Mechanics - hkage.org. Mechanics 1 Vector 1.1 Any vector x 0 y 0 z A A x A y A z 0 r r r r = + +...

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Transcript of Mechanics - hkage.org. Mechanics 1 Vector 1.1 Any vector x 0 y 0 z A A x A y A z 0 r r r r = + +...

  • 1

    Mechanics

    1 Vector

    1.1

    Any vector 000 zAyAxAA zyxrrrr

    ++= (1.1)

    Addition of two vectors:

    000 )()()( zBAyBAxBABAC zzyyxxrrrrrr

    +++++=+= (1.2)

    Dot product of two vectors

    zzyyxx BABABAABBA ++== cosrr

    (1.3)

    It is also referred to as the projection of Ar

    on Br

    , or vise versa.

    Amplitude of the vector

    AAAAAA zyx

    rrr=++ 222|| (1.4)

    1.2

    Position vector of a particle: 000 zzyyxxrrrrr

    ++= (1.5).

    If the particle is moving, then x, y, and z are function of time t.

    Velocity:

    000000 zvyvxvzdt

    dzy

    dt

    dyx

    dt

    dx

    dt

    rdv zyx

    rrrrrrr

    r++=++= (1.6)

    Acceleration

    00002

    2

    02

    2

    02

    2

    000 zayaxazdt

    zdy

    dt

    ydx

    dt

    xdz

    dt

    dvy

    dt

    dvx

    dt

    dv

    dt

    vda zyx

    zyx rrrrrrrrrr

    r++=++=++= (1.7)

    1.3 Uniform circular motion

    Take the circle in the X-Y plane, so z = 0,

    )cos( += tRx , )sin( += tRy (1.8)

    is the angular speed. is the initial phase. Both are constants.

    Using the above definition of velocity (1.6),

    00 )cos()sin( ytRxtRvrrr

    +++ (1.9),

    vr

    is always perpendicular to rr

    .

    Its amplitude is Rv = (1.10)

    The acceleration is:

    rytRxtRarrrr 2

    0

    2

    0

    2 )sin()cos( =++ (1.11).

    Its amplitude is R

    vRa

    22 == (1.12)

    Ar

    Br

    Ar

    Br

    x

    y

    += t

    x

    y

    += t

    (x,y,z)

    x

    y

    z

    rr

    o

    (x,y,z)

    x

    y

    z

    rr

    (x,y,z)

    x

    y

    z (x,y,z)

    x

    y

    z

    rr

    o

    Cr

    Br

    Ar

    Cr

    Br

    Ar

  • 2

    2 Relative Motion

    A reference frame is needed to describe any motion of an object.

    Consider two such reference frames S and S with their origins at

    O and O, respectively. The X-Y-Z axes in S are parallel to the X-

    Y-Z axes in S. One is moving relative the other.

    Note: Rrrrrr

    += ' . (2.1)

    So the velocity is: uvdt

    Rd

    dt

    rd

    dt

    rdv

    rrrrr

    r '

    '+=+== (2.2)

    Similar for acceleration: '

    'dv dv du

    a a Adt dt dt

    = + = +

    r r rrr r

    (2.3)

    This is the classic theory of relativity. If ur

    is constant, then 0A =r

    , and 'aarr

    = , i. e., Newtons Laws work in all inertia reference frames.

    Properly choosing a reference frame can sometimes greatly simplify the problems.

    3 Forces

    Pull through a rope, push, contact forces (elastic force and friction force), air resistance, fluid

    viscosity, surface tension of liquid and elastic membrane, gravity, electric and magnetic,

    strong interaction, weak interaction. Only the last four are fundamental. All the others are the

    net effect of the electric and magnetic force.

    3.1 Tension

    Pulling force (tension) in a thin and light rope:

    Two forces, one on each end, act along the rope direction. The

    two forces are of equal amplitude and opposite signs because

    the rope is massless. It is also true for massless sticks.

    3.2 Elastics

    Elastic contact forces are due to the deformation of solids.

    Usually the deformation is so small that it is not noticed. The

    contact force is always perpendicular to the contact surface. In

    the example both 1Fr

    and 2Fr

    are pointing at the center of the

    sphere.

    3.3 Friction

    The friction force between two contact surfaces is caused by the relative motion or the

    tendency of relative motion. Its amplitude is proportional to the elastic contact force, so a

    friction coefficient can be defined.

    When there is relative motion, the friction force is given by f =

    kN, where k is the kinetic friction coefficient. The direction of the friction is always opposite to the direction of the

    relative motion.

    O

    O

    rr

    'rr

    Rr

    O

    O

    rr

    'rr

    Rr

    Tr

    Tr

    Tr

    Tr

    1Fr

    2Fr

    1Fr

    2Fr

    vr

    Nr

    fr

    vr

    Nr

    fr

  • 3

    When there is no relative motion but a tendency for such

    motion, like a block is being pushed by a force Fr

    , the

    amplitude of f is equal to F until it reaches the limiting value

    of sN if F keeps increasing, where s is the static friction coefficient.

    Once the block starts to move, the friction becomes f = kN. Usually k < s.

    3.4 Viscosity

    Air resistance and fluid viscous forces are proportional to the relative motion speed and the

    contact area. A coefficient called viscosity is used in these cases.

    3.5 Inertial force

    Inertial force is a fake force which is present in a reference frame (say S) which itself is

    accelerating. Recall that Aaarrr

    += ' . Assume frame-S is not accelerating, then according to

    Newtons Second Law, )'( AamamFrrrr

    +== . So in the S-frame, if one wants to correctly

    apply Newtons Law, she will get 'amAmFrrr

    = , i. e., there seems to be an additional force

    AmFrr

    =int (3.1)

    acting upon the object.

    Example 3.1

    A block is attached by a spring to the wall and placed on the

    smooth surface of a cart which is accelerating. According to

    the ground frame, the force Fr

    acting on the block by the spring

    is keeping the block accelerating with the cart, so amFrr

    = . In the reference frame on the cart, one sees the block at rest but

    there is a force on the block by the spring. This force is

    balanced by the inertial force amFrr

    =int

    3.6 Gravity

    Between two point masses M and m, the force is

    rr

    GMmF

    )r

    2= (3.2)

    The gravitation field due to M is rr

    GMg

    )r

    2= (3.3).

    The field due to a sphere at any position outside the

    sphere is equal to that as if all the mass is concentrated at

    the sphere center. (Newton spent nearly 10 years trying to

    proof it.)

    Potential energyr

    GMU

    = (3.4).

    Nr

    fr

    Fr

    Nr

    fr

    Fr

    M

    M

    =

    M

    M

    =

    rr

    Mm

    rr

    Mm

    amFrr

    =

    Ground frame

    ar

    Cart frame

    amFrr

    =amFrr

    =int

    amFrr

    =

    Ground frame

    ar

    Cart frame

    amFrr

    =amFrr

    =int

  • 4

    Application of superposition: Uniform density

    larger sphere with a smaller spherical hole.

    Near Earth surface, because the large radius

    of Earth, the gravitation field of Earth can be

    taken as constant, and its amplitude is

    2

    E

    E

    R

    GMg = = 9.8 m/s2 (3.5).

    Its direction is pointing towards the center of Earth, which in practice can be regarded as

    downwards in most cases.

    Example 3.2

    The weight, or the force of the ground on a person on

    different places on Earth. Take the radius of Earth as R, and

    the rotation speed being (= 2/86400 s-1). On the Equator, we have mg - N = ma = m2R, so N = mg - m2R = m(g - 2R)

    At the South Pole (or North Pole), we have N = mg

    At latitude , by breaking down the forces along the X-direction and Y-direction, we have

    mafNmg = sincos)( , and

    0cossin)( =+ fNmg .

    Solving the two equations, we get cos)( maNmg = , sinmaf = , where cos2 Ra = .

    The negative sign of f means that its direction is the opposite of what we have guessed.

    One can also break down the forces along the tangential and radial directions to obtain the

    same answers. One can also take Earth as reference frame and introduce the inertia force to

    account for the rotational acceleration.

    3.7 Buoyancy

    In a fluid (liquid or gas) of mass density at depth H, consider a column of it with cross section area A, then the

    total mass of the column is AH, and the gravity on it is AHg. The gravity must be balanced by the supporting force from below, so the force of the column on the rest of the

    liquid is F = AHg and pointing downwards. The pressure

    =

    1rr

    2rr

    1rr

    2rr

    O O O

    -=

    1rr

    2rr

    1rr

    2rr

    O O O

    -

    y

    x

    gmr

    gmr

    Nr

    Nr

    Nr

    gmr f

    rar

    y

    x

    gmr

    gmr

    Nr

    Nr

    Nr

    gmr f

    rar

    gmrHgmrH

  • 5

    P = F/A = Hg (3.6).

    Now consider a very small cubic of fluid with all six side area of A at depth H. The force on

    its upper surface is AHg and pointing down, the force on its lower surface is AHg but pointing upwards so the cubic is at rest. However, for the cubic not to be deformed by the two

    forces on its upper and lower surfaces, the forces on its side surfaces must be of the same

    magnitude. This leads to the conclusion that the pressure on any surface at depth H is AHg, and its direction is perpendicular to the surface. One can then easily prove that the net force of

    the fluid (buoyancy) on a submerged body of volume V is equal Vg. (See the HKPhO 2003 paper.) The buoyancy force is acting on the center of mass of the submerged portion of the

    object.

    3.8 Torque

    When two forces of equal amplitude and opposite directions

    acting upon the two ends of a rod, the center of the rod remains

    stationary but t