Hoofstuk 10 - Rotasie• In hierdie hfst. bestudeer ons die rotasie beweging

van starre liggame om ‘n vaste as. Om hierdie tipe beweging te beskryf gaan ons met die volgende nuwe konsepte kennis maak:

• Hoekverplasing (simbool: Δθ)• Hoeksnelheid (simbool: ω )• Hoekversnelling (simbool: α )• Traagheidsmoment (simbool I )• Kineties rotasie energie;• Draaimoment (symbol τ )• Ons gaan probleme moet oplos wat handel oor

bogenoemde konsepte.

Die Rotasie Veranderlikes• In hierdie hoofstuk bestudeer ons die rotasie

beweging van starre voorwerpe om vaste asse.• ‘n Starre voorwerp is een wat kan roteer met al sy

dele vas aanmekaar sonder verandering van vorm.• ‘n Vaste as beteken ‘n as waarom die rotasie

plaasvind sonder dat die as beweeg, ook genoem die rotasie-as.

• Elke deel van die liggaam beweeg in ‘n sirkelwaarvan die middelpunt op die rotasie-as geleë is en elke punt beweeg deur dieselfde hoek tydens ‘n sekere tydinterval.

Beskou die volgende starre figuur:Die voorwerp is in suiwer rotasie omdie z-as. Die posisie van dieverwysingslyn is arbitrêr gekies, maardit is wel loodreg op die rotasie-as. Ditis vas in die voorwerp en roteer saamdaarmee.Die hoek oriëntasie van die lyn is die hoek tussen die lyn en ‘n vaste rigting wat as die nul(zero) hoekposisie gekies word.

In Fig. 10-3, word die hoek oriëntasie relatief tot die positiewe rigting van die x-as gemeet.

rs

Van meetkunde weet ons dat gegee word deur:

(in radiale) (10-1) Hier is s die booglengte langs die sirkel en tussen die x-as en die verwysings-lyn; r is die radius van die sirkel.

Hierdie hoek θ word in radiale (rad) eerder as in omwentelings (rev) of grade gemeet. Die radiaal, wat die verhouding tussen twee lengtes is, is dimensieloos. Omdat die omtrek van ‘n sirkel met radius r gelyk is aan 2r, is daar 2 rad in ‘n volle sirkel: rad

rrrev _223601 0

(10-2)

1 rad = 57,30 = 0,159 rev (11-3)

Hoekverplasing • Indien die liggaam van Fig. 10-3 om die

rotasie as draai soos in Fig. 10-4 en die hoek posisie van die verwysingslyn verander van van 1 na 2, ondergaan die voorwerp ‘n hoekverplasing ,wat gegee word deur: Δθ = θ2 – θ1 (10-4)

• Hierdie definisie van hoekverplasing is geldig vir die liggaam in sy geheel, sowel as vir elke deeltjie van die liggaam.

• Die hoekverplasing in ‘n antikloksgewyse rigting is positief en in ‘n kloksgewyse rigting is negatief.

Hoeksnelheid • Gestel dat die roterende voorwerp

op die hoekposisie 1 tydens tyd t1 is en by hoekposisie 2 tydens tyd t2. Dan word die gemiddelde hoeksnelheid () van die liggaam tydens tydinterval t van t1 tot t2 as volg gedefinieer :

tttavg

12

12 (10-5)

Oombliklike hoeksnelheid, ω:

dtd

tt

0

lim (10-6)

Eenheid v ω: rad/s of rev/s

Hoekversnelling • Indien die hoeksnelheid v ‘n roterende voorwerp

verander met tyd, het die voorwerp ‘n hoekversnelling, α

• Laat 2 en 1 die hoeksnelhede op tye t2 en t1 respektiewelik wees. Die gemiddelde hoekversnelling ( α) van t1 tot t2 , word gedefinieer as: tttavg

12

12 (10-7)

• Die oombliklike hoekversnelling, α:

dtd

tt

0

lim (10-8)

Doen Sample Problems 10-1 & 10-2, p244-246

Eenheid: rad/s2 of rev/s2

Verband tussen liniêre en hoekveranderlikes

• Uit vgl.10-1: s = θr (10-17) • Diff. Vgl.10-17, met r = konstant:

rdtd

dtds

v = ωr (10-18)

• Diff. Vgl.10-18, met r = konstant:r

dtd

dtdv

(10-21)

Consider a point on a rigid body rotating about a fixed axis. At 0 the reference line that connects the origin with point is on the -axis (point ).

Pt

O P x A

Relating the Linear and Angular Variables :

During the time interval , point moves along arc and covers a distance . At the same time, the reference line rotates by an angle .

t P APs

OP

Aθs

O

The arc length and the angle are connected by the equation

where is the distance . The speed of point is , .

sd rds dr OP P v rs r

dt dt dt

Relation between Angular Velocity and Speed :

v rcircumference 2 2 2The period of revolution is given by .

speedr rT T

v r

2T

1Tf

2 f

(10-17)

(10-18)

(10-19) (10-20)

• at = αr met α ≠ 0 (10-22)• Tangeniale versnellingskomponent,

versnelling is raaklyn tot bewegingspad.• ar = v2/r = ω2r met ω ≠ 0 (10-23)• Radiale versnellingskomponent, langs radius

gerig inwaarts tot middelpunt.• Liniêre versnelling van ‘n starre roterende

voorwerp het dus twee komponente

Verband tussen liniêre en hoekveranderlikes

rO

The acceleration of point is a vector that has twocomponents. The first is a "radial" component along the radius and pointing toward point . We have enountered this component in

P

O

The Acceleration :

Chapter 4 where we called it "centripetal" acceleration. Its magnitude is

22

rva rr

The second component is along the tangent to the circular path of and is thus known as the "tangential" component. Its magnitude is

The magnitude of the acceleration vector is

t

P

d rdv da r rdt dt dt

2 2 .t ra a a

ta r

(10-8)

(10-22)

(10-23)

When the angular acceleration is constant we can derive simple expressionsthat give us the angular velocity and the angula

r posi

Rotation with Constant Angular Acceleration :

tion as a function of time. We could derive these equations in the same way we did in Chapter 2. Instead we will simply write the solutions by exploiting the analogy between translational and rotat

Translational Motion Rotational Motion

ional motion using the following correspondence b

etween the two motions.

x

0

2

0

0

2

0

2 20

2 20

(

eq. 1)

(eq. 2)

2

2

2

2

o

o

t

tt

va

v v at

atx x v t

v v a x x

0 (eq. 3)(10-6)

Ori

iv

mi

1 2 3

Consider the rotating rigid body shown in the figure. We divide the body into parts of masses , , ,..., ,....The part (or "element") at has an index and mass .

i

i

m m m mP i m

Kinetic Energy of Rotation :

2 2 21 1 2 2 3 3

The kinetic energy of rotation is the sum of the kinetic 1 1 1energies of the parts: ....2 2 2

K m v m v m v

22

2 2 2 2

1 1 The speed of the th element .2 2

1 1 The term

rotational i

is known as

nertia moment of inerti

2 2

or about the axis of rotation. Tha e axis

i i i i i ii i

i i i ii i

K m v i v r K m r

K m r I I m r

of

rotation be specified because the value of for a rigid body depends on its mass, its shape, as well as on the position of the rotation axis. The rotationalinertia of an object describ

mu

e o

st

s h

I

w the mass is distributed about the rotation axis.2

i ii

I m r212

K I 2 I r dm

(10-31)

(10-32) (10-33)(10-34)

(10-35)

Berekening van die rotasie traagheidsmoment, I2

i ii

I m rHierdie uitdrukking kan gebruik word vir ‘n starre liggaam wat ‘n diskrete massaverspreiding het.

Vir ‘n kontinue massaverspreiding, word die som die integraal van massa inkrimente dm as volg:

2I r dm

In the table below we list the rotational inertias for some rigid bodies.2 I r dm

(10-10)

2comI I Mh (10-36)

Parallelle-As Teorie:

I hang af van die posisie van die rotasie-as.Vir ‘n nuwe rotasie-as, moet die integraal vir I dus oor bereken word. ‘n Eenvoudiger metode maak gebruik van die Parallelle-As Teorie.

Beskou die starre liggaam M getoon in die figuur. Ons aanvaar dat ons weet wat die traagheidsmoment, Icom is vir die rotasie-as wat deur die massamiddelpunt by punt O loodreg op die papier beweeg.

Die traagheidsmoment, I, langs ‘n as deur punt P, parallel aan die as deur O, ‘n afstand h vanaf O, word gegee deur die volgende vergelyking:

A

com

We take the origin to coincide with the center of mass of the rigid body shownin the figure. We assume that we know the rotational inertia

for an axis that i

O

I

Proof of the Parallel - Axis Theorem :

s perpendicular to the page and passes through . O

We wish to calculate the rotational inertia about a new axis perpendicular to the page and passing through point with coordinates , . Consider

an element of mass at point with coordinates

IP a b

dm A x

2 2

2 22

2 2 2 2

, . The distance

between points and is .

.

2 2 . The second

and third integrals are zero. The first

y r

A P r x a y b

I r dm x a y b dm

I x y dm a xdm b ydm a b dm

Rotational Inertia about :P

2 2 2com

2 2

integral is . The term .

Thus the fourth integral is equal to

I a b h

h dm Mh

2

comI I Mh

Doen Sample Problems 10-6 to 10-8, p254-256

In fig. a we show a body that can rotate about an axis through

point under the action of a force applied at point a distance

from . In fig. b we resolve into two com

Tor

ponents, radia

que:

O F P

r O F

l and

tangential. The radial component cannot cause any rotationbecause it acts along a line that passes through . The tangentialcomponent sin on the other hand causes the rotation of the

o

r

t

FO

F F

bject about . The ability of to rotate the body depends on themagnitude and also on the distance between points and .Thus we define as torque The distance is

s

ink n

.now

t

trF rF r

O FF

rF

r P A

as the and it is the

perpendicular distance between point and the vector .The algebraic sign of the torque is assigned as follows:

If a force tends to rotate an objec

moment a

t in the count

rm

O F

F

erclockwise

direction the sign is positive. If a force tends to rotate an object in the clockwise direction the sign is negative.

F

r F (10-13)

(10-39)

(10-41)(10-40)

Draaimoment

For translational motion, Newton's second law connects the force acting on a particle with the resulting acceleration.There is a similar relationship between the torqu

Newton's Second Law for Rotation :

e of a forceapplied on a rigid object and the resulting angular acceleration.

This equation is known as Newton's second law for rotation. We will explore this law by studying a simple body that consists of a point mass at the end

of a massless rod of length . A force is

m

r F

applied on the particle and rotates

the system about an axis at the origin. As we did earlier, we resolve into a tangential and a radial component. The tangential component is responsible for the

F

2

rotation. We first apply Newton's second law for . (eq. 1)The torque acting on the particle is (eq. 2). We eliminate

between equations 1 and 2: .

t t t

t t

t

F F maF r F

ma r m r r mr I

(compare with: )F ma I (10-45)

We have derived Newton's second law for rotation for a special case: a rigid body that consists of a pointmass at the end of a massless rod of length . We willnow

m r

Newton's Second Law for Rotation :

derive the same equation for a general case. O

12

3

i

ri

net

Consider the rod-like object shown in the figure, which can rotate about an axisthrough point under the action of a net torque . We divide the body into parts or "elements" and label them. The

O

1 2 3

1 2 3

1 1 2 2

3 3

elements have masses , , ,..., and they are located at distances , , ,..., from . We apply Newton's secondlaw for rotation to each element: (eq. 1), (eq. 2),

(eq

n

n

m m m mr r r r O

I II

21 2 3 1 2 3

1 2 3 net

. 3), etc. If we add all these equations we get

... ... . Here is the rotational inertia

of the th element. The sum ... is the net torque applied.T

n n i i i

n

I I I I I m r

i

1 2 3he sum ... is the rotational inertia of the body. Thus we end up with the equation

nI I I I I

net I (10-15)(10-42)

Rotational Motion Translational Motion

xv

Analogies Between Translational and Rotational Motion

0 0

2

0 0

2

2

0

2 20

20 0

2

2

2

2

o

o

av v at

atx x v t

v v a x x

t

tt

2 2

2

2

mvK

mF ma

FP Fv

IK

II

P (10-18)