Mdm dA

A

22 dA r dr and A R

2 2

22

M Mdm rdr rdr

R R2

2 32 0

2

2

RM MRI r dm r dr

R

AP Physics C 29

Moment of Inertia con’t

AP Physics C 30

Parallel Axis Theorem

2|| CMI I Md

AP Physics C 31

||-axis Theorem Proof

22 2' ' 2 'I x dm x d dm x dm d x dm d dm 2

CMI I Md

AP Physics C 32

Rotational Kinetic Energy

2 21 11 1 2 22 2 ...rotK m v m v

2 2 2 21 11 1 2 22 2 ...rotK m r m r

2 2 21 12 2i i

i

m r I

212rotK I

212mech cmE I Mgy

AP Physics C 33

Rotational Dynamics

t tF =ma =mrα2

trF =mr α2 =mr α

2Net i i

i

= mr α

Net Iα

AP Physics C 34

Rotation About a Fixed Axis

AP Physics C 35

Bucket Problem

A 2.0 kg bucket is attached to a mass-less string that is wrapped around a 1.0 kg, 4.0 cm diameter cylinder, as shown. The cylinder rotates on an axel through the center. The bucket is released from rest 1.0 m above the floor, How long does it take to reach the floor

AP Physics C 36

Bucket Problem con’t

B yF =ma =T-mg

net =αI=TR

Net =TR

2

TR TR 2Tα= = =

I .5MR MR

y

2T 2Ta =-αR =- R =-

MR M

AP Physics C 37

Bucket Problem Con’t

y

2Ta =-

M 2y-a M

T=

yy

Mama =- -mg

2

2m

y sa =-7.84

21y2Δy= a Δt Δt=0.50s

y

2mga =-

2m+M

AP Physics C 38

Static Equilibrium

NetF =0

Net=0

AP Physics C 39

Statics Problem

A 3.0 m ladder leans against a frictionless wall at an angle of 60°. What is the minimum value of μs, that prevents the ladder from slipping?

x 2 sF =n -f =0

y 1F =n -mg=0

0

Net 1 G 2 2

o o12 2

322

τ =dF -d n

= Lcos60 Mg- Lsin60 n =0

=.5( .75) Mg- 3n

AP Physics C 40

Statics Problem

2 s s 1n =f =μn1n =Mg 2 o

Mgn =

2tan60

s s 1 so

Mgf = =μn μ

2tan60Mg

s o

1μ =

2tan60

sμ ≥0.29

s o

Mgμ =

2tan60Mg

AP Physics C 41

Balance and Stability

θc

h

t/2

θc h

t/2

AP Physics C 42

Rolling Motion

cm cmΔx =2πR =v T cm

2πRv = =Rv

T

AP Physics C 43

Rolling Motion con’t

i cm i,relr =r +r

i cm i,relv =v +v

AP Physics C 44

Rolling Motion con’t

AP Physics C 45

Rolling Motion con’t

AP Physics C 46

Rolling Kinetic Energy

P1

rot,P 2K = I ω2

P cmI =I +MR

221 1cm2 2K= I ω + M Rω

AP Physics C 47

Great Downhill Race

AP Physics C 48

Downhill Race con’t

2 21 1cm cm2 2I ω + Mv =Mgh

2cmI =cMR

2

2 cmv

R

21 12 2cMR ω + M =Mgh

cm

2ghv =

1+c

Particlec =0Spherec =2/5Cylinderc =1/2Hoopc =1

AP Physics C 49

Angular Momentum

p=mv

L=Iω=mrv

L=r×p=r mv sin

AP Physics C 50

Angular Momentum

AP Physics C 51

Angular Momentum

z tL =mrv =r×p

dL d dr dp= r×p = p+r

dt dt dt dtnet=v×p+r×F

net

dL=

dt

AP Physics C 52

Conservation of Angular Momentum

fiL =L2 21 1ffi i2 2m ω = m ωl l2 2ffi iω = ωl l

Two equal masses are at the ends of a mass-less 500 cm long rod. The rod spins at 2.0 Rev/s about an axis through its midpoint. If the rod lengthens to 160 cm, what is the angular velocity

2 2

ifi

f

50ω ω = 2=.20

160ll

revs

AP Physics C 53

Testing Understanding

There is no torque on the buckets so angular momentum is conserved.Increased mass in buckets increases inertia so angular velocity must decrease.

AP Physics C 54

Problem 1

An 18 cm long bicycle crank arm with a pedal at one end is attached to a 20 cm diameter sprocket, the toothed disk around which the chain moves. A cyclist riding this bike increases her pedaling rate from 60 rpm to 90 rpm in 10 s.

a. What is the tangential acceleration of the pedal?b. Want length of chain passes over the top of the sprocket during this interval?

AP Physics C 55

Problem 1 con’t

2Δω 9.43 rad-6.28 rad/sα= = =0.314 rad/s

Δt 10 s

a. Since at = rα, find α first. With 60 rpm = 6.28 rad/s and 90 rpm = 9.43 rad/s:

2

fi i

1θ =θ +ωΔt+ α Δt

2

22fi

1θ -θ =Δθ= 6.28 rad/s 10 s + 0.314 rad/s 10 s

2=78.5 rad

b. Since L = r∆θ, find r∆θ.

The angular acceleration of the sprocket and chain are the same.

2 2ta =rα= 0.18 m 0.314 rad/s =0.057 m/s

The length of chain which has passed over the top of the sprocket is

Length=0.10 m 78.5 rad =7.9 m

AP Physics C 56

Problem 2

A 100 g ball and a 200 g ball are connected by a 30 cm mass-less rigid rod. The balls rotate about their center of mass at 120 rpm. What is the speed of the 100g ball?

cm

100 g 0 cm + 200 g 30 cmx = =20 cm

100 g+200 g

1 cm

2π rad minv =rω=x ω= 0.20 m 120 rev/min =2.5 m/s

rev 60 s

AP Physics C 57

Problem 3

A 300 g ball and a 600 g ball are connected by a 40 cm mass-less rigid rod, The structure rotates about its center of mass at 100 rpm. What is its rotational kinetic energy?

cm

( 300 g) ( 0 cm)+( 600 g) ( 40 cm)x = =26.67 cm

300 g+600 g

2 2cm cm

2 2 2

I=( 300 g) ( x ) +( 600 g) ( 40 cm-x )

=( 0.300 kg) ( 0.2667 m) +( 0.600 kg) ( 0.1333 m) =0.032 kg m

22 2

rot

1 1 100×2πK = Iω = ( 0.032 kg m ) rad/s =1.75 J

2 2 60

AP Physics C 58

Problem 4

A 25 kg solid door is 220 cm tall, 91 cm wide. What is the door’s moment of inertia for (a) rotation on its hinges (b) rotation about a vertical axis inside the door, 15 cm from one edge.

2 21I= 25 kg 0.91 m =6.9 kg m

3

0.91 m

d= -0.15 m =0.305 m.2

2 22 2cm

1I=I +Md = 25 kg 0.91 m + 25 kg 0.305 cm =4.1 kg m

12

AP Physics C 59

Problem 5

An athlete at the gym holds a 3.0 kg steel ball in his hand. His arm is 70 cm long and has a mass of 4.o kg. What is the magnitude of the torque about his shoulder if he holds the ball (a) straight out to his side, parallel to the floor and (b) straight, but 45 degrees below horizontal.

ball arm b b a aτ=τ +τ =mgrsin90°+mgrsin90°

2 2=3.0 kg 9.8 m/s 0.70 m +4.0 kg 9.8 m/s 0.35 m

=34 N m

AP Physics C 60

Problem 5 con’t

ball arm b b a aτ=τ +τ =mgrsin45°+mgrsin45°

2 2=3.0 kg 9.8 m/s 0.70 m 0.707 +4.0 kg 9.8 m/s 0.35 m 0.707

=24 N m

AP Physics C 61

Problem 6

Starting from rest, a 12 cm diameter compact disk takes 3.0 s to reach its operating angular velocity of 2000 rpm. Assume that the angular acceleration is constant. The disk’s moment of inertia is 2.5x10-5 kgm2. (a) How much torque is applied? (b) How many revolutions does it make before reaching full speed?

AP Physics C 62

Problem 6 con’t

a. Using the rotational kinematic equation 1 0 1 0ω =ω +α( t -t )

2π

( 2000 rpm) rad/s=0 rad+α 3.0 s -0 s60

2200πα= rad/s

9

-5 2 2200πτ=Iα=( 2.5×10 kg m ) rad/s

9-3=1.75×10 N m

22 21 0 0 1 0 1 0

1 1 200πθ =θ +ω( t -t )+ α( t -t ) =0 rad+0 rad+ rad/s 3.0 s-0 s

2 2 9100π

=100π rad= revolutions=50 rev2π

b.

AP Physics C 63

Problem 7

The two objects in the figure are balanced on the pivot. What is the distance d? G G1 2

- F - F +P=0 N

G G1 2P= F + F

2 2=1.0 kg 9.8 m/s +4.0 kg 9.8 m/s

=49 N

net =0 Nm

1 2Pd-w( 1.0 m)-w( 1.5 m)=0 Nm

2 249 Nd-1.0 kg 9.8 m/s 1.0 m -4.0 kg 9.8 m/s 1.5 m =0 N

d=1.40 m