Download Problem Solving Tip Sheet

Classical Mechanics Tip Sheet
The following is a tip sheet that should help highlight some useful strategies and common
pitfalls encountered by students studying classical mechanics. It should not be used as a
substitute for the textbook, lecture, your notes, or anything else.
Problem Solving Tips
If you do not know when a formula is useful (correct and pertinent), then the formula is useless.
Becoming a good physicist does not come from memorizing a bunch of formulas and practicing
math. Strive to understand which formulas are valid under a wide variety of conditions (ex: Newton’s
laws) or definitions and which formulas only apply in special cases (ex: N = mg). Also, strive to
understand when certain problem solving methods are more useful than others.
Solve all problems symbolically before plugging in numbers. This will provide insight into which
variables affect your answer. It will also help avoid excessive rounding error.
Units
We will use a subset of the metric system called the International System of Units (SI). Other sets
and subsets of units are perfectly valid, so this decision is somewhat arbitrary. A good system of units
should be universal (reproducible by anyone, anywhere), precise, stable, and simple (easy to convert
and providing reasonable numbers for common use).
Dimension
Unit
Past
Present
Distance
Meter
1/10,000,000 the
distance between the
equator and the north
pole
Length of the path
travelled by light in
vacuum during a time
interval of
A metal bar in France
1/299,792,458 of a
second
1,650,763.73 wavelengths
of the orange-red
emission line in the
electromagnetic spectrum
of the krypton-86 atom
in a vacuum
Mass
Time
Kilogram Mass of 1000 cm3 of
water at 0° C
A metal cylinder in
France
Second
9,192,631,770 periods of
the radiation
corresponding to the
transition between the
1/86,000 of a day
Future?
Watt balance
Precise spheres of
silicon
1/31,556,925.9747 of a
year
two hyperfine levels of
the ground state of the
cesium-133 atom
Temperature Kelvin
Temperature in °C plus
273.
1⁄273.16 of the
temperature of the triple
point of water
The temperature
scale for which
Boltzmann's
constant is
1.3806505E-23
J/K
Current
The current that would
deposit 0.001118000
grams of silver per
second from a silver
nitrate solution
The constant current
that will produce an
attractive force of 2E–7
newton per meter of
length between two
straight, parallel
conductors placed one
meter apart in a vacuum
Flow of
6.2415093E18
elementary
charges (such as
electrons)
Ampere
Coordinate System
In this class, we will default to the coordinate system where +x is to the right, +y is up, and
+z is out of the page. Be very clear in your work if you decide to use an unconventional
coordinate system.
Vector Arithmetic
Vectors can be reported as x-y components (Cartesian) or magnitude and direction (polar).
Cartesian vectors can be recorded many ways. The following vectors are the same:
A = 3î - 4ĵ
A = (3, -4)
A = 3x – 4y
Ax = 3, Ay = -4
Polar vectors also can be reported in different ways. The following vectors are the same:
A = (5, -53°)
A = 5, θ = -53°
When finding x and y components of vectors (force, velocity, acceleration, momentum, etc.) from
magnitude and direction, the following formulas are used when the angle is measured
counterclockwise from the positive-x axis (this is by convention):
Ax = A*cosθ
Ay = A*sinθ
The signs of the x-y components will normally take care of themselves with the trig functions. The
common exception to these formulas is when you use a rotated coordinate system such as in a sliding
block problem. In this case, the angle is often measured from the negative y axis, so the trig functions
get switched and you must put in a + or – sign manually.
The reverse of these formulas can be used to find the magnitude and direction of a vector when the
x and y components are known:
A = √(Ax2 + Ay2)
θ = tan-1(Ay/Ax) Note: add 180 degrees to your answer when Fx < 0.
Kinematics
The most abused formulas in kinematics are the definitions vaverage = Δx/Δt and v = Δx/Δt in the
limit of Δt goes to zero. The first formula can be used correctly, though vaverage doesn’t appear
anywhere in the other kinematics formulas and is therefore not particularly useful. It is useful when
acceleration is zero. The second formula can only be used correctly when using calculus. Use
kinematics equations from your textbook or in the table below instead of the above equations.
The symbol g is used for the magnitude of the free-fall acceleration due to gravity. It is always
positive since it is a magnitude, so don’t ever write, “g = -9.8 m/s2.” It varies from place to place, but
the typical textbook uses 9.8 or 9.81 m/s2. We use 9.806 m/s2 in labs in Salem, Oregon. The
acceleration (vector) of an object under the influence of gravity alone using a conventional
coordinate system will be ax = 0 and ay = -9.8 m/s2.
The equations below are all you need to solve constant acceleration 1-d or 2-d kinematics problems.
Every other formula in your textbook is a special case derived from these:
x
y
vfx = vix + axt
vfy = viy + ayt
Δx = vixt + ½axt2
Δy = viyt + ½ayt2
vfx2 = vix2 + 2axΔx
vfy2 = viy2 + 2ayΔy
Δx = ½(vix + vfx)t
Δy = ½(viy + vfy)t
Δx = x-component of displacement vector (also
written xf – xi)
Δy = y-component of displacement vector (also
written yf – yi)
vix = x-component of initial velocity vector (also
written v0x)
viy = y-component of initial velocity vector (also
written v0y)
vfx = x-component of final velocity vector (also
written vx)
vfy = y-component of final velocity vector (also
written vy)
ax = x-component of acceleration vector
ay = y-component of acceleration vector
t = time interval (also written Δt)
t = time interval (also written Δt)
Note that all of the above equations have four variables. This means that you need three known
quantities to solve a kinematics equation and you should start every kinematics problem by gathering
information. When it doesn’t appear you have enough information to solve one, consider the
following possibilities:




a velocity of zero (such as stopping, starting, or a turnaround point)
a known acceleration such as free-fall (ax = 0 and ay = -g) or frictionless inclined plane (a =
±g*sinθ)
conversion of a magnitude and direction of a vector to x and y components (see below)
an acceleration that can be calculated using Newton’s second law (see below)
There is one special case formula that is particularly useful. It is the range formula for projectile
motion when initial and final heights are the same:
R = vi2sin(2θ)/g
A common misunderstanding is that positive acceleration is speeding up and negative acceleration is
slowing down. This is not necessarily true. You need to look at the relationship between the velocity
and acceleration vectors to determine this. If the velocity and acceleration vectors are in the same
direction, then the object is speeding up. A dropped ball is an example of negative acceleration and
increasing speed. If the velocity and acceleration vectors are in the opposite direction, then the object
is slowing down. A car driving to the left and applying the brakes has a negative velocity and a
positive acceleration, yet is slowing down. If the velocity and acceleration are consistently
perpendicular, then there is no change in speed. This occurs in uniform circular motion.
Forces
Newton’s 1st law states that a body at rest remains at rest and a body in linear motion remains in
motion with constant velocity until a net external force is applied to it: ΣF = 0.
Newton’s 2nd law is a mathematical relationship between net (or total) force on an object, mass of
an object, and the acceleration of an object: ΣF = ma.
Newton’s 3rd law states that for every action there is an equal and opposite reaction: Fab = -Fba. The
source and receiver of the force are switched and the direction of the force is reversed. The
magnitudes (strengths) of the two forces are the same. The types of the forces are the same. The two
forces are called an action-reaction pair.


When calculating the net force on an object, use vector (not scalar) arithmetic.
Conventional notation for forces is to specify the type of force with a single letter (or F with
a subscript), the proximate, external source of a force with the first subscript, the receiver of


Name
the force with the second subscript, the direction with an arrow, and the relative strength of
the force indicated by the length of the arrow. For example, a weight force exerted by the
Earth on a person directed down would be indicted this way: WEp↓.
When all the forces on an object are indicated this way, it is called a “free body diagram”. Do
not include accelerations, velocities, net forces, or forces exerted by an object in a free body
diagram.
Do not include an action-reaction pair in a free-body diagram for a single object. If there are
two or more free-body diagrams with an action-reaction pair present, then indicate the
presence by drawing a dotted line connecting the two forces.
Symbol Magnitude
Direction
Notes
gravitational Fgrav
force
Gm1m2/r2
attractive
G = 6.67E-11 Nm2/kg2. This formula
only works for spheres or objects
separated by distances much bigger than
their individual sizes.
weight
W
mg
down, towards
the center of
the Earth
Source is almost always Earth. Use g =
9.8 m/s2 unless the specific location is
known. Weight and gravitational force are
the same thing.
normal
N
variable
perpendicular
to surface;
repulsive
Most “pushing” forces are normal forces.
Do not set N = mg or mg*cos(θ) unless
this can be proven with Newton’s laws.
tension
T
variable; the
same
magnitude at
both ends
(unless the
rope is
massive or a
pulley has
mass or
friction)
parallel to
rope, chain,
etc.; attractive
Most “pulling” forces are tension forces.
Do not set T = mg unless this can be
proven with Newton’s laws. Source of
tension is usually the rope, chain, etc.
kinetic
friction
fk
μkN
parallel to
surface;
opposite
relative
motion
The coefficient of kinetic friction, μk,
must be distinguished from the force of
kinetic friction, fk. The coefficient of
kinetic friction can be taken from a table
if the two substances are known.
static
friction
fs
0 ≤ fs ≤ μsN
parallel to
surface;
prevents
relative
The coefficient of static friction, μs, must
be distinguished from the force of
friction, fs. The coefficient of static
friction can be taken from a table if the
motion
two substances are known. Wheels that
aren’t skidding have static friction.
spring
Fs or S
kx
opposite
displacement
from
equilibrium
The variable k represents the stiffness of
the spring. The location of x = 0 must be
at equilibrium. The formula can be
written with vectors as follows: Fs = -kx.
drag (air)
D
≈¼Av2
opposite
velocity with
respect to air
A = cross sectional area. The formula is
approximate and only applies to
macroscopic objects moving at
reasonable speeds through air near the
surface of the Earth.
drag
(sphere in
liquid)
D
≈6πηrv
opposite
velocity with
respect to
liquid
η = viscosity of liquid
thrust
Fth
r = radius of sphere
vexhaust*∂m/∂t opposite
velocity of
exhaust
Friction Problem Solving
1) Does the problem state definitively if the surface of the object is moving in relation to the other
surface? If it is moving in relation to the other surface, then use kinetic friction (3). If it is not
necessarily moving, then use static friction (2). Note that the surface of an object that is rolling
without sliding is not moving in relation to the surface of the ground.
2) Test if the other forces can “break” the static friction. Calculate (free body diagrams, Newton’s
first or second law, and vector addition) the magnitude the friction needs to be to hold the object
in place relative to the surface. Then compare that value to the maximum possible static friction,
μsN. If the static frictional force can hold the object in place, then the static force of friction will
take the magnitude and direction necessary to do so. If the static frictional force cannot hold the
object in place, then the object will move, so you must use kinetic friction (3).
3) If the other forces break the static friction or the object is already known to be moving relative
to the surface, simply use the formula fk = μkN for the magnitude. The direction will be opposite
the motion of the object’s surface in relation to the other surface.
Circular Motion
If an object moves in a circle with constant speed, then it is characterized as having uniform circular
motion. The object will be accelerating towards the center of the circle (centripetal acceleration) with
the following magnitude: acp = v2/r. The so-called centripetal force is the net force on the object
which will be macp by Newton’s second law. The centripetal force is not a new force to add to the list
of forces such as N, W, etc.; it is a characteristic of a single force, force component, or combination
of forces whose vector sum points towards the center of a circle.
If an object or point on an object moves in a circle with changing speed, then things get more
complicated. If the angular acceleration (α) is constant, then the rotational kinematics equations as
listed in the text apply.
There are specific mathematical relationships between the angular and linear quantities. The
tangential speed of a point on a rotating object is v = rω. The centripetal acceleration of a point on a
rotating object is the same as that for uniform circular motion: acp = v2/r. This formula can be
rewritten as acp = rω2. This is only one component of the acceleration vector. The other component
is the tangential acceleration atan = rα. Be careful with signs as the center of a circle may be down
(negative acceleration), depending upon your choice of coordinate system.
Momentum
The definition of momentum is p = mv. This is a useful quantity to analyze when objects interact
with one another with brief, strong forces that cannot easily be calculated, such as in a collision or
explosion.
Total momentum for an object or system of objects will be conserved (no net change) if the net
external force is zero or insignificant compared to the internal forces. The relationship can be written
mathematically as: Σpi = Σpf. Since momentum is a vector, this should be broken up into x and y
components when the problem has 2 dimensions. The sign of the velocity should indicate the
direction of motion in each dimension. Here is conservation of momentum for two objects in two
dimensions:
m1v1ix + m2v2ix = m1v1fx + m2v2fx
m1v1iy + m2v2iy = m1v1fy + m2v2fy
There are certain types of collisions which are special. A collision (not an object) can be characterized
as completely inelastic, elastic, explosive, or none of these. If the collision is completely inelastic, then
the velocities of all objects will be the same after the collision (the objects stick together). This is also
the maximum possible decrease in total kinetic energy. If the collision is elastic, then the total kinetic
energy of the system does not change: ΣKi = ΣKf. There are some formulas for velocity available
for this special case. You should only use these when you know for a fact that the collision is elastic.
Do not assume that a collision must be completely inelastic or elastic as there are other possibilities.
Type
Momentum Kinetic Energy
Notes
Completely
inelastic
Σpi = Σpf
ΣKi > ΣKf
Objects stick together after collision
ΣKf is minimum possible
while still consistent with
conservation of momentum
Some sound, permanent deformation,
or change in temperature
(transformation of kinetic energy into
other types)
In between
Σpi = Σpf
ΣKi > ΣKf
Objects do not stick together after
collision
Some sound, permanent deformation,
or change in temperature
(transformation of kinetic energy into
other types)
Completely
elastic
Σpi = Σpf
ΣKi = ΣKf
Minimal sound, permanent
deformation, or change in temperature
Explosive
Σpi = Σpf
ΣKi < ΣKf
Addition of kinetic energy from loss
of chemical, nuclear, electric, spring,
or other potential energy
Energy
Many problems that are difficult to solve or cannot be solved with Newton’s laws can often be
solved using an energy calculation. This is often when forces vary in magnitude or direction as a
function of time. The use of energy methods can come at the expense of not knowing time or a
direction of a velocity.
1) Specifically identify the system.
2) Specifically identify the starting and ending events for the system.
3) Write down the “macroscopic energy accounting” equation:
ΣUi + ΣKi + ΣW = ΣUf + ΣKf.
4) Specifically identify the external forces (type and source) affecting the objects at any time
between the starting and ending events.
5) Decide how to take into account the effect of each force on the equation. Each force must
be categorized as having a work function of zero (perpendicular forces), a potential energy
function (typically gravity and springs), or a non-zero work function (typically friction or
some external agent pushing or pulling).
6) Calculate the work done by all non-conservative forces (see below)
7) Insert formulas for U (depends on the type of force, U = mgy or U = ½ks2) and K (½mv2).
8) Solve for the variable of interest.
Do make a clear distinction between when the problem asks for the work done by a particular force
and the total work. The formula W = F*d*cosφ refers to the work done by a particular force. The
formula Wtotal = Fnet*d*cosφ refers to the total work. The formula W = -ΔU refers to the relationship
between the work done by a particular force and the change in potential energy associated with that
same force. It is not a useful formula unless you are deriving a potential energy function (you
probably won’t do this). The formula Wtotal = ΔK refers to the total work by all forces when potential
energy functions are not used.
Do think very carefully about the angle in the formula W = F*d*cosφ. This is the angle measured
between the force and the displacement. This is neither the angle of the force nor the angle of the
displacement. The numbers F and d are magnitudes and they are always positive. The term cosφ will
determine the sign of the work. Do not use W = F*d unless you know for a fact that the force and
motion are in the same direction. Here are some special cases for calculating work:
Angle
Description
Work
φ = 0°
Force and motion are in the
same direction
F*d
0° < φ < 90°
F*d*cosφ >0
φ = 90°
Force and motion are
perpendicular to one another
90° < φ < 180°
0
F*d*cosφ < 0
φ = 180°
Force and motion are opposite
one another
-F*d
Do not omit or double count anything from the formula: ΣUi + ΣKi + ΣW = ΣUf + ΣKf. Any of
the five terms might be zero, or they might have more than one term.
Sometimes an energy calculation will need to include thermal energy (microscopic kinetic and
potential energy, denoted Eth) or heat (microscopic energy transfer, denoted Q). The energy equation
will be expanded to be the following:
ΣUi + ΣKi + ΣEth,i+ ΣW + ΣQ = ΣUf + ΣKf + ΣEth,f
This can be written more concisely by grouping together initial and final values into changes:
ΣW + ΣQ = ΔU + ΔK + ΔEth
When potential and kinetic energy do not change, then the energy calculation is written as follows
(this is often how the first law of thermodynamics is stated):
ΣW + ΣQ = ΔEth
Common Sense
Sometimes it is a good idea to compare your answers to what you know is “reasonable”. This
generally doesn’t work with galaxies or subatomic particles, but with things like people and cars, you
can make good comparisons. A rough rule is that a speed in m/s can be doubled to obtain a speed in
miles/hour. Accelerations can be compared to g.
Event
Maximum
acceleration as
multiple of g
Event
Speed (m/s)
Roller coaster
3-5
Human sprinter
(maximum)
10.5
Racecar driver
5
Hurricane
33
Blackout possible without
training or special
equipment
5
Cyclist (maximum)
37
Sound (depends a bit on
atmospheric conditions)
340
Fighter pilot or astronaut
in anti-G suit
9-12
Bullets
300-1000
Death possible (note:
depends on duration)
50-100
Escape velocity from Earth
11,200
Light in vacuum (by
definition)
299,792,458
Things that are never true:
g = -9.8 m/s2 (Weight force vectors and free fall acceleration vectors are generally negative)
Δx = at2 or a = Δx/t2 (these are dimensionally correct, but just plain false)
Things to use only if you are absolutely sure they are true and useful:
vave = Δx/Δt (True, but seldom useful. I refer to this formula as “6th grade kinematics”.)
ωave = Δθ/Δt (True, but seldom useful. This would be “6th grade rotational kinematics.”)
N = mg (True only in a special case)
N = mg*cos(theta) (True only in a special case)
T = mg (True only in a special case)
Work = Fd (True only in a special case)
fs = μsN (True only in a special case)