Physics: Classical Mechanics
Physics 61L: 61L: Spring 2011 A Summary Dated 5 May 2011
Classical Mechanics and Newtonian Dynamics Duke University University Edmund Edmund T. Pratt, Pratt, Jr., School School of Engineering Engineering
ABSTRACT A summary summary of the major concepts concepts and equations equations relevant relevant to a first-sem first-semester, ester, calculus-bas calculus-based ed course course in introduct introductory ory mechanics. mechanics. Particul Particular ar emphasis is placed placed on the dynamism dynamism of system system elements elements and the applicati application on of Newton’s Newton’s laws to nontrivi nontrivial al systems. Select Select derivatio derivations ns of formulae formulae are included included on the basis of the amount of insight gained through their inclusion.
The final expression above—a result of algebraic manipulation of the standard particulate position equation with constant 1 acceleration (x (xf = xi + vi t + at2 )—is of great practical use; 2 said manipulation is given below, and involves solving for the time variable in the velocity equation...
Kinematics
Kinematics is a language through which the motions of bodies (particles) and systems (discrete sets of particles) can be described without regard for the forces originally responsible for the motion. More specifically, changes to the vector quantities of position, velocity, and acceleration constitute the means by which the motion of these mobile elements is understood.
vf = vi + at =⇒ t =
...and then eliminating the parameter through substitution:
The dynamic particle with variable or non-constant acceleration requires an integral treatment, since all three of the above vector quantities vary with time:
1 xf = xi + vi t + at2 2 1 2 (xf − xi ) = vi t + at 2 vf − vi 1 vf − vi 2 = vi + a a 2 a 2 2 vf − 2vf vi + vi2 vi vf vi 1 = − + a a a 2 a2
Position: x(t) = x0 +
v dt
Velocity: v(t) =
d x = v0 + dt
d v d2 x = 2 dt dt
2
2
=
A particle moving with either zero or constant acceleration can be described via the following equations: vx/y = vx/z + vz/y vi + vf xf = xi + t 2 1 xf = xi + vi t + at2 2 1 xf = xi + vf t − at2 2 vf = vi + at
2 vf vi vf vi2 vf vi v2 − + − + i a a 2a a 2a 2 2 2 vf 2v v =− i + + i 2a 2a 2a 2 vf vi2 (xf − xi ) = − + 2a 2a 2 2 2a(xf − xi ) = −vi + vf
adt
Acceleration: a(t) =
vf − vi a
2 vf = vi2 + 2a 2a(xf − xi )
Note that this equation is time-independent , time-independent , since we found it by eliminating the variable t from the kinematic equations. For the specific case of an object in free fall, Constant acceleration
ay = g = 9. 9 .8 m/s2 vy = vi,y − gt yf = yi + vi,y t −
vf = vi + 2a 2a(xf − xi )
1
1 2 gt 2
• Law of Reaction (Third Law): Law): If object object A exer exerts ts a forc forcee F A→B on object B along a line connecting the two objects, then object B exerts an equal and opposite reaction force B→A = −F A→B on object A. The implications are as of F follows:
Projectile Motion
Key concept : in projectile motion, gravitational forces in the y -direction have no affect on vx . Because of the vector nature of motion and force, a complex two-dimensional motion can be "uncoupled" or resolved to motion relative to the two coordinate axes. x-component
i→j = −F j →i F
y-component
ax = 0
i,j
ay = −g
vf,x = vi,x (constant v)
(The latter form means that the sum of all internal forces between particles in a closed system of particles cancel.)
vf,y = vi,y + ay t
xf = xi + vi,x t
1 yf = yi + vi,y t + ay t2 2 F y = −(mg) mg) j
x = 0 F i
Statics
In a system that exists in static equilibrium, the following two conditions must necessarily hold:
Fluids
net F net = 0
Pressure Pressure is, by definition, definition, a distribute distributed d force. force. In static fluids, only normal stress is exerted (force perpendicular to surfaces). The equation for pressure is as follows: P =
P = P 0 + ρgd
i
Archimedes’ Principle : the buoyant force on an object is equal and opposite to the force that gravity would exert on the fluid displaced displaced by the object.
F y,i 0 , and y,i = 0,
i
F z,i z,i = 0
i
The gravitational force near the Earth’s surface: a downward-directed, constant, fundamental force; in vector g = −mg j (in a coordinate frame where up is +y ). form, F
Here, ρ represents fluid density (in mass per unit volume), which is a known quantity in most cases. Therefore, the expression ρV can be interpreted to be equivalent to mass, m.
F g = mg
Dynamics
The spring force (Hooke’s Law) in one dimension: a force directed back toward the spring’s natural point of equilibrium (a “restoring force" =⇒ direction opposes displacement).
Newton’s laws:
F x = −k ∆x The normal force : a force directed directed perpendicularly perpendicularly and away away from contact surface, with magnitude sufficient to exactly oppose the force of contact (a “force of restraint").
• Law of Inertia (First Law): Law) : Objects Objects at rest rest or in uniform uniform motion (at a constant velocity) remain so unless acted upon by an unbalanced (net) force. d v i = 0 = m F a=m dt
=⇒
F ⊥ = N
v = constant vector
Tension in a massless, unbreakable string: a force directed away from the object at point of connection.
• Law of Dynamics (Second Law): Law) : The net force force applied applied to an object is directly directly proportional proportional to its acceleration acceleration.. The constant of proportionality is called the mass of mass of the object. = F
F x,i 0, x,i = 0,
Force Rules
F b = ρV g
i
τ τnet et = 0 n
An effective approach to the static equilibrium condition is to declare the validity of the above equations, and to then break them down into further equations by reducing the force and tension vectors into their respective components. For example, a net force of zero implies
F A
Pascal’s Law : pressure in a fluid depends only on depth, with P 0 the surface pressure (which in most cases is the atmospheric pressure).
i,j F i,j = 0
i
F s = T
d(m v) d p i = m F a= = , dt dt
Static friction : a force directed opposite to the net force and parallel to surface; localized at the point of contact between the static object and the surface.
where the momentum of a particle, p = m v , is discussed below.
f s ≤ µs N 2
Kinetic friction : a force directed opposite to the direction of motion of the moving object; localized at the point of contact between the static object and the surface.
This is because a nonzero net force is necessary for acceleration to be present at all; the logic is that, if a given particle is lacking acceleration, then it has either zero or constant velocity, which means that there is no change in particulate velocity and thus no change in momentum (a quantity dependent on velocity) This leads directly to a more formal statement...
F k = µk N Drag force : a force directed opposite to the relative direction of motion through a given fluid. It arises because the surface of an object moving through a fluid is literally bouncing fluid particles off in the leading direction while moving away from particles in the trailing direction, so that there is a differential pressure on the two surfaces.
Law of conservation of momentum: d p net When F = 0, 0 , then p = constant and p i = p f . net = dt
F d = −bvn , 1 ≤ n ≤ 2 Case Study: Study: A Collision Uniform Circular Motion
For the specific case of two objects, mA and mB , colliding in an inelastic (or “sticky") collision, we can conclude that Uniform circular motion is a very specific, simplified instance of momentum is conserved so long as the two objects fall within radial radial motion. motion. In such motion, motion, there exists some net, constant constant our definition of the system. Then, we are free to solve for the force (whether from tension, applied, etc.) directed antiparallel velocity of the combined two-object system after collision: to the r-axis (radial) and no force in the direction of increasing p i = p f angular displacement, ∆θ. This force acts to constantly change the direction of the velocity vector of the circulating particle, mAvA,i + mAvB,i = (m ( mA + mB )vsys,f producing a constant radial acceleration known as “centripetal mAvA,i + mAvB,i acceleration," vsys,f = mA + mB v2 aradial = . Most collision problems in introductory mechanics will be r restricted to a single dimension, in which case the vectors above Newton’s second law thus requires the net force on the particle reduce to the single particular component of interest. of mass m to equal the aforementioned force—referred to as the Two-dimensional collisions will involve resolving the momentum “centripetal force" vector p into its components and then solving for the x- and v2 . r (Remember: this force accelerates a body by acting ⊥ to motion and thereby changing only the direction of v and not the magnitude. In other words, if there is no centripetal force producing a radial acceleration, then the particle is traveling linearly and not in circular motion)
y -components of the velocity vectors, usually by way of appending the terms cosθ and sinθ to the magnitude of the velocity vector, respectively respectively.. These types of problems can get pretty messy, but who doesn’t like messy?
F ext ext = F cent cent = maradial = m
Impulse
In general terms, impulse is a measure of the transmittance of a force to a rigid body over a given time (and specifically, the integral of a force with respect to time), with the end result being an incurred change in the momentum of that rigid body.
Momentum
Momentum can be thought of as the assigned “importance” of a force, force, insofar insofar as a high velocity velocity becomes becomes considerab considerably ly more significant (and dangerous) when it is associated with a progressively larger mass.
The formal definition of the impulse transmitted by a variable force (a force that varies with time) is
Translational momentum:
= J
p i
d p net net F =⇒ F p, net = net dt = d dt
Newton’s second law (as above):
i
we can equivalently write, in a form known as the impulse-momentum theorem ,
d v d(m v) d p i = m F a=m = = dt dt dt =⇒ when
net F net (t)dt,
but because
i
= F
tf
ti
p = m v p sys =
i
d p i = 0, F 0, =0 dt
= J
p f
pi
3
d p = p f − p i ,
or Newton’s second law reformulated:
= ∆ J ∆ p
MaCM = M
When the force is non-variable (or constant), then the expression for impulse no longer contains any differentials and is considerably easier to solve—especially if we know the net average force, F net , over a time interval: net F net =
d2 d2 ( R ) = CM dt2 dt2
miri =
i
iext, F
i
which, combined with the definition of momentum, yields the following:
J p f − p i ∆ p net = = =⇒ J = ∆ ∆ p = F net ∆t ∆t ∆t ∆t
ext F tot = M tot ext = 0 =⇒ p tot V CM CM = constant where CM V CM =
Center of Mass
vi mi vA mA + vB mB + · · · = m mA + mB + · · · i i
Key concept: the above outcome implies that it is the external The process of computing the center of mass can be viewed as force which determines translation of the center of mass. If an attempt to condense the entirety of the mass of a rigid body there is no external force, then there is no CM acceleration and (or a system of bodies) into a point mass located at a single no change in CM velocity, producing a similar lack of change in point in space, in order for us to treat the previously complex CM momentum. body as a straightforward particle. Note that a body’s center of mass need not necessarily be confined to the body itself (see the horseshoe, or the infamous doughnut which physicists so readily cite). Work and Energy
For a system of discrete particles: CM = R
XCM = Y CM CM = Z CM CM =
i
Kinetic energy: energy:
miri = XCM i + Y CM CM j + Z CM CM k, m i i
K =
where
Work-kinetic energy theorem:
mi xi mA xA + mB xB + · · · = m mA + mB + · · · i i mA yA + mB yB + · · · i mi yi = m mA + mB + · · · i i mA zA + mB zB + · · · i mi zi = m mA + mB + · · · i i i
Whereas the impulse-momentum theorem assigned force to be a function of time (with impulse being the integral of force with respect to time), the work-kinetic work-kinetic energy energy theorem theorem instead assigns force to be a function of displacement, ∆x (with work being the integral of force with respect to displacement). We can actually derive the expression for the theorem in one dimension by returning to the time-independent kinematic equation for the final velocity of a particle, 2 2 vf = vi + 2a 2a(xf − xi ), and hammer it algebraically into a more relevant form. Multiplying both sides by the term 1/2 and the mass m gives:
For a solid body: CM = R
rdm = XCM i + Y CM CM j + Z CM CM k dm where XCM = Y CM CM = Z CM CM =
but when we substitute F = ma by Newton’s second law,
ydm dm
1 1 2 mvf = mvi2 + F ∆ F ∆x 2 2 1 1 2 F ∆ F ∆x = mvf − mvi2 . 2 2
zdm dm
include:
dm =
1 1 2 mvf = mvi2 + ma( ma(xf − xi ), 2 2
xdm dm
and where common methods of resolving the term
dm =
ρdv =
λds =
1 mv2 2
Since work W is analogous to a force exerted on a body over a 1 distance, F ∆ F ∆x, and since kinetic energy K = mv2 , then we 2 are finished:
dm
ρdxdydz
incurred change in kinetic energy
1 1 W = ∆K ∆ K = mvf 2 − mvi2 2 2
m ds
work done by net force
4
Key concept: since we derived the work-kinetic energy theorem from an equation which was itself derived through elimination of the time parameter, the theorem has the property of being time-independent as well; applying the principles of work-energy or conservation of energy (further below) might be good choices for situations in which time is either not given or not relevant. relevant.
For the specific case of a system moving from an initial position of rest to a final position of rest, W net ∆ K = 0 since vi = vf = 0. 0. net = ∆K
Power
Work equations:
Power in physics is the rate at which a force does work on a particle. The average power exerted is simply the net work divided by the time interval during which the force acted:
• Work → variable force in one dimension :
xf
W =
F x (x)dx
xi
P P =
• Work → constant force in one dimension :
xf
W =
F x dx = xF x
xi
If we take the limit as ∆t → ∞, we find the (generally) much more useful differential power expression:
xf
= F x ∆x xi
dw dt |cosθ) (|F | F cosθ)|dx| dx| = dt dx | = |F | F cosθ dt || = |F || F v|cosθ
P =
• Work → variable force in three dimensions :
rf
W =
(t) · d F ( F r(t)
ri
• Work → constant force in three dimensions :
rf
W =
W ∆t
· v = F
· d ||∆ F r = |F || F ∆r|cosθ
ri
where θ is the angle between the force vector and the direction of displacement. The significance of the cosine function is that a force ⊥ to the displacement of an object π thus does no work on that object (i.e. cos = 0). 0) . 2
Work, Energy, and Power
• Work → gravitational force (conservative) :
yf
W grav grav =
F grav grav dy −mgdy
yi
yf
= −mgy
=
W
=
K =
yi yf
=
W
= −mg( mg(yf − yi )
∆U g
=
P avg avg
=
P
=
F ∆ F ∆x cos θ
· d F s
1 mv2 2 mg∆ mg∆h W ∆t F v cos θ
yi
= −mg∆ mg∆y
Gravity
• Work → spring force force (conservative) (conservative)::
F grav grav
=
U G
=
xf
W spring spring =
F spring spring dx
xi xf
=
−kxdx
xi
1 = − kx2 2 =
xf xi
1 2 − xi2 ) = − k(xf 2
1 2 1 2 kx − kxf 2 i 2
• Work → friction force (non-conservative) : W = −f k |∆x| 5
Gm1 m2 r2 Gm1 m2 − r −
Radial Motion
ω
=
α
=
v
=
T
=
atot
=
Waves
k is wave # in rad/unit length
dθ = 2πf 2 πf dt d ω dt ωr 1 f aradial + atangential
y (x, t)
=
A cos(ωt cos(ωt − kx + φ)
k
=
2π/λ
ω
=
2πf
v
=
λf = ω/k =
v
=
λ
=
λ
=
2L/n {bothopen/ bothopen/closed} closed}
2
aradial
=
atan
=
v inward r αr⊥
ωf
=
ωi + αt
θf
=
ωf 2
=
1 θi + ωi t + αt2 2 2 ωi + 2α 2α(θf − θi )
I =
F/µ
tension/ tension/linear linear density density
4L/(2 L/(2n n − 1) {harm : 1open} 1open}
Trigonometry
Sides a, b, and c are across from angles α, β , and γ respectively. sin α a c2
Moment of Inertia and Torque
= =
sin β sin γ = b c 2 2 a + b − ab cos γ
2
mi ri
i
I =
2 I cm cm + mh
τ =
K rot rot
=
W L
=
+rF sin θ forCCW = dL r × F dt Iα 1 2 Iω 2 τ ∆ τ ∆θ = ∆K ∆ K rot rot
=
I ω = r × p
I ring ring
=
mr2
I disk disk
=
1/2 mr2
I rod rod
=
1/12 m2
I solid solid sphere sphere
=
2/5 mr2
I hallow hallow sphere sphere
=
2/3 mr2
τ τ = τ =
Variables
Simple Harmonic Motion
s F
=
−kx
T s
=
U s
=
T p
=
2π m/k 1 2 kx 2 2π /g
Constants
Vectors
|a|
=
x ˆ × yˆ = a · b = a × b = a × b = |a × b| =
a= zˆ
ax2 + ay2 + az2
|a| | b| cos θ = ax bx + ay by + az bz (ax x ˆ + ay yˆ + az zˆ) × (bx x ˆ + by yˆ + bz zˆ) x ˆ(ay bz − az by ) + yˆ(az bx − ax bz ) + zˆ(ax by − ay bx ) |a| | b| sin θ
6
a = acceleration F = force f = frequency h = height I = moment of inertia J = impulse K = kinetic energy k = spring constant = length m = mass N = normal force P = power p = momentum r = radius or distance T = period t = time U = potential energy v = velocity or speed W = work done on system x = position α = angular acceleration µ = coefficient of friction θ = angle τ = torque ω = angular velocity
g ≡ 9.80665 m/s2 ≈ 9.81 m/s c ≡ 299 792 458 m/s ≈ 3.00 × 108 m/s G ≈ 6.67 × 10−11 m3 /kg· /kg·s2
