dynamics aviso

Upload: cha-castillo

Post on 02-Jun-2018

223 views

Category:

Documents


0 download

TRANSCRIPT

  • 8/10/2019 Dynamics AVISO

    1/32

    Dynamicsis that branch of mechanics which deals

    with the state of motion of particles and bodiesunder the action of forces.

    A particleis a physical body or portion of a physical

    body which has mass, the dimensions of which arenegligible in terms of its surroundings and its motion.

    A rigid body is any quantity of matter, the particles of

    which do not move relative to each other.

  • 8/10/2019 Dynamics AVISO

    2/32

  • 8/10/2019 Dynamics AVISO

    3/32

    Motion of a Particle

    A) Rectilinear translation- a particle constrained

    moving along a straight-line path.

    B) Curvilinear translationa particle

    constrained moving along a curved path.

    P

  • 8/10/2019 Dynamics AVISO

    4/32

    General Rectilinear Motion

    The linear displacement of a particle at any

    time is its change in position with reference to

    same fixed point.

    Q S

    R

    Linear displacement from Q

    is represented by the vector

    QS.

    -S +S

    0 P

    Sp

    a b

  • 8/10/2019 Dynamics AVISO

    5/32

    Linear Velocity

    Defined to be the first derivative of displacement

    with respect to time, which can be thought of as the

    time rate of change of displacement or the time rate

    of travel.

    -S +S

    0 P

    Sp

    a bP

    S

    Sp

    Velocity is therefore a vector quantity, while speed is a scalar quantity.

  • 8/10/2019 Dynamics AVISO

    6/32

    Case I. Given Displacement as a Function of Time, s=f(t)

    The velocity and acceleration functions can be

    obtained by successive differentiation of s with

    respect to time. Therefore,

  • 8/10/2019 Dynamics AVISO

    7/32

    Case II. Given Velocity as a Function of Time, v = f(t)

    Substitution of the function into v = ds/dt or f(t) = ds/dt,

    which can be solved by separation of variables. Therefore,

    or

    The acceleration can be obtained be differentiating the given

    function with respect to time. Therefore,

  • 8/10/2019 Dynamics AVISO

    8/32

    Case III. Given Acceleration as a Function of Time, a = f(t)

    Substitution of the function into a= dv/dt or f(t) = dv/dt,

    which can be solved by separation of variables. Therefore,

    or

    The displacement function can be obtained by integrating v=

    ds/dt which would be

    or

  • 8/10/2019 Dynamics AVISO

    9/32

    Case IV. Given Velocity as a Function of Displacement, v = f(s)

    Substituting the function into v= ds/dt gives f(s) =

    dv/dt, which can be solved by separation of variables.

    Therefore,

    This would yield t as a function of s; it would therefore

    be necessary to solve for s as a function of t to obtains = g(t). Since s is now a function of t, the velocity and

    acceleration functions can be obtained as in Case I

  • 8/10/2019 Dynamics AVISO

    10/32

    Case V. Given Acceleration as a Function of Velocity, a = f(v)

    Substituting the function into a= ds/dt or f(v) = dv/dt,which can be solved by separation of variables.Therefore,

    This would give t as a function of v and it would benecessary to solve for v as a function of t to obtain v

    =g(t). Since v is now a function of t, the displacementand acceleration functions can be obtained as in CaseII.

  • 8/10/2019 Dynamics AVISO

    11/32

  • 8/10/2019 Dynamics AVISO

    12/32

    Example

    A particle moves along a straight line so that after t seconds itsdisplacement s in meters from a fixed reference point O on the

    line is given by The particle is 4m to

    the left of the origin at t =2s. Determine :

    a)The acceleration of the particle when t = 3s

    b)The displacement during the interval from t =2s to t =5s,

    c)The total distance travelled during the interval from t = 2s to

    t =5s and ;

    a)The average velocity during the interval from t = 2s to t = 5s.

  • 8/10/2019 Dynamics AVISO

    13/32

    Applying case I

    12 m4

    t=2s

    +S

    When t = 2s,

    Therefore, the displacement is positive to the left of the

    origin, and all measurements will be positive when directed

    to the left.

  • 8/10/2019 Dynamics AVISO

    14/32

    (a) The acceleration when t = 3s can be obtained

    from the acceleration expression. Therefore,

    (b) When t=2s;

    When t=5s;

    The displacement during this interval is :

    mS 412)2(48)2(30)2(4 232

    mS 2212)5(48)5(30)5(4 23

    5

    leftthetomsss 1842225

    rightthetodirectedsm

    sm

    a

    2

    2

    /12

    /12

    60)3(24

  • 8/10/2019 Dynamics AVISO

    15/32

    (c) For rectilinear translation, the total distance is

    equal to the magnitude of the displacement

    unless the particle reverses direction of motionduring the time interval. A particle reverses

    motion during an interval if velocity becomes

    zero during the time interval. In this example,the time when velocity is zero can be obtained

    from the velocity expression :

    sorst

    tt

    tt

    tt

    4,1

    0)4)(1(

    045

    04860122

    2

  • 8/10/2019 Dynamics AVISO

    16/32

    (c) cont. During the interval from t=2s to t=4s,

    the magnitude of the velocity is positive; that is,

    the particle is moving to the left, and from t=4sto t=5s, the velocity is negative; that is the

    particle is moving to the right. When t =4s, the

    position of the particle is :mS 4412)4(48)4(30)4(4 234

    ,62

    2240

    4422444

    4524

    SSSSd

    The total distance travelled is the sum of the

    distance travelled in the two directions; that is,

  • 8/10/2019 Dynamics AVISO

    17/32

    (d) The average velocity during t =2s to t=5s can be

    obtained by using :

    25

    25;tt

    ssv

    t

    Sv aveave

    leftthetos

    m625

    422

  • 8/10/2019 Dynamics AVISO

    18/32

    Example

    The horizontal motion of the shaft and plunger is arrested by the

    resistance of the attached disk. The disk moves through the oilbath. The velocity of the plunger is vo in the position P where s

    =0 and t =0, and the velocity varies accordingly to the relations

    v=vo-ks where k is a constant. Derive the expressions for the the

    positions coordinates s, velocity v, and acceleration a in terms ofthe time t.

    dt

    dsksv

    dt

    dsv

    o

    ;

    disk

    shaft

    plunger

    S

    v

  • 8/10/2019 Dynamics AVISO

    19/32

    tvk

    ksvk

    tksvk

    tcuk

    dtuk

    duk

    duds

    kdsdu

    ksvuLet

    cudu

    u

    RECALL

    dtksv

    ds

    o

    ts

    oo

    s

    o

    t

    o

    s

    o

    t

    oo

    )ln1

    ()ln(1

    ln1

    ln1

    )(

    ln1

    :

    0

    0

    0

    kt

    o

    kt

    o

    kto

    o

    kt

    o

    o

    kt

    o

    kt

    oo

    kt

    o

    o

    u

    u

    o

    o

    ekva

    evv

    ek

    v

    vevk

    s

    kvevks

    evksv

    ev

    ksv

    ue

    ueRECALL

    ktv

    ksv

    1

    1

    1

    )ln(

    ;:

    )ln(

    ln

  • 8/10/2019 Dynamics AVISO

    20/32

    Example

    The resistance to motion of a particle in air is

    approximately proportional to the square of its velocityv for speeds not exceeding 150 m/s. Thus, the

    deceleration is given by the expression a = -kv2, where k

    is taken to be a constant whose numerical value

    depends on the prevailing air conditions and the shape,

    roughness, and mass. If a particle which moves in a

    horizontal straight line is fired with an initial velocity of

    50m/s for a condition where k =1/1000m-1, in whatdistance and elapsed time after firing will the velocity

    be reduced to 8 m/s?

  • 8/10/2019 Dynamics AVISO

    21/32

    dt

    dvv

    dt

    dva 2

    100

    1;

    50

    15050100

    1001

    5050

    100

    1

    50

    11

    5011

    1001

    1

    100

    1

    100

    1

    500

    50 20

    vtv

    tv

    v

    tv

    vt

    vt

    v

    dvdt

    vt

    vt

    tv

    tv

    vtv

    vtv

    vtv

    vtv

    2

    100

    )2(100

    2100

    2100

    )50(2

    50

    15050100

  • 8/10/2019 Dynamics AVISO

    22/32

    Displacement therefore ;

    mt

    s

    ts

    ts

    dtt

    ds

    dt

    ds

    t

    dt

    dsv

    ts

    ts

    ,2

    2ln100

    2ln100)2ln(100

    )2ln(100

    2

    100

    2

    100

    00

    00

    When speed has been reduced to 8 m/s

    s5.1012

    1008

    Displacement in this time is : ms 26.1832

    5.102ln100

  • 8/10/2019 Dynamics AVISO

    23/32

    Rectilinear Motion with Uniform Acceleration

    (Equation 1.6)

    (Equation 1.7)

  • 8/10/2019 Dynamics AVISO

    24/32

  • 8/10/2019 Dynamics AVISO

    25/32

    Three Fundamental Equations in Rectilinear

    Motion with Uniform Acceleration

    For a Free falling body:

    Neglect the frictional resistance effects of air on the particles

    Acceleration due to gravity varies for different locations It can be observed that the value of g decreases as the

    elevation above sea level increases

    Assume that g has a constant value of 9.8m/s^2 directed

    downward

  • 8/10/2019 Dynamics AVISO

    26/32

    Case 1 Case 2

    a =-g

    So =0

    Vo

    +S

    So =0

    Vo

    +S

  • 8/10/2019 Dynamics AVISO

    27/32

    Case 3

    a =-g

    So =0

    Vo

    +S

  • 8/10/2019 Dynamics AVISO

    28/32

    Example

    A point moving with constant acceleration travels 33m

    in the half second which elapses after the second of itsmotion and 198m in the eleventh second of its motion.

    Find its initial velocity and acceleration.

    Solution:

    t=0s t=2s t=2.5s t11=11s

    S11

    S10 198 m

    S2 33 m

    S2.5

  • 8/10/2019 Dynamics AVISO

    29/32

  • 8/10/2019 Dynamics AVISO

    30/32

    Example

    A stone is dropped from a balloon which rises vertically

    at a constant rate of 4 seconds from the ground. Thestone reaches the ground in 10 seconds. Find the

    velocity and the height of the balloon when the stone is

    dropped

    Solution:

    a =0

    So =0

    Vo

    +S h

  • 8/10/2019 Dynamics AVISO

    31/32

    a =g

    So =0

    Vo

    +S

  • 8/10/2019 Dynamics AVISO

    32/32

    Flight of a Projectile