dyna - beer

download dyna - beer

of 130

  • date post

    03-Apr-2018
  • Category

    Documents

  • view

    221
  • download

    0

Embed Size (px)

Transcript of dyna - beer

  • 7/28/2019 dyna - beer

    1/130

  • 7/28/2019 dyna - beer

    2/130

    Chapter 11 KINEMATICS OF PARTICLES

    x

    PO

    x

    The motion of a particle along astraight line is termed rectilinearmotion. To define the positionPof the particle on that line, we

    choose a fixed origin O and apositive direction. The distancex from O toP, with the

    appropriate sign, completely defines the position of the particle

    on the line and is called the position coordinateof theparticle.

  • 7/28/2019 dyna - beer

    3/130

    x

    PO

    x

    The velocity v of the particle is equal to the time derivative ofthe position coordinatex,

    v =dx

    dtand the accelerationa is obtained by differentiating v withrespect to t,

    a =dv

    dt

    or a =d2x

    dt2

    we can also express a as

    a = vdv

    dx

  • 7/28/2019 dyna - beer

    4/130

    x

    PO

    x

    v =dx

    dta =

    dv

    dt

    or a = d2

    xdt2 a = v dvdxor

    The velocity v and acceleration a are represented by algebraic

    numbers which can be positive or negative. A positive value for

    v indicates that the particle moves in the positive direction, anda negative value that it moves in the negative direction. A

    positive value fora, however, may mean that the particle is truly

    accelerated (i.e., moves faster) in the positive direction, or that

    it is decelerated (i.e., moves more slowly) in the negativedirection. A negative value fora is subject to a similar

    interpretation.

    +-

  • 7/28/2019 dyna - beer

    5/130

    Two types of motion are frequently encountered: uniform

    rectilinear motion, in which the velocity v of the particle is

    constant and

    x =xo + vt

    and uniformly accelerated rectilinear motion, in which the

    acceleration a of the particle is constant and

    v = vo + at

    x =xo

    + vot + at2

    1

    2

    v2 = vo + 2a(x -xo )2

  • 7/28/2019 dyna - beer

    6/130

    x

    O

    xA

    xB

    xB/A

    A B

    When particlesA andB move along the same straight line, the

    relative motion ofB with respect toA can be considered.

    Denoting byxB/Athe relative position coordinate ofB with respect

    toA , we have

    xB =xA +xB/A

    Differentiating twice with respect to t, we obtain

    vB = vA + vB/A aB = aA + aB/A

    where vB/A and aB/A represent, respectively, the relative velocityand the relative acceleration ofB with respect toA.

  • 7/28/2019 dyna - beer

    7/130

    A

    B

    C

    xA

    xB

    xC

    When several blocks are are connected by inextensible cords,

    it is possible to write a linear relation between their position

    coordinates. Similar relations can then be written between

    their velocities and their accelerations and can be used toanalyze their motion.

  • 7/28/2019 dyna - beer

    8/130

    Sometimes it is convenient to use a graphical solution for

    problems involving rectilinear motion of a particle. The graphical

    solution most commonly involvesx - t, v - t, and a - tcurves.

    a

    tv

    tx

    t

    t1 t2

    v1

    v2

    t1 t2

    v2 - v1 = a dtt1

    t2

    x1

    x2

    t1 t2

    x2 -x1 = v dtt1

    t2

    At any given time t,

    v = slope ofx - tcurve

    a = slope ofv - tcurve

    while over any given time interval

    t1 to t2,

    v2 - v1 = area undera - tcurve

    x2 -x1 = area underv - tcurve

  • 7/28/2019 dyna - beer

    9/130

    x

    y

    r

    P

    Po

    O

    v

    s

    The curvilinear motion of a particle

    involves particle motion along a

    curved path. The positionPof the

    particle at a given time is definedby theposition vectorr joining the

    origin O of the coordinate system

    with the pointP.

    The velocity v of the particle is defined by the relation

    v =dr

    dtThe velocity vector is tangent to the path of the particle, and

    has a magnitude v equal to the time derivative of the lengths ofthe arc described by the particle:

    v =ds

    dt

  • 7/28/2019 dyna - beer

    10/130

    x

    y

    r

    P

    Po

    O

    v

    s

    v =dr

    dt

    In general, the acceleration aof the particle is not tangent

    to the path of the particle. It

    is defined by the relation

    v =ds

    dt

    a =dv

    dtx

    y

    r P

    Po

    O

    a

    s

  • 7/28/2019 dyna - beer

    11/130

    x

    y

    zi

    j

    k

    vx

    vy

    vz

    xiyj

    zk

    P

    x

    y

    z

    i

    j

    k

    r

    ax

    ay

    az

    P

    Denoting byx,y, andzthe rectangular

    coordinates of a particleP, the

    rectangular components of velocity and

    acceleration ofPare equal, respectively,to the first and second derivatives with

    respect to tof the corresponding

    coordinates:

    vx =x vy =y vz=z. . .

    ax =x ay =y az=z.. .. ..

    r

    The use of rectangular components is

    particularly effective in the study of the

    motion of projectiles.

  • 7/28/2019 dyna - beer

    12/130

    x

    y

    z

    x

    y

    z

    A

    B

    rA

    rB rB/A

    For two particles A andB moving

    in space, we consider the

    relative motion ofB with respect

    toA , or more precisely, withrespect to a moving frame

    attached toA and in translation

    withA. Denoting by rB/Athe

    relative position vector ofB with

    respect toA , we have

    rB = rA + rB/A

    Denoting by vB/Aand aB/A, respectively, the relative velocityand

    the relative acceleration ofB with respect toA, we also have

    vB = vA + vB/A

    aB = aA + aB/Aand

  • 7/28/2019 dyna - beer

    13/130

    x

    y

    C

    P

    an = en

    O

    v 2

    r

    at = etdv

    dt

    It is sometimes convenient to

    resolve the velocity and acceleration

    of a particlePinto components other

    than the rectangularx,y, andz

    components. For a particlePmoving

    along a path confined to a plane, we

    attach toP the unit vectors et

    tangent to the path and en normal to

    the path and directed toward thecenter of curvature of the path.

    The velocity and acceleration are expressed in terms of tangential

    and normal components. The velocity of the particle is

    v = vet

    The acceleration is

    a = et + env2

    r

    dv

    dt

  • 7/28/2019 dyna - beer

    14/130

    v = vet

    In these equations, v is the speed of the particle and r is theradius of curvature of its path. The velocity vectorv is directed

    along the tangent to the path. The acceleration vectora

    consists of a component at directed along the tangent to thepath and a component an directed toward the center of

    curvature of the path,

    a = et + env2r

    dvdt

    x

    y

    C

    P

    an = en

    O

    v 2

    r

    at = etdv

    dt

  • 7/28/2019 dyna - beer

    15/130

    x

    P

    O

    eq

    q

    r = rer

    erWhen the position of a particle moving

    in a plane is defined by its polar

    coordinates r and q, it is convenient to

    use radial and transverse componentsdirected, respectively, along the

    position vectorr of the particle and in

    the direction obtained by rotating r

    through 90

    o

    counterclockwise. Unitvectors er and eq are attached toPand are directed in the radial

    and transverse directions. The velocity and acceleration of the

    particle in terms of radial and transverse components is

    v = rer+ rqeq. .

    a = (r- rq2)er + (rq + 2rq)eq... .. . .

  • 7/28/2019 dyna - beer

    16/130

    x

    P

    O

    eq

    q

    r = rer

    erv = rer+ rqeq

    . .

    a = (r- rq2)er + (rq + 2rq)eq... .. . .

    In these equations the dots represent differentiation withrespect to time. The scalar components of of the velocity

    and acceleration in the radial and transverse directions are

    therefore

    vr= r vq= rq. .

    ar= r- rq2 aq = rq + 2rq... .. . .

    It is important to note that ar is not equal to the time derivative

    ofvr, and that aq is not equal to the time derivative ofvq.

  • 7/28/2019 dyna - beer

    17/130

  • 7/28/2019 dyna - beer

    18/130

    Chapter 12 KINETICS OF PARTICLES:

    NEWTONS SECOND LAW

    Denoting by m the mass of a particle, by S F the sum, orresultant, of the forces acting on the particle, and by a the

    acceleration of the particle relative to a newtonian frame of

    reference, we write

    S F = maIntroducing the linear momentum of a particle,L = mv,

    Newtons second law can also be written as

    S F = L.

    which expresses that the resultant of the forces acting on a

    particle is equal to the rate of change of the linear momentum

    of the particle.

  • 7/28/2019 dyna - beer

    19/130

    To solve a problem involving the motion of a

    particle, S F = ma should be replaced byequations containing scalar quantities. Using

    rectangular components ofF and a, we have

    SFx = max SFy= may SFz= maz

    x

    y

    P

    an

    O

    at

    x

    y

    z

    ax

    ay

    az

    P

    x

    P

    aq

    O

    ar

    Using tangential and normal components,

    SFt = mat= m dvdtv2

    rUsing radial and transverse components,

    ...

    .. . .SFr = mar= m(r- rq2)

    qr

    SFn = man= m

    SFq = maq= m(rq + 2rq)

  • 7/28/2019 dyna - beer

    20/1