Lecture 5 Body Kinetics OK PDF

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Prof Dinh Van Phong...

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Chapter 5. Planar Kinetics of a Rigid Body

111

Chapter 5

Planar Kinetics of a Rigid Body Chapter Objectives •

To introduce the methods used to determine the mass moment of inertia of a body.

•

To develop the planar kinetic equations of motion for a symmetric rigid body

•

To discuss applications of these equations to bodies undergoing translation, rotation about a fixed axis, and general plane motion.

•

To develop formulations for the kinetic energy of a body, and define the various ways a force and couple do work.

•

To apply the principle of work and energy to solve rigid-body planar kinetic problems that involve force, velocity, and displacement.

•

To show how the conservation of energy can be used to solve rigid-body planar kinetic problems.

•

To develop formulations for the linear and angular momentum of a body.

•

To apply the principles of linear and angular impulse and momentum to solve rigidbody planar kinetic problems that involve force, velocity, and time.

•

To discuss application of' the conservation of momentum.

5.1 Introduction In this chapter we will study the kinetics of rigid bodies. The kinetics of rigid bodies treats the relationships between the forces which act upon the bodies from sources external to their boundaries and the corresponding translational and rotational motions of the bodies. For this purpose a body, which can be approximated as a thin slab with its motion confined to the plane of the slab, will be treated under category of plane motion. The plane of motion will contain the mass center, and all forces which act on the body will be projected onto the plane of motion. In order to determine the state of rigid-body plane motion we will need two force equations and one moment equation. Recall that two equations of motion were required to define the plane motion of a particle whose motion has only two linear components. Similarly to the particle kinetics we will develop several approaches for study the kinetics of bodies. Basis is the isolation of the body or system to be analyzed. For problems involving the instantaneous relationships among forces, mass and acceleration or momentum, the body or

Chapter 5. Planar Kinetics of a Rigid Body

112

system should be explicitly defined by isolating it with its free-body diagram. When the principles of work and energy are employed, an active force diagram which shows only those external forces that do work on the system may be used in lieu of the free-body diagram. In following we will investigate three methods. Section 5.3 relates the forces and moments to the instantaneous linear and angular accelerations. Section 5.4 treats the solution of problems by the method of work and energy. Section 5.5 covers the methods of impulse and momentum. Also, the three types of motion, namely, translation, fixed-axis rotation, and general plane motion will be treated. In the kinetics of rigid bodies which have angular motion, it is necessary to introduce a property of the body which accounts for the radial distribution of its mass with respect to a particular axis of rotation normal to the plane of motion. This property is known as the mass moment of inertia of the body and it is the topic of section 5.2.

5.2 Moment of Inertia of a Rigid Body 5.2.1 Moment of Inertia Since a body has a definite size and shape, an applied nonconcurrent force system may cause the body to both translate and rotate. The translational aspects of the motion were studied with the motion of particles in Chapter 3. These aspects are governed by the equation F = ma. It will be shown later in this chapter that the rotational aspects, caused by a moment M, are governed by an equation of the form M = Iα. The symbol I in this equation is termed the moment of inertia. By comparison, the moment of inertia is a measure of the resistance of a body to angular acceleration (M = Iα) in the same way that mass is a measure of the body's resistance to acceleration (F = ma). In order to solve rotational problems engineers should be able to calculate this property. The flywheel on the engine of this tractor has a large moment of inertia about its axis of rotation. Once it is set into motion, it will be difficult to stop, and this in turn will prevent the engine from stalling and instead it will allow it to maintain a constant power.

Fig. 5.1 We define the moment of inertia as the integral of the "second moment" about an axis of all the elements of mass dm which compose the body. Therefore, if r is the perpendicular distance from the z axis to the arbitrary element dm, we can calculate the body's moment of inertia about the z axis as

Chapter 5. Planar Kinetics of a Rigid Body

2 I =  r dm

113

(5.1)

m

It is clear that the value of I is different for each axis about which it is computed since the formulation involves r. In the study of planar kinetics, the axis which is generally chosen for analysis passes through the body's mass center G and is always perpendicular to the plane of motion. The moment of inertia computed about this axis will be denoted as IG. Note that the mass moment of inertia is always a positive quantity. Common unit used for its measurement is kgm2. If the body consists of material having a variable density, ρ= ρ(x, y, z), the elemental mass dm of the body may be expressed in terms of its density and volume as dm = ρdV. Hence, the body's moment of inertia is computed using volume elements for integration as I =  r 2 dV

(5.2)

V

In the special case of ρ being a constant, this term may be factored out of the integral, and the integration is then purely a function of geometry, I =   r 2dV

(5.3)

V

When the elemental volume chosen for integration has infinitesimal dimensions in all three directions, e.g., dV= dxdydz, Fig. 5.2, the moment of inertia of the body must be determined using triple integration. The integration process can, however, be simplified to a single integration provided the chosen elemental volume has a differential size or thickness in only one direction. Shell or disk elements are often used for this purpose.

Fig. 5.2 Now, in order to show the algorithm we will consider only symmetric bodies having surfaces which are generated by revolving a curve about an axis. An example of such a body which is generated about the z axis is shown in Fig. 5.2. Two types of differential elements can be chosen. If a shell element having a height z, radius r = y, and thickness dy is chosen for integration. Fig. 5.3a, then the volume is dV = (2πy)(z)dy. This element may be used in Eq. (5.2) or (5.3) for determining the moment of inertia I of the body about the z axis, since the entire element, due to its thinness, lies at the same perpendicular distance r = y from the z axis. If a disk element having a radius r = y and a thickness dz is chosen for integration, Fig. 5.3b, the volume is dV = (πy2)dz. This element is finite in the radial direction, and consequently its parts do not all lie at the same radial distance r from the axis. As a result,

Chapter 5. Planar Kinetics of a Rigid Body

114

Eq. (5.2) or (5.3) cannot be used to determine I directly. Instead, to perform the integration it is first necessary to determine the moment of inertia of the element about the z axis and then integrate this result. z

y x

y

y

dy

(a)

(b) Fig. 5.3

It is worth to make a remark here that another property of the body, which measures the symmetry of the body's mass with respect to a coordinate system, is the product of inertia. This property applies to the three dimensional motion of a body. The product of inertia for a differential element dm is defined with respect to a set of two orthogonal planes as the product of the mass dm and the perpendicular distances from the planes to the element. For examples, with respect to the x-z and y-z planes, the product of inertia dIxy for the element is dIxy = xydm By integrating over the entire mass the product of inertia of the body with respect to the x-z and y-z planes may be express as I xy = I yx =

 xydm

(5.4)

m

We can define the product of inertia of the body for other combinations of planes Ixz, Iyz. Occasionally, the moment of inertia of a body about a specified axis is reported in handbooks using the radius of gyration k. This value has units of length, and when it and the body's mass m are known, the body's moment of inertia is determined from the equation I = mk2

(5.5)

k = I /m

(5.6)

or

Note the similarity between the definition of k in this formula and r in the equation dI = r2dm which defines the moment of inertia of an elemental mass din of the body about an axis. EXAMPLE 5.1 Determine the moment of inertia of the cylinder shown in Fig. 5.4 about the z axis. The density of the material, ρ, is constant.

Chapter 5. Planar Kinetics of a Rigid Body

115

Solution This problem may be solved using the shell element with a height z, radius r and thickness dr. This leads to the single integration. Really, the volume of the element is dV = (2πr)(h)dr. Therefore the mass of the element is dm = ρ (2πr)(h)dr. Since the entire element lies at the same distance r from the z axis, the moment Fig. 5.4

of inertia of the element about the z axis is dIz = r2 dm = ρ 2πhr3dr Integrating over the entire region of the cylinder yields R

Iz =

2 3  r dm =  2 h  r dr = 0

m

 2

R4 h

It is easy to show that the mass of the cylinder is R

m=

2  dm =  2 h  rdr =  R h 0

m

Therefore we get the result Iz =

1 mR 2 2

Fig. 5.5

5.2.2 Parallel-Axis Theorem If the moment of inertia of the body about an axis passing through the body's mass center is known, then the moment of inertia about any other parallel axis may be determined by using the parallel-axis theorem. This theorem can be derived by considering the body shown in Fig. 5.5. The z axis passes through the mass center G, whereas the corresponding parallel z axis lies at a constant distance d away. Selecting the differential element of mass dm which is located at point (x, y), and using the Pythagorean theorem, we can express the moment of inertia of the body about the z axis as Iz =

 r dm =  [(d + x ) 2

2

m

=

 (x  m

+ y 2 ]dm

m 2

+ y  2)dm + 2d  x dm + d 2  dm m

m

Clearly, the first integral represents IG. The second integral equals zero, since the z axis passes through the body's mass center, i.e.,

 x dm

= x G m = 0 since x G = 0. Finally. the

m

third integral represents the total mass m of the body. Hence, the moment of inertia about the z

Chapter 5. Planar Kinetics of a Rigid Body

116

axis is Iz = IG + md2

(5.7)

In this expression IG is the moment of inertia about the z axis passing through the mass center G, m is the mass of the body and d is the perpendicular distance between the parallel axes z and z. If a body is constructed of a number of simple shapes such as disks, spheres, and rods, the moment of inertia of the body about any axis z can be determined by adding algebraically the moments of inertia of all the composite shapes computed about the z axis. Algebraic addition is necessary since a composite part must be referred as a negative quantity if it has already been counted as a piece of another part, for example, a "hole" subtracted from a solid plate. The parallel-axis theorem is needed for the calculations if the center of mass of each composite part does not lie on the z axis. In such case we can calculate I =

 (I

Gi

+ m id i2 )

(5.8)

i

Here IGi for each of the composite parts is computed by integration or can be determined from a table, commonly given in engineering handbooks.

5.3 Planar Kinetics of Rigid Bodies: Force and Acceleration 5.3.1 General Equations of Motion In the following analysis we will limit our study of planar kinetics to rigid bodies which, along with their loadings, are considered to be symmetrical with respect to a fixed reference plane. In this case the path of motion of each particle of the body is a plane curve parallel to a fixed reference plane. Since the motion of the body may be viewed within the reference plane, all the forces (and couple moments) acting on the body can then be projected onto the plane. Consider a rigid body acted upon by several external forces. These external forces represent the effect of gravitational, electrical, magnetic, or contact forces between adjacent bodies. Since this force system has been considered previously in Chapter 3 for the analysis of a system of particles, we may assume the body to be made of a large number of particles and apply the results obtained in Chapter 3. We recall Eq. (3.6) and write ΣF = maG

(5.9)

Here ΣF is the sum of all external forces, m is the mass of body and aG is the acceleration of the mass center. This equation is referred to as the translational equation of motion for the mass center of a rigid body. It states that the sum of all the external forces acting on the body is equal to the body's mass times the acceleration of its mass center G.

Chapter 5. Planar Kinetics of a Rigid Body

117

For motion of the body in the x-y plane, the translational equation of motion may be written in the form of two independent scalar equations, namely, ΣFx = m(aG)x

(5.10)

ΣFy = m(aG)y

(5.11)

(a)

(b)

(c)

Fig. 5.6 Now we will determine the effects caused by the moments of the external force system computed about an axis perpendicular to the plane of motion (the z axis) and passing through point P. As shown on the free-body diagram of the ith particle, Fig. 5.6a, Fi represents the resultant external force acting on the particle, and fi is the resultant of the internal forces caused by interactions with adjacent particles. If the particle has a mass mi and at the instant considered its acceleration is ai. then the kinetic diagram is constructed as shown in Fig. 5.6b. If moments of the forces acting on the particle are summed about point P, we require r x Fi + r x fi = r x mi ai or (MP)i = r x mi ai The moments about P can be expressed in terms of the acceleration of point P, Fig.5.6c. If the body has an angular acceleration α, angular velocity ω, then we can write

( M P )i

(

= mir  a P + α  r −  2 r

)

2 = mi  r  a P + r  ( α  r ) −  ( r  r ) 

The last term is zero, since r  r = 0 . Expressing the vectors with Cartesian components and carrying out the cross-product operations yields

( M P ) i k = m i ( x i + y j)  ( a P ) x i + ( a P ) y j + ( x i + y j)   k  ( x i + y j)  2 2 = mi − y ( a P ) x + x ( a P ) y +  x +  y  k  

(M P )i

= m i − y ( a P ) x + x ( a P ) y +  r 2   

Chapter 5. Planar Kinetics of a Rigid Body

118

Letting mi → dm and integrating with respect to the entire mass m of the body, we obtain the resultant moment equation

M

P

 = −   ydm m

  (a P )x +   xdm  m

  2  (a P )y +   r dm  m

  

Here ΣMP represents only the moment of the external forces acting on the body about point P. The resultant moment of the internal forces is zero, since for the entire body these forces occur in equal and opposite collinear pairs and thus the moment of each pair of forces about P cancels The integrals in the first and second terms on the right are used to locate the body's center of mass G with respect to P, Fig. 5.6c, since

 ydm

= ym ,

m

 xdm

= xm

m

Also, the last integral represents the body’s moment of inertia IP computed about the z axis. Therefore we can write ΣMP= - ym (aP)x+ xm (aP)y+ IP α

(5.12)

If point P coincides with the mass center G we can reduce this equation to a simpler form. Since x and y are zero in this case we have ΣMG = IG α

(5.13)

It means that the sum of the moment of all the external forces computed about the body’s mass center G is equal to the product of the moment of inertia of the body about an axis passing through G and the body’s angular acceleration. In Eq. (5.12) we use the acceleration components of point P, however, we can rewrite it in terms of the components of aG and the body’s moment of inertia IG. Using the parallel-axis theorem and the vector relation between aP and aG we have after some derivation: ΣMP = - ym (aG)x + xm (aG)y + IGα

(5.14)

Keep in mind that x , y are the coordinates of the mass center G with respect to P. Eq. (5.14) states that the sum of moments of the external forces about point P is equal to the sum of the kinetic moment of the components of maG about P, plus the kinetic moment of IGα. In a more general form we can write Eq. (5.14) as ΣMP = Σ(MK)P

(5.15)

In this form Σ(MK)P represent the kinetic moments of all components, i.e. maG and IGα. To summarize the analysis of general equations of motion for the plane motion, three independent scalar equations may be written to describe the motion of a symmetrical rigid body.

Chapter 5. Planar Kinetics of a Rigid Body

119

ΣFx = m(aG)x

(5.16a)

ΣFy = m(aG)y

(5.16b)

ΣMG = IG α

(5.16c)

Instead of the last equation we can also use Eq. (5.15). Also, we can use other coordinate system, for example, in many cases the normal and tangential coordinate system are very convenient for the first equations. When applying these equations, one should always draw a free-body diagram. In some problems it may also be helpful to draw the kinetic diagram for the body. This diagram graphically accounts for the terms maG and IGα.

5.3.2 Equations of Motion: Translation When a rigid body undergoes a translation, Fig. 5.7a, all the particles of the body have the same acceleration, so that aG = a. Furthermore, α = 0, in which case the rotational equation of motion applied at point G reduces to a simplified form, namely, ΣMG = 0. Application of this and the translational equations of motion will now be discussed for each of the two types of translation: rectilinear and curvilinear translation. F1 F4

G

M1

F3

d F2

M2

maG

A

G

(a)

(b) Fig. 5.7

As we know when a body is subjected to rectilinear translation, beside of the fact that the velocity and acceleration of particles are the same, all the particles of the body travel along parallel straight-line paths. Therefore we can write the equations of motion in this case as ΣFx = m(aG)x

(5.17a)

ΣFy = m(aG)y

(5.17b)

ΣMG = 0

(5.17c)

The last equation requires that the sum of the moments of all the external forces (and couple moments) computed about the body's center of mass be equal to zero. It is poss...