Mecânica GeralAula 02

Prof. Ettore Baldini-Netobaldini@uninove.br

Centro de Massa

Lançamento de uma bola Lançamento de um taco

• Embora o movimento do taco seja mais complicado devido a cada parte descrever uma trajetória distinta, o taco possui um ponto chamado Centro de Massa, que descreve a trajetória parabólica mais simples.

• As outras partes do taco movem-se ao redor do seu centro de massa

• Formalmente, o centro de massa de um sistema de partículas é o ponto que se move como:

• se toda massa do sistema estivesse concentrada nele

• se todas as forças externas estivessem aplicada nele

Centro de massa de um sistema de duas partículas

xCM =m1x1 +m2x2

m1 +m2

xCM =m1x1 +m2x2

m1 +m2

Se m1 = 0 xCM = x2

Se m2 = 0 xCM = x1

Se m1 = m2 xCM =x1 + x2

2

Este resultado pode ser generalizado para um sistema de N partículas

xCM =m1x1 +m2x2 +m3x3 + ...+mNxN

m1 +m2 +m3 + ...+mN

ou em notação mais compacta,

xCM =1

M

NX

i=1

mixi M =NX

i=1

mi

Exemplo 1: Uma massa de 4.0kg encontra-se na origem do eixo x, enquanto uma massa de 2.0kg está em x=6.0cm. Encontre a posição do centro de massa.

Exemplo 2: Três partículas de massas 1.2kg, 2.5kg e 3.4kg formam um triângulo equilátero de lado a=140cm. Onde fica o centro de massa deste sistema?

Caso as partículas estejam distribuídas em três dimensões podemos escrever

xCM =1

M

NX

i=1

mixi yCM =1

M

NX

i=1

miyi zCM =1

M

NX

i=1

mizi

Usando a linguagem vetorial.

~rCM = xCM ı+ yCM |+ zCM k

~rCM =1

M

NX

i=1

mi~ri

Exemplo 3. Uma molécula de água consiste de um átomo de O e dois átomos de H. A massa do O é de 16u e cada hidrogênio tem massa de 1u. A distância média entre os átomos de H e o átomo de O é de 96pm. Os H’s são separados por um ângulo de 104,5o . Encontre o CM da molécula de água.

The Center of Mass S E C T I O N 5 - 5 | 151

2. Check to see if the mass distribution is composed of highly symmetricsubsystems. If so, then calculate the centers of mass of the individualsubsystems, and then calculate the center of mass of the system bytreating each subsystem as a point particle at its center of mass.

3. If the system contains one or more point particles, place the origin at thelocation of a point particle. (If the particle is at the origin, then )

CHECK Make sure your center-of-mass determinations make sense. In manycases, the center of mass of an object is located near the more massive andlarger part of the object. The center of mass of a multi-object system or anobject such as a hoop may not be located within or on any object.

rSi ! 0.i th

Example 5-14 The Center of Mass of a Water Molecule

A water molecule consists of an oxygen atom and two hydrogen atoms. An oxygen atom hasa mass of 16.0 unified mass units (u), and each hydrogen atom has a mass of 1.00 u. Thehydrogen atoms are each at an average distance of 96.0 pm from the oxygenatom, and are separated from one another by an angle of 104.5°. Find the center of mass of awater molecule.

PICTURE We can simplify the calculation by selecting a coordinate system such that the ori-gin is located at the oxygen atom, with the x axis bisecting the angle between the hydrogenatoms (Figure 5-42). Then, given the symmetries of the molecule, the center of mass will beon the x axis, and the line from the oxygen atom to each hydrogen atom will make an angleof

SOLVE

52.25°.

(96.0 : 10!12 m)

1. The location of the center of mass is given by itscoordinates, and (Equations 5-15 and 5-16):ycmxcm

xcm !gmixiM

, ycm !gmiyiM

+y

+xmO = 16.0 u

96.0 pm

104.5°

52.25°

mH = 1.00 u

mH = 1.00 u1

2

F I G U R E 5 - 4 22. Writing these out explicitly gives:

ycm !mH1yH1 " mH2yH2 " mOyO

mH1 " mH2 " mO

xcm !mH1xH1 " mH2xH2 " mOxO

mH1 " mH2 " mO

3. We have chosen the origin to be the location of theoxygen atom, so both the x and y coordinates of theoxygen atom are zero. The x and y coordinates ofthe hydrogen atoms are calculated from the angle each hydrogen makes with the x axis:

52.25°

yH2 ! 96.0 pm sin (#52.25°) ! #75.9 pm

yH1 ! 96.0 pm sin52.25° ! 75.9 pm

xH2 ! 96.0 pm cos (#52.25°) ! 58.8 pm

xH1 ! 96.0 pm cos52.25° ! 58.8 pm

xO ! yO ! 0

4. Substituting the coordinate and mass values intostep 2 gives :xcm

! 0.00 pmxcm !(1.00 u)(75.9 pm) " (1.00 u)(#75.9 pm) " (16.0 u)(0)

1.00 u " 1.00 u " 16.0 u

! 6.53 pmxcm !(1.00 u)(58.8 pm) " (1.00 u)(58.8 pm) " (16.0 u)(0)

1.00 u " 1.00 u " 16.0 u

5. The center of mass is on the x axis: 6.53 pm in " 0.00jnrScm ! xcmin " ycm jn !

CHECK That can be seen from the symmetry of the mass distribution. Also, the cen-ter of mass is very close to the relatively massive oxygen atom, as expected.

TAKING IT FURTHER The distance 96 pm is read “ninety six picometers,” where pico is theprefix for 10#12.

ycm ! 0

Objetos extensos I

• Para objetos extensos como bolas, tacos, etc, podemos imaginá-los como se fossem constituídos por um número muito grande de partículas.

• Do ponto de vista de cálculo, este fato dá origem a uma distribuição contínua de massa

• Para objetos altamente simétricos, a localização do CM coincide com o centro de simetria do objeto.

• Os CMs de um cilindro ou de uma esfera uniformes coincidem com o centro geométrico da figura.

• Para objetos com um eixo ou plano de simetria, o centro de massa encontra-se em algum lugar sobre este eixo ou plano.

Centro de massa de uma área

O CM de uma área define o centro geométrico dela, através do ponto P(x,y).

Caso esta área tenha uma forma arbitrária com na figura abaixo, temos que x e y são definidos como:

784

A.1 Centroid of an Area

The centroid of an area refers to the point that defines the geometriccenter for the area.If the area has an arbitrary shape,as shown in Fig.A–1a,the x and y coordinates defining the location of the centroid C aredetermined using the formulas

(A–1)

The numerators in these equations are formulations of the “moment” ofthe area element dA about the y and the x axis, respectively, Fig. A–1b;the denominators represent the total area A of the shape.

x = LA x dA

LA dA

y = LA y dA

LA dA

Geometric Propertiesof an AreaA

APPENDIX

Fig. A–1

(a)

y

x

_y

_x

C

A

(b)

y

xx

y

dA

• Em alguns casos, a localização do CM pode ser feita a partir de propriedades de simetria da figura plana.

• Quando a área tiver um eixo de simetria, o CM necessariamente estará localizado sobre este eixo.

• Quando houverem dois eixos de simetria, o CM localiza-se na interesecção dos dois.

Centro de Massa de uma área

A.1 CENTROID OF AN AREA 785

A

The location of the centroid for some areas may be partially orcompletely specified by using symmetry conditions. In cases where thearea has an axis of symmetry, the centroid for the area will lie along thisaxis. For example, the centroid C for the area shown in Fig. A–2 must liealong the y axis, since for every elemental area dA a distance to theright of the y axis, there is an identical element a distance to the left.The total moment for all the elements about the axis of symmetry willtherefore cancel; that is, (Eq. A–1), so that In caseswhere a shape has two axes of symmetry, it follows that the centroid liesat the intersection of these axes, Fig. A–3. Based on the principle ofsymmetry, or using Eq. A–1, the locations of the centroid for commonarea shapes are listed on the inside front cover.

Composite Areas. Often an area can be sectioned or divided intoseveral parts having simpler shapes. Provided the area and location ofthe centroid of each of these “composite shapes” are known, onecan eliminate the need for integration to determine the centroid for theentire area. In this case, equations analogous to Eq. A–1 must be used,except that finite summation signs replace the integrals; i.e.,

(A–2)

Here and represent the algebraic distances or x, y coordinates for thecentroid of each composite part, and represents the sum of the areasof the composite parts or simply the total area. In particular, if a hole, ora geometric region having no material, is located within a compositepart, the hole is considered as an additional composite part having anegative area. Also, as discussed above, if the total area is symmetricalabout an axis, the centroid of the area lies on the axis.

The following example illustrates application of Eq. A–2.

©Ay~x~

x = ©x'A©A y =

©y'A©A

x = 0.1x dA = 0

-x+x

y

x

dA

C

dA

!x "x

C

Fig. A–3

Fig. A–2

A.1 CENTROID OF AN AREA 785

A

The location of the centroid for some areas may be partially orcompletely specified by using symmetry conditions. In cases where thearea has an axis of symmetry, the centroid for the area will lie along thisaxis. For example, the centroid C for the area shown in Fig. A–2 must liealong the y axis, since for every elemental area dA a distance to theright of the y axis, there is an identical element a distance to the left.The total moment for all the elements about the axis of symmetry willtherefore cancel; that is, (Eq. A–1), so that In caseswhere a shape has two axes of symmetry, it follows that the centroid liesat the intersection of these axes, Fig. A–3. Based on the principle ofsymmetry, or using Eq. A–1, the locations of the centroid for commonarea shapes are listed on the inside front cover.

Composite Areas. Often an area can be sectioned or divided intoseveral parts having simpler shapes. Provided the area and location ofthe centroid of each of these “composite shapes” are known, onecan eliminate the need for integration to determine the centroid for theentire area. In this case, equations analogous to Eq. A–1 must be used,except that finite summation signs replace the integrals; i.e.,

(A–2)

Here and represent the algebraic distances or x, y coordinates for thecentroid of each composite part, and represents the sum of the areasof the composite parts or simply the total area. In particular, if a hole, ora geometric region having no material, is located within a compositepart, the hole is considered as an additional composite part having anegative area. Also, as discussed above, if the total area is symmetricalabout an axis, the centroid of the area lies on the axis.

The following example illustrates application of Eq. A–2.

©Ay~x~

x = ©x'A©A y =

©y'A©A

x = 0.1x dA = 0

-x+x

y

x

dA

C

dA

!x "x

C

Fig. A–3

Fig. A–2

Centro de Massa de uma área composta

Muitas vezes, uma área pode ser dividida em várias partes. Conhecendo a posição do CM de cada uma das partes podemos determinar o CM da área total através de:

x =

PxAPA

; y =

PyAPA

;

Os termos dentro da somatória representam as posições dos CMs de cada área composta e o denominador representa a soma das áreas de cada parte, ou a área total.

Atenção, se houver um furo ou uma região geométrica onde não exista material, estes devem ser incluídos como uma parte composta adicional mas com área negativa.

Exemplo 4: Determine o CM da placa de madeira abaixo.

152 | C H A P T E R 5 Additional Applications of Newton’s Laws

cm1

m1

Rod 1

Rod 2

cm2

m2

cmtotal

0.60 m0.20 m

0.80 m

0.40 m

0.60 m

0.80 m

0.40 m

+x

1

2

0.20 m

0.20 m

+y

(b)

(a)

F I G U R E 5 - 4 5

1. Write the x and y coordinates of the centerof mass in terms of and m2 .m1

ycm !1M

(m1ycm1 " m2ycm2)

xcm !1M

(m1xcm1 " m2xcm2)

F I G U R E 5 - 4 4

PICTURE The sheet can be divided into two symmetrical parts (Figure 5-45b). Thecenter of mass of each part is at that part’s geometric center. Let be the mass ofpart 1 and be the mass of part 2. The total mass is The masses areproportional to the areas, where and are the areas of the pieces of mass

and respectively.

SOLVE

Steps

M,m1 , m2 ,AA1 , A2 ,

M ! m1 " m2 .m2

m1

2. Substitute area ratios for the mass ratios.

ycm !A1

Aycm1 "

A2

Aycm2

xcm !A1

Axcm1 "

A2

Axcm2

3. Calculate the areas and the ratios of theareas, using the values from Figure 5-45b. A1

A!

8.09.0

A2

A!

1.09.0

A1 ! 0.32 m2; A2 ! 0.040 m2

4. Write the x and y coordinates of the center-of-mass coordinates for each part byinspection of the figure.

xcm2 ! 0.70 m, ycm2 ! 0.50 mxcm1 ! 0.40 m, ycm1 ! 0.20 m

5. Substitute these results into the step-2result to calculate and ycm.xcm

0.23 m, ycm !0.43 mxcm !

Note that we could also have solved Example5-14 by first finding the center of mass of just thetwo hydrogen atoms. For a system of three parti-cles Equation 5-18 is

The first two terms on the right side of this equa-tion are related to the center of mass of the firsttwo particles :

The center of mass of the three-particle systemcan then be written

So we can first find the center of mass for two of the particles, the hydrogen atoms,for example, and then replace them with a single particle of total mass atthat center of mass (Figure 5-43).

The same technique enables us to calculate centers of mass for more complex sys-tems, for instance, two uniform rods (Figure 5-44). The center of mass of each rodseparately is at the center of the rod. The center of mass of the two-rod system canbe found by modeling each rod as a point particle at its individual center of mass.

m1 " m2

MrScm ! (m1 " m2)rS#cm " m3rS3

m1rS1 " m2rS2 ! (m1 " m2)rS#cm

rS#cm

MrScm ! m1rS1 " m2rS2 " m3rS3

Example 5-15 The Center of Mass of a Plywood Sheet

Find the center of mass of the uniform sheet of plywood shown in Figure 5-45a.

+x

H

H

m1 + m2 = 2.00 u

+yF I G U R E 5 - 4 3 Example 5-14 with the two hydrogenatoms replaced by a single particle of mass

on the x axis at the center of mass ofthe original atoms. The center of mass then falls betweenthe oxygen atom at the origin and the calculated center ofmass of the two hydrogen atoms.

m1 " m2 ! 2.00 u

CHECK As expected, the center of mass of the system is very near the center of mass ofpart 1 (because )

TAKING IT FURTHER Drawing an axis through the geometric centers of parts 1 and 2 andplacing the origin at the geometric center of part 1, would have made locating the center ofmass considerably easier.

m1 ! 8m2 .

152 | C H A P T E R 5 Additional Applications of Newton’s Laws

cm1

m1

Rod 1

Rod 2

cm2

m2

cmtotal

0.60 m0.20 m

0.80 m

0.40 m

0.60 m

0.80 m

0.40 m

+x

1

2

0.20 m

0.20 m

+y

(b)

(a)

F I G U R E 5 - 4 5

1. Write the x and y coordinates of the centerof mass in terms of and m2 .m1

ycm !1M

(m1ycm1 " m2ycm2)

xcm !1M

(m1xcm1 " m2xcm2)

F I G U R E 5 - 4 4

PICTURE The sheet can be divided into two symmetrical parts (Figure 5-45b). Thecenter of mass of each part is at that part’s geometric center. Let be the mass ofpart 1 and be the mass of part 2. The total mass is The masses areproportional to the areas, where and are the areas of the pieces of mass

and respectively.

SOLVE

Steps

M,m1 , m2 ,AA1 , A2 ,

M ! m1 " m2 .m2

m1

2. Substitute area ratios for the mass ratios.

ycm !A1

Aycm1 "

A2

Aycm2

xcm !A1

Axcm1 "

A2

Axcm2

3. Calculate the areas and the ratios of theareas, using the values from Figure 5-45b. A1

A!

8.09.0

A2

A!

1.09.0

A1 ! 0.32 m2; A2 ! 0.040 m2

4. Write the x and y coordinates of the center-of-mass coordinates for each part byinspection of the figure.

xcm2 ! 0.70 m, ycm2 ! 0.50 mxcm1 ! 0.40 m, ycm1 ! 0.20 m

5. Substitute these results into the step-2result to calculate and ycm.xcm

0.23 m, ycm !0.43 mxcm !

Note that we could also have solved Example5-14 by first finding the center of mass of just thetwo hydrogen atoms. For a system of three parti-cles Equation 5-18 is

The first two terms on the right side of this equa-tion are related to the center of mass of the firsttwo particles :

The center of mass of the three-particle systemcan then be written

So we can first find the center of mass for two of the particles, the hydrogen atoms,for example, and then replace them with a single particle of total mass atthat center of mass (Figure 5-43).

The same technique enables us to calculate centers of mass for more complex sys-tems, for instance, two uniform rods (Figure 5-44). The center of mass of each rodseparately is at the center of the rod. The center of mass of the two-rod system canbe found by modeling each rod as a point particle at its individual center of mass.

m1 " m2

MrScm ! (m1 " m2)rS#cm " m3rS3

m1rS1 " m2rS2 ! (m1 " m2)rS#cm

rS#cm

MrScm ! m1rS1 " m2rS2 " m3rS3

Example 5-15 The Center of Mass of a Plywood Sheet

Find the center of mass of the uniform sheet of plywood shown in Figure 5-45a.

+x

H

H

m1 + m2 = 2.00 u

+yF I G U R E 5 - 4 3 Example 5-14 with the two hydrogenatoms replaced by a single particle of mass

on the x axis at the center of mass ofthe original atoms. The center of mass then falls betweenthe oxygen atom at the origin and the calculated center ofmass of the two hydrogen atoms.

m1 " m2 ! 2.00u

CHECK As expected, the center of mass of the system is very near the center of mass ofpart 1 (because )

TAKING IT FURTHER Drawing an axis through the geometric centers of parts 1 and 2 andplacing the origin at the geometric center of part 1, would have made locating the center ofmass considerably easier.

m1 ! 8m2 .

Objetos extensos - II

~rCM =1

M

Z~rdm

150 | C H A P T E R 5 Additional Applications of Newton’s Laws

PRACTICE PROBLEM 5-7A 4.0-kg mass is at the origin and a 2.0-kg mass is on the x axis at Find

We can generalize from two particles in one dimension to a system of many par-ticles in three dimensions. For N particles in three dimensions,

Using more concise notation, this is written

5-15

where again is the total mass of the system. Similarly, in the y and zdirections,

5-16

and

5-17

In vector notation, is the position vector of the ith particle.The position of the center of mass, is defined by

5-18

DEFINITION: CENTER OF MASS

where Now let us consider extended objects, such as balls, baseball bats, and even cars.

We can think of objects such as these as a system containing a very large numberof particles, with a continuous distribution of mass. For highly symmetric objects,the center of mass is at the center of symmetry. For example, the center of mass of auniform sphere or a uniform cylinder is located at its geometric center. For an objectwith a line or plane of symmetry, the center of mass lies somewhere along that lineor plane. To find the position center of mass of an object, we replace the sum inEquation 5-18 with an integral:

5-19

CENTER OF MASS, CONTINUOUS OBJECT

where dm is a small element of mass located at position , as shown in Figure 5-41.(We will examine in detail how this integral is set up after Example 5-15.)

PROBLEM-SOLVING STRATEGY

Solving Center-of-Mass Problems

PICTURE Determining centers of mass often simplifies determinations of themotions of an object or system of objects. Drawing a sketch of the object orsystem of objects is useful when trying to determine a center of mass.

SOLVE1. Check the mass distribution for symmetry axes. If there are symmetry

axes, the center of mass will be located on them. Use existing symmetryaxes as coordinate axes where feasible.

rS

MrScm ! ! rS dm

rScm ! xcmin " ycmjn " zcmkn.

MrScm ! m1rS1 " m2rS2 " Á ! ai

mirSi

rScm,rSi ! xi in " yi jn " zikn

Mzcm ! ai

mizi

Mycm ! ai

miyi

M ! ai

mi

Mxcm ! ai

mixi

Mxcm ! m1x1 " m2x2 " m3x3 " Á " mNxN

xcm.x ! 6.0 cm.

r

y

z

x

dm

F I G U R E 5 - 4 1 A mass element dmlocated at position is used for finding thecenter of mass by integration.

rSo símbolo ∫ representa a integral ( ou uma soma ) sobreos elementos de massa infinitesimais dm

Exemplo 5: Calcule o CM de um bastão fino localizado sobre o eixo x.

The Center of Mass S E C T I O N 5 - 5 | 153

FINDING THE CENTER OF MASS BY INTEGRATIONIn this section, we find the center of mass by integration (Equation 5-19):

We will start by finding the center of mass of a uniform thin rod to illustrate thetechnique for setting up the integration.

Uniform rod We first choose a coordinate system. A good choice fora coordinate system is one with an x axis through the length of therod, with the origin at one end of the rod (Figure 5-46). Shown on thefigure is a mass element dm of length dx a distance x from the origin.Equation 5-19 thus gives

The mass is distributed on the x axis along the interval Tointegrate dm along the mass distribution means the limits of the inte-gral are 0 and L. (We integrate in the direction of increasing x.) The ratio dm/dx isthe mass per unit length so

5-20

where

5-21

Because the rod is uniform, is constant and can be factored from each of theintegrals in Equations 5-20 and 5-21, giving

5-22

and

5-23

Solving Equation 5-23 for gives Thus, for a uniform rod the mass perunit length is equal to the total mass divided by the total length. Substituting for M in Equation 5-22, we complete the calculation and obtain the expected result

Semicircular hoop In calculating the center of mass of a uniform semicircularhoop of radius R, a good choice of coordinate axes is one with the origin at the cen-ter and with the y axis bisecting the semicircle (Figure 5-47). To find the center ofmass, we use (Equation 5-19,) where The semicircularmass distribution suggests using polar coordinates,* for which and

The distance of the points on the semicircle from the origin is With these substitutions, we have

rScm !1M !(xin " yjn)dm !

1M !R(cosuin " sinujn)dm

r ! R.y ! r sinu.x ! r cosu

rS ! xin " yjn.MrScm ! " rS dm

rScm !1lLlinL2

2!

12Lin

lLl !M/L.l

M ! l!L

0dx ! lL

rScm !1Mlin !

L

0xdx !

1MlinL2

2

l

M ! !dm ! !L

0l dx

rScm !1M

in !xdm !1M

in !L

0xldx

dm ! l dx:l,

0 # x # L.

rScm !1M ! rS dm !

1M !xin dm

rScm !1M ! rS dm

r

xdx

dm = λ dx

+x

+y

x = L+z

F I G U R E 5 - 4 6

*

y

x

R θdθ

(a)

cm

(b)

ds = R d θ

y

x

θ

x = R cos θ

y = R sin θ

dm = ds = R dθλλ

R

rcm

F I G U R E 5 - 4 7 Geometry for calculatingthe center of mass of a semicircular hoop byintegration.

* In polar coordinates the coordinates of a point are r and were r is the magnitude of the position vector and is theangle that the position vector makes with the direction."x

urSu,