Mecânica Geral Aula 03- Momento de Inérciamecanicageral2015.weebly.com/uploads/3/7/5/7/37571293/aula03-mec... · Mecânica Geral Aula 03- Momento de Inércia Bibliograﬁa e Figuras:

Mecânica GeralAula 03- Momento de Inércia

Bibliografia e Figuras: Halliday, Resnick e Walker, vol 1, 8a Ed. LTC Tipler e Mosca, vol 1, 6a Ed.

Prof. Ettore [email protected]

mailto:[email protected]

• Nas aulas anteriores aprendemos:

• A calcular a resultante de um sistema bidimensional de forças e sua orientação por:

• decomposição

• geometricamente: leis dos cosenos e senos

• Calcular o Centro de Massa

• de um sistema com poucas partículas

• de sistemas geométricos simples

• de sistemas onde a massa é continuamente distribuída (integração).

Rotações

• Nos movimentos de translação um corpo move-se em linha reta ou descrevendo curvas

• Nos movimento de rotação, um objeto move-se em torno de um eixo.

• Exemplos: Terra girando ao redor de seu eixo, um cd no seu tocador, uma patinadora no gelo.

• Vamos considerar o movimento de rotação ao redor de um eixo fixo no espaço ou ao redor de um eixo que se move mas sem mudar sua direção

290 | C H A P T E R 9 Rotation

F I G U R E 9 - 1

dsi

riReference line

dθ

Pi

θ i

F I G U R E 9 - 2

(Fred Habegger/Grant Heilman Photography, Inc.)

9-1 ROTATIONAL KINEMATICS: ANGULAR VELOCITY AND ANGULAR ACCELERATION

Every point of a rigid object rotating about a fixed axis moves in a circle whose cen-ter is on the axis and whose radius is the radial distance from the axis of rotationto that point. A radius drawn from the rotation axis to any point on the bodysweeps out the same angle in the same time. Imagine a disk spinning about a fixedaxis perpendicular to the disk and through its center (Figure 9-1). Let be the dis-tance from the center of the disk to the ith particle (Figure 9-2), and let be theangle measured counterclockwise from a fixed reference line in space to a radialline from the axis to the particle. As the disk rotates through an angle the par-ticle moves through a circular arc of directed length such that

9-1

where is measured in radians. If counterclockwise is designated as the positivedirection, then and shown in Figure 9-2, are all positive. (If clockwise isdesignated the positive direction, they are all negative.) The angle the directedlength and the distance vary from particle to particle, but the ratio called the angular displacement is the same for all particles of the disk. For onecomplete revolution, the arc length is and the angular displacement is

The time rate of change of the angle is the same for all particles of the disk, andis called the angular velocity of the disk. The instantaneous angular velocityis an angular displacement of short duration divided by the time. That is,

9-2

DEFINITION—ANGULAR VELOCITY

so is positive if is positive and negative if is negative. All points on the diskundergo the same angular displacement during the same time, so they all have thesame angular velocity. The SI units of are Because radians are dimension-less, the dimension of angular velocity is that of reciprocal time, The magni-tude of the angular velocity is called the angular speed. We often use revolutionsper minute ( or RPM) to specify the angular speed. To convert betweenrevolutions, radians, and degrees, we use

PRACTICE PROBLEM 9-1A compact disk is rotating at What is its angular speed in radians persecond?

Angular acceleration is the rate of change of angular velocity. If the rotation rate ofa rotating object increases, the angular speed increases. (If is increasing, andif the angular velocity is clockwise, then the change in the angular velocity isalso clockwise.) The average angular acceleration vector

9-3

DEFINITION—AVERAGE ANGULAR ACCELERATION

aav !¢v¢t

¢vv

ƒv ƒƒv ƒ

3000 rev>min.

1 rev ! 2p rad ! 360°

rev>min

T "1.rad>s.v

duduv

v !dudt

vv

¢u !siri

!2priri

! 2p rad ! 360° ! 1 rev

¢u2prisi

du,dsi>ri ,ridsi ,

ui ,dsi ,du, ui ,

du

dsi ! ri du

dsi ,du,

ui

ri

• Todo ponto de um corpo rígido que gira ao redor de um eixo fixo descreve um círculo com o centro sobre o eixo e cujo raio é a distância do eixo até onde o ponto se encontra. O elemento de arco descrito pelo ponto P é:

290 | C H A P T E R 9 Rotation

F I G U R E 9 - 1

dsi

riReference line

dθ

Pi

θ i

F I G U R E 9 - 2

(Fred Habegger/Grant Heilman Photography, Inc.)

9-1 ROTATIONAL KINEMATICS: ANGULAR VELOCITY AND ANGULAR ACCELERATION

Every point of a rigid object rotating about a fixed axis moves in a circle whose cen-ter is on the axis and whose radius is the radial distance from the axis of rotationto that point. A radius drawn from the rotation axis to any point on the bodysweeps out the same angle in the same time. Imagine a disk spinning about a fixedaxis perpendicular to the disk and through its center (Figure 9-1). Let be the dis-tance from the center of the disk to the ith particle (Figure 9-2), and let be theangle measured counterclockwise from a fixed reference line in space to a radialline from the axis to the particle. As the disk rotates through an angle the par-ticle moves through a circular arc of directed length such that

9-1

where is measured in radians. If counterclockwise is designated as the positivedirection, then and shown in Figure 9-2, are all positive. (If clockwise isdesignated the positive direction, they are all negative.) The angle the directedlength and the distance vary from particle to particle, but the ratio called the angular displacement is the same for all particles of the disk. For onecomplete revolution, the arc length is and the angular displacement is

The time rate of change of the angle is the same for all particles of the disk, andis called the angular velocity of the disk. The instantaneous angular velocityis an angular displacement of short duration divided by the time. That is,

9-2

DEFINITION—ANGULAR VELOCITY

so is positive if is positive and negative if is negative. All points on the diskundergo the same angular displacement during the same time, so they all have thesame angular velocity. The SI units of are Because radians are dimension-less, the dimension of angular velocity is that of reciprocal time, The magni-tude of the angular velocity is called the angular speed. We often use revolutionsper minute ( or RPM) to specify the angular speed. To convert betweenrevolutions, radians, and degrees, we use

PRACTICE PROBLEM 9-1A compact disk is rotating at What is its angular speed in radians persecond?

Angular acceleration is the rate of change of angular velocity. If the rotation rate ofa rotating object increases, the angular speed increases. (If is increasing, andif the angular velocity is clockwise, then the change in the angular velocity isalso clockwise.) The average angular acceleration vector

9-3

DEFINITION—AVERAGE ANGULAR ACCELERATION

aav !¢v¢t

¢vv

ƒv ƒƒv ƒ

3000 rev>min.

1 rev ! 2p rad ! 360°

rev>min

T "1.rad>s.v

duduv

v !dudt

vv

¢u !siri

!2priri

! 2p rad ! 360° ! 1 rev

¢u2prisi

du,dsi>ri ,ridsi ,

ui ,dsi ,du, ui ,

du

dsi ! ri du

dsi ,du,

ui

ri

A rotação é positiva no sentido anti-horário

dsi = rid✓A velocidade angular instantânea, medida em rad/s e a aceleração instantânea medida em rad/s2 são, respectivamente :

! =d✓

dt

↵ =d!

dt=

d2✓

dt2

A velocidade linear de uma partícula do corpo rígido é tangente à trajetória circular, e é escrita como:

vi =dsidt

Como dsi = rid✓

vi = rid✓

dt= ri!

Analogamente, a aceleração tangencial desta partícula é:

at =dvidt

= rid!

dt= ri↵

E cada partícula também tem a componente centrípeta da aceleração, dirigida para o centro da trajetória.

acp =v2iri

= ri!2

vi = rid✓

dt= ri!

at =dvidt

= rid!

dt= ri↵

acp =v2iri

= ri!2

Energia Cinética de Rotação

• Podemos entender a energia cinética de um corpo rígido rodando em torno de um eixo fixo como sendo a soma de todas as energias cinéticas individuais das partículas que o constituem.

K =X

i

1

2miv

2i K =

X

i

1

2mi(ri!)

2

K =1

2

X

i

mir2i

!!2

K =1

2

X

i

mir2i

!!2

O termo entre parênteses é definido como o momento de inércia do corpo rígido em relação ao eixo de rotação.

I =X

i

mir2i

O momento de inércia nos diz como a massa do corpo que está rodando está distribuída em torno do eixo de rotação. Seu valor é constante para um dado corpo rígido e para um dado eixo de rotação. Este eixo deve sempre ser especificado.

K =1

2

X

i

mir2i

!!2

I =X

i

mir2i

Podemos escrever

com o auxílio de

Finalmente, a energia cinética de rotação é escrita como:

K =1

2I!2

Lembre-se que r é a distância perpendicular da partícula até o seu eixo de rotação.

[kg.m2]

A rotação de um longo bastão é mais fácil ocorrer ao longo de seu eixo longitudinal do que ao longo de um eixo perpendicular que passa pelo seu centro pois sua massa está distribuída mais perto do eixo longitudinal.

A inércia rotacional ( momento de inércia ) envolve não somente a massa, mas a forma como ela se distribui em um corpo.

Exemplo 1: Calcule os momentos de inércia dos sistemas abaixo. Qual deles tem a maior inércia rotacional?

Exemplo 2: Um objeto é formado por quatro partículas pontuais de mesma massa conectadas por um bastões rígidos de massas desprezíveis. O sistema forma um retângulo de lados 2a e 2b conforme mostrado e roda com velocidade angular em torno de um eixo que passa pelo seu centro conforme mostrado. Calcule a energia cinética de rotação do sistema.

Rotational Kinetic Energy S E C T I O N 9 - 2 | 293

Summing over all the particles and using gives

The sum in the expression farthest to the right is theobject’s moment of inertia I for the axis of rotation.

9-11

MOMENT OF INERTIA DEFINED

The kinetic energy is thus

9-12

KINETIC ENERGY OF ROTATING OBJECT

Example 9-2 A Rotating System of Particles

An object consists of four point particles, each of mass m, connected by rigid massless rodsto form a rectangle of edge lengths 2a and 2b, as shown in Figure 9-3. The system rotates withangular speed about an axis in the plane of the figure through the center, as shown.(a) Find the kinetic energy of this object using Equations 9-11 and 9-12. (b) Check your resultby individually calculating the kinetic energy of each particle and then taking their sum.

PICTURE Because we are given that the objects are point particles, we use Equation 9-11 tocalculate I and then use Equation 9-12 to calculate K.

SOLVE

v

K ! 12 Iv2

I ! ai

mir2i

K ! ai

(12miv

2i ) ! 1

2 a (mir2iv

2) ! 12 (a

i

mir2i )v

2

vi ! riv

F I G U R E 9 - 3

2

r1

m1

m3

m2

m4r3

r2

r4

a

2b

Axis of rotation

(a) 1. Apply the definition of moment of inertia (Equation 9-11), where is the radial

distance from the rotation axis to the particle ofmass mi :

riI ! ©mir2i! m1r

21 " m2r

22 " m3r

23 " m4r

24I ! a

i

mir2I

2. The masses and the distances are given:rimir1 ! r2 ! r3 ! r4 ! a

m1 ! m2 ! m3 ! m4 ! m

3. Substitution gives the moment of inertia: I ! ma2 " ma2 " ma2 " ma2 ! 4ma2

4. Using Equation 9-12, solve for the kinetic energy: 2ma2v2K ! 12 Iv2 ! 1

2 4ma2v2 !

(b) 1. To find the kinetic energy of the ith particle, wemust first find its speed:

Ki ! 12miv

2i

2. The particles are all moving in circles of radius a.Find the speed of each particle:

vi ! riv ! av (i ! 1, Á , 4)

3. Substitute into the Part-(b) step-1 result: Ki ! 12miv

2i ! 1

2ma2v2

4. Each particle has the same kinetic energy. Sum the kinetic energies to get the total:

5. Compare with the Part-(a) result:

CHECK The fact that the two calculations give the same result is a plausibility check.

TAKING IT FURTHER Notice that I is independent of the length b. The moment of inertia de-pends only upon how far from the axis the masses are, and not where along the axis they are.

PRACTICE PROBLEM 9-5 Find the moment of inertia of this system for rotation about anaxis parallel to the first axis but passing through two of the particles, as shown in Figure 9-4. F I G U R E 9 - 4

2 a

2 b

Axis of rotation

! 4(12ma2v2) ! 2ma2v2

K ! a4i!1Ki ! 1

2m1v21 " 1

2m2v22 " 1

2m3v23 " 1

2m4v24

The two calculations give the same result.

The Crab Pulsar is one of the fastest-rotating neutron stars known, but it isslowing down. It appears to blink on (left) and off (right) like the rotatinglamp in a lighthouse, at the fast rate of about 30 times per second, but theperiod is increasing by about This rate of loss in rotational energyis equivalent to the power output of The lost kinetic energyappears as light emitted by electrons accelerated in the magnetic field of thepulsar. (David Malin/ Anglo-Australian Telescope Board.)

100,000 suns.10#5 s>y.

Exemplo 3: Calcule aproximadamente o momento de inércia de um bastão fino de comprimento L e massa M através de um eixo perpendicular ao bastão em uma de suas extremidades. Considere o bastão como constituído por 3 massas pontuais, cada uma dela representando 1/3 do bastão.

Calculando o momento de inércia

• A expressão que deduzimos para o momento de inércia pode ser utilizada quando um sistema é constituído por um número pequeno (que pode ser contado facilmente) de partículas.

• Caso o corpo rígido seja composto por um número incontável de partículas de tal modo que sua distribuição de massa seja contínua, devemos recorrer ao cálculo integral para calcular seu momento de inércia.

I =

Zr2dm

Exemplo 4: Calcule o momento de inércia de um bastão fino de comprimento L e massa M através de um eixo perpendicular ao bastão em uma de suas extremidades

2

Thin cylindrical shell aboutaxis

S ut diameter

parallelepipedh center face

H

Thin cylindrical shell aboutdiameter through center

Thin rod aboutperpendicular line

Thin spherical shell aboutdiameter

1 2 12 ( )4 12( 1 2)

Calculating the Moment of Inertia S E C T I O N 9 - 3 | 295

Example 9-4 Moment of Inertia of a Thin Uniform Rod

Find the moment of inertia of a thin uniform rod of length L and mass M about anaxis perpendicular to the rod and through one end.

PICTURE Use (Equation 9-13) to calculate the moment of inertia aboutthe specified axis. The rod is uniform, which means that for any segment of the rod,the mass per unit length is equal to

SOLVE

1. Draw a sketch (Figure 9-6) showing the rod along the axis with its end at the origin. To calculate I about the y axis, we choose a mass element dm at a distance x from the axis:

!x

M>L.l

I " !r2dm

y

dx

dm = dx

x

LML

x

F I G U R E 9 - 62. The moment of inertia about the y axis is given by the integral: I " "x2 dm

3. To compute the integral, we first relate dm to dx. Express dm interms of the linear mass density and dx:l

dm " ldx "MLdx

4. Substitute and perform the integration. We choose integrationlimits so that we integrate through the mass distribution in thedirection of increasing x: 1

3ML2"

ML

13x3 ` L

0"MLL3

3"

I " "x2 dm " "L

0x2 MLdx "

ML "

L

0x2 dx

CHECK This result is in good agreement with the approximate result obtained in Example 9-3.

TAKING IT FURTHER The moment of inertia about the z axis is also and that aboutthe x axis is zero (assuming that all of the mass is a negligible distance from the x axis).

We can calculate I for continuous objects of various shapes, again usingEquation 9-13 (Table 9-1). Some of these calculations are done here.

13ML2,

Table 9-1 Moments of Inertia of Uniform Bodies of Various Shapes

*A disk is a cylinder whose length L is negligible. By setting the above formulas for cylinders hold for disks.L " 0,

Teorema dos eixos paralelos (Teorema de Steiner)

• O problema é encontrar o momento de inércia I de um corpo de massa M em relação a um dado eixo.

• Sempre podemos utilizar a definição utilizando integração

• Conhecendo o momento de inércia ICM do corpo em relação a um eixo paralelo que passa pelo centro de massa do corpo, podemos facilitar o cálculo.

I = ICM +Mh2

Onde ICM é o momento de inércia em relação ao eixo que passa pelo CM e h é a distância do eixo considerado ao eixo que passa pelo CM.

Try It Yourself


THE PARALLEL-AXIS THEOREMWe can often simplify the calculation of moments of inertia for various objects byusing the parallel-axis theorem, which relates the moment of inertia about an axisthrough the center of mass to the moment of inertia about a second, parallel axis(Figure 9-10). Let I be the moment of inertia, and let be the moment of inertiaabout a parallel axis through the center of mass. In addition, let M be the total massof the object and let h be the distance between the two axes. The parallel-axistheorem states that

9-14PARALLEL-AXIS THEOREM

Example 9-2 and the Practice Problem following it illustrate a special case of thistheorem with and

Example 9-5 Applying the Parallel-Axis Theorem

A thin uniform rod of mass M and length L on the x axis (Figure 9-11) has one end at the ori-gin. Using the parallel-axis theorem, find the moment of inertia about the axis, which isparallel to the y axis, and through the center of the rod.

PICTURE Here you know that about one end (see Example 9-4) and want to findUse the parallel-axis theorem with

SOLVE

Cover the column to the right and try these on your own before looking at the answers.

h ! 12L.Icm.

I ! 13ML2

y"

Icm ! 4ma2.h ! a,M ! 4m,

I ! Icm #Mh2

Icm

h

cmcm

F I G U R E 9 - 1 1

y

x

L2

cm

y ’

Steps Answers

1. Apply the parallel-axis theorem to write I about the end interms of Icm. Iy ! Iy" #M(1

2L)2

I ! Icm #Mh2

2. Substitute, using for for and solve for Icm.Iy ,Iy , Icm13ML2 1

12ML2!Icm ! Iy $Mh2 ! 13ML2 $M(1

2L)2

CHECK Calculate the moment of inertia by direct integration. This calculation is the sameas the calculation in Example 9-4 except that the integration limits are from to Theresult is

which is the same as the step-2 result.

TAKING IT FURTHER The step-2 result is only 25% of the result gotten in Example 9-4,where the uniform rod is rotated about an axis through one end.

PROOF OF THE PARALLEL-AXIS THEOREMTo prove the parallel-axis theorem, we start with an object (Figure 9-12) that is ro-tating about a fixed axis, one that does not pass through the center of mass. The ki-netic energy K of such an object is given by (Equation 9-12), where I is the mo-ment of inertia about the fixed axis. We saw in Chapter 8 (Equation 8-7) that the ki-netic energy of a system can be written as the sum of its translational kinetic en-ergy ( ) and the kinetic energy relative to the center of mass. For an objectthat is rotating, the kinetic energy relative to its center of mass is where Icm

12 Icmv

2,

12Mv2

cm

12 Iv2

I ! !x2 dm !ML !

#L>2$L>2 x2 dx !

ML

13x3 ` #L>2

$L>2 !M3LaL3

8#L3

8b !

112ML2

# 12L.$ 1

2L

*

F I G U R E 9 - 1 2

h

cmvcm

ω

ω

vcm = h

F I G U R E 9 - 1 0 An object rotating aboutan axis parallel to an axis through the centerof mass and a distance h from it.

Exemplo 5: Um corpo consiste de duas partículas de massa m conectadas por um bastão de comprimento L de massa desprezível.

(a) Calcule o momento de inércia do corpo em relação a um eixo perpendicular ao bastão e que passa pelo centro de massa do sistema

(b) Calcule o momento de inércia do corpo em relação a um eixo que passa pela extremidade esquerda e que seja paralelo ao primeiro.

Alguns momentos de inércia- I

2


S ut diameter


H




1 2 12 ( )4 12( 1 2)





SOLVE


!x

M>L.l

I " !r2dm

y

dx

dm = dx

x

LML

x



dm " ldx "MLdx


3ML2"

ML

13x3 ` L

0"MLL3

3"

I " "x2 dm " "L

0x2 MLdx "

ML "

L

0x2 dx




13ML2,



2


S ut diameter


H




1 2 12 ( )4 12( 1 2)





SOLVE


!x

M>L.l

I " !r2dm

y

dx

dm = dx

x

LML

x



dm " ldx "MLdx


3ML2"

ML

13x3 ` L

0"MLL3

3"

I " "x2 dm " "L

0x2 MLdx "

ML "

L

0x2 dx




13ML2,



2


S ut diameter


H




1 2 12 ( )4 12( 1 2)





SOLVE


!x

M>L.l

I " !r2dm

y

dx

dm = dx

x

LML

x



dm " ldx "MLdx


3ML2"

ML

13x3 ` L

0"MLL3

3"

I " "x2 dm " "L

0x2 MLdx "

ML "

L

0x2 dx




13ML2,



2


S ut diameter


H




1 2 12 ( )4 12( 1 2)





SOLVE


!x

M>L.l

I " !r2dm

y

dx

dm = dx

x

LML

x



dm " ldx "MLdx


3ML2"

ML

13x3 ` L

0"MLL3

3"

I " "x2 dm " "L

0x2 MLdx "

ML "

L

0x2 dx




13ML2,



2


S ut diameter


H




1 2 12 ( )4 12( 1 2)





SOLVE


!x

M>L.l

I " !r2dm

y

dx

dm = dx

x

LML

x



dm " ldx "MLdx


3ML2"

ML

13x3 ` L

0"MLL3

3"

I " "x2 dm " "L

0x2 MLdx "

ML "

L

0x2 dx




13ML2,



2


S ut diameter


H




1 2 12 ( )4 12( 1 2)





SOLVE


!x

M>L.l

I " !r2dm

y

dx

dm = dx

x

LML

x



dm " ldx "MLdx


3ML2"

ML

13x3 ` L

0"MLL3

3"

I " "x2 dm " "L

0x2 MLdx "

ML "

L

0x2 dx




13ML2,



Alguns momentos de inércia- II

2


S ut diameter


H




1 2 12 ( )4 12( 1 2)





SOLVE


!x

M>L.l

I " !r2dm

y

dx

dm = dx

x

LML

x



dm " ldx "MLdx


3ML2"

ML

13x3 ` L

0"MLL3

3"

I " "x2 dm " "L

0x2 MLdx "

ML "

L

0x2 dx




13ML2,



2


S ut diameter


H




1 2 12 ( )4 12( 1 2)





SOLVE


!x

M>L.l

I " !r2dm

y

dx

dm = dx

x

LML

x



dm " ldx "MLdx


3ML2"

ML

13x3 ` L

0"MLL3

3"

I " "x2 dm " "L

0x2 MLdx "

ML "

L

0x2 dx




13ML2,



2


S ut diameter


H




1 2 12 ( )4 12( 1 2)





SOLVE


!x

M>L.l

I " !r2dm

y

dx

dm = dx

x

LML

x



dm " ldx "MLdx


3ML2"

ML

13x3 ` L

0"MLL3

3"

I " "x2 dm " "L

0x2 MLdx "

ML "

L

0x2 dx




13ML2,



2


S ut diameter


H




1 2 12 ( )4 12( 1 2)





SOLVE


!x

M>L.l

I " !r2dm

y

dx

dm = dx

x

LML

x



dm " ldx "MLdx


3ML2"

ML

13x3 ` L

0"MLL3

3"

I " "x2 dm " "L

0x2 MLdx "

ML "

L

0x2 dx




13ML2,



2


S ut diameter


H




1 2 12 ( )4 12( 1 2)





SOLVE


!x

M>L.l

I " !r2dm

y

dx

dm = dx

x

LML

x



dm " ldx "MLdx


3ML2"

ML

13x3 ` L

0"MLL3

3"

I " "x2 dm " "L

0x2 MLdx "

ML "

L

0x2 dx




13ML2,



Documents

Mecânica Geral Aula 03- Momento de Inérciamecanicageral2015.weebly.com/uploads/3/7/5/7/37571293/aula03-mec... · Mecânica Geral Aula 03- Momento de Inércia Bibliograﬁa e Figuras: