Physics 121 Concepts - Summary
Last updated March 11, 2010, 1:40 pm
Note: Although I have done my best to check for typos and list the formulas correctly, you
should verify the formulas are correct before using them.
- Dr. Nazareth
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Know/Understand the information in Appendices of your textbook
o Powers of Ten and Scientific Notation
o Significant Figures
o Algebra
o Exponents and Logarithms
o Geometry and Trigonometry
o Sine, cosine, tangent only apply to right triangles (where one of the angles = 90°)
o Tables of typical values, planetary data, elements, isotopes
Units (1.2)
Dimensional Analysis (1.3)
Conversion of Units (1.5)
o Units should always be written out and carried throughout calculations.
o Units obey the same algebraic rules that numbers do because they are mathematical
entities.
o In a conversion, if the units do not combine algebraically to desired result, then the
conversion was not done correctly.
Order of Magnitude Calculations
Scalars and Vectors (1.7)
o A scalar quantity is described completely by its size or magnitude.
o A vector quantity has a magnitude (size or strength) and a direction.
Problem Solving in Physics (1.8)
o Read the problem carefully
o Sketch the system
o Visualize the physical process
o Strategize
o Identify appropriate equations
o Solve the equations (use variables to solve and input numerical values at the end)
o Check your answer to see if it makes sense/is reasonable/is what you expected
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Position, Distance, and Displacement (2.1)
o Start by defining a coordinate system
o Distance – total length of travel
o Position – where you are relative to origin
o Displacement - a vector that points from an object’s initial position (usually taken
to be the origin or 0,0) to its final position.
 The magnitude (size) of the displacement vector equals the shortest distance
between the initial and final positions.
 Remember there are positive and negative directions (point opposite way)
Average Speed and Velocity (2.2)
o Average speed = distance traveled/elapsed time
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 Speed is a scalar quantity = magnitude only
 ex: car travels 200 m in 10 sec; average speed = 20 m/s
o Average velocity = displacement/time
 vector quantity = magnitude and direction
 direction is the same as the displacement (can be positive or negative)
 Graphical interpretation of average velocity – x-versus t graph
o Average velocity is slope of straight line connecting two points
corresponding to given time interval.
Instantaneous Velocity (2.3)
o Instantaneous velocity – indicates the speed of the object and the direction of
motion at any moment in time
 instantaneous velocity can change in magnitude or direction or both
 vector quantity = magnitude and direction
 direction is the same as the displacement (can be positive or negative)
 In this class we use the word velocity to mean the instantaneous velocity.
 Graphical interpretation - – x-versus t graph
o Instantaneous velocity is slope of tangent line at given instant of time
Acceleration (2.4)
o Acceleration is the change in velocity over a certain interval of time.
o Acceleration changes the velocity (in magnitude, direction, or both).
o Acceleration can speed you up or slow you down (often called deceleration).
o Constant velocity means the acceleration is zero and visa versa.
o Average acceleration = change in velocity/elapsed time
 vector quantity = magnitude and direction
 direction is the same as the change in velocity (can be positive or negative)
o Instantaneous acceleration – indicates the acceleration of the object and the
direction of motion at any moment in time
 instantaneous acceleration can change in magnitude or direction or both
 vector quantity = magnitude and direction
 direction is the same as the change in velocity (can be positive or negative)
 In this class we use the word acceleration to mean the instantaneous
acceleration.
o In this class, we usually assume the acceleration is constant.
o Make sure you know the difference between initial velocity, final velocity, and
average velocity. The final velocity only equals the average velocity when the
acceleration is zero.
Equations of Motion for Constant Acceleration (2.5)
v = v0 + at
Assumptions
x = displacement
vave = ½(v0 + v)
&
a = acceleration (= constant)
x = x0 + ½(v0 + v)t
Definitions
v = final velocity at time t
2
x = x0 + v0t + ½ at
v0 = initial velocity at time t0 = 0
v2 = v02 + 2aΔx
t = time elapsed since t0 = 0
Applications of the Equations of Motion (2.6)
Strategy for Solving Physics Problems
1. Read the problem carefully.
2. Visualize the motion.
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3. Draw a picture to represent the situation.
4. Choose a convenient coordinate system if one is not specified for you to use. Decide
positive and negative direction.
5. Write down the values given to you and identify what they represent (e.g., the 2 m/s
given is the final velocity).
6. Identify and write down what variables/quantities you are being asked to determine.
7. Identify the physics of the problem. This will point you to the equation or set of
equations that are suitable for the problem.
8. Strategize. Look at what variables/quantities you have been given and think about
which equations you know that might be applied to the problem. If you think you
don’t have enough known variables, remember that the motion of two or more objects
might be related by a common variable like time or the distance between the two
objects.
9. Use algebra to solve the equation(s) for the unknown using variables. After the
unknown variable is solved for, then stick the actual numbers (with units!) into the
equation to solve for the final numerical answer. Remember to include your units in
the final equation and make sure they come out right.
Freely Falling Objects (2.7)
o Free-fall – idealized motion where air resistance is negligible and acceleration
approximately constant
o Acceleration of a freely falling body called acceleration due to gravity
 Magnitude at Earth’s surface = g = 9.80 m/s2 = 32.3 ft/s2
 Always directed downward toward center of the Earth
o In absence of air resistance, book and feather fall with same acceleration and land
at same time
o Any object moving under influence of gravity only is a freely falling object
 motion can be downward or upward
 always experiences downward acceleration due to gravity
o time symmetry – object thrown upward takes same time to reach maximum height
as to return from there to starting point
 at maximum height, the velocity of the object thrown upward is 0 (zero)
Scalars and Vectors (3.1)
o A scalar quantity is described completely by its size or magnitude.
o A vector quantity has a magnitude (size or strength) and a direction.
The Components of a Vector (3.2)
o Components add together to equal original vector.
o Components must be perpendicular to each other.
o The choice of components is a matter of convenience – which axes are better to
use for the particular problem.
o Use trigonometry to resolve a vector into its components.
Vector Addition and Subtraction (3.3)
o When adding (or subtracting) vectors, you must take into account both the
magnitude and the direction of the vectors.
o Can add (subtract) vectors by two methods
 Graphical method
 draw vectors head to tail
 lengths drawn to scale (use ruler)
 angles drawn accurately (use protractor)
 Analytic method – add the components of the vectors
 C x and C y are the vector components and are perpendicular to each
other.
 C  Cx  C y
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Magnitude, C  C x2  C y2
 Cy 
 *** don’t forget to specify what the
Direction: angle   tan 1 
C
 x
angle, θ, is relative to (north, south, east, west, vertical, horizontal, etc)
If C  A  B , then C x  Ax  Bx and C y  Ay  B y
Position, Displacement, Velocity, and Acceleration (in 2-D) (3.5)
o Position
o Displacement
o Average velocity
o Instantaneous Velocity
o Average acceleration
o Instantaneous acceleration
o These concepts that we first discussed for motion along a line (one dimension, x)
work just as well in a plane (two dimensions, x and y)
Relative Velocity (3.6)
o Remember pattern of velocity subscripts
o Review examples in book
v AB = v AG + vGB
and
vGB = -vBG
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Motion in Two Dimensions (4.1)
o Motions in x-direction and y-direction are independent of each other and occur
simultaneously
x–component:
y-component:
vx = v0x + axt
vy = v0y + ayt
x = x0 + ½(v0x + vx)t
y = y0 + ½(v0y + vy)t
2
x = x0 + v0xt + ½ axt
y = y0 + v0yt + ½ ayt2
2
2
vx = v0x + 2axΔx
vy2 = v0y2 + 2ayΔy
*** Make sure you understand the difference between initial velocity, v0, final velocity, vf,
and average velocity, vave. The final velocity ONLY equals the average velocity when the
acceleration is 0. ***
(Note: bold variables are vectors; otherwise, variables are scalars or magnitudes of vectors)
 Projectile Motion (4.2)
o Assume no air resistance.
o In the x-direction (horizontal): moving object doesn’t slow down (change
velocity) if no air resistance.
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 v0x = vx
 ax = 0
In the y-direction (vertical): object is subject to gravity.
 vy changes
 ay = g (acceleration due to gravity)
Resulting motion is a curved trajectory (parabolic in shape)
Remember x and y motions are independent of each other. This means that a
package dropped from a hot air balloon at height H, will take exactly the same
time to fall as a package dropped from an airplane traveling horizontally at height
H. The y-components of the velocities are also the same for the two packages at
all points on the way down.
Speed of the projectile at any location along its path v = (vx2 + vy2)1/2
 horizontal and vertical components of velocity contribute to the speed or
magnitude of velocity.
At maximum height, vy = 0.
Zero launch angle (v0 is horizontal)
 v0y = 0
 x depends on v0x and time to fall (use y-component to get time)
Range – horizontal distance before landing (y = y0)
Maximum range of projectile occurs when angle of initial velocity vector,  = 45
(if there is no air resistance)
Projectiles have symmetries like freely falling bodies did.
 Time for projectile to reach maximum height = time for projectile to fall
from maximum height back to original height.
 (General launch angle) Max height occurs at x = ½R
 v0x = vx (because ax = 0)
 magnitude of v0y = magnitude of vy when at same height (y0 = y)
- direction of v0y and vy NOT opposite
- but θ above horizontal for v0 = θ below horizontal for v (when at
same height)
The Concepts of Force and Mass (5.1)
o Force is a vector quantity
o Mass is a scalar quantity
Newton’s First Law of Motion (5.2)
o An object stays at rest (velocity = 0), or in a constant state of motion at a constant
speed in a straight line, changed by a net force.
o A net force applied to a body changes an object’s velocity – it doesn’t sustain it.
o Inertia is the tendency of an object to remain in motion in a straight line (or at rest
where velocity = 0).
o The inertia of an object is measured by its mass.
Newton’s Second Law of Motion (5.3)
o F = ma
o As Components (5.5)
 Fx = max
 Fy = may
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o F = 0 implies a = 0, and visa versa.
Newton’s Third Law of Motion (5.4)
o “For every action, there is an equal, but opposite, reaction.”
o FA on B = -FB on A
“Types of Forces” in mechanics
o Common forces
 Gravitational (Weight – at/near Earth’s surface)
 Normal
 Frictional forces
 Tension force
 Restoring force of a spring (Hooke’s Law)
Weight (5.6)
o Weight = gravitational force that the earth (or another astronomical body like the
moon) exerts on an object
 Weight always acts downward, toward the center of the earth
 An object has weight regardless of whether or not it is resting on the
surface of the earth
 Weight depends on the acceleration at the surface of the Earth (or other
astronomical body like the moon), while the mass of an object does not
change
The Normal Force (4.7)
o The component of force that a surface exerts on an object with which it is in
contact
o This component is perpendicular to the surface
o The normal force is NOT always equal to the weight of the object lying on the
surface – it depends on what other forces are present.
o Apparent weight (see textbook for discussion)
Static and Kinetic Frictional Forces (6.1)
o Even highly polished surfaces are rough when looked at on a microscopic scale
o Static frictional force – doesn’t exist until you try to slide an object
 fs  fsmax (balances out force trying to slide an object to a certain point, a
maximum that occurs just before the object begins to slide)
 fsmax = sFn
 s = coefficient of static friction (typical values: 0.01 for smooth surfaces,
and 1.5 for rough surfaces)
 FN = magnitude of the normal force
 Direction – opposes sliding motion
 Independent of surface area
o Kinetic frictional force
 fk = kFn
 k = coefficient of kinetic friction
 FN = magnitude of the normal force
 Direction – opposes sliding motion
 k typically less than s
The Tension Force (6.2 – Strings)
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o A force is applied to right side of rope that is tied to a block (on the ropes left
side) – every particle of the rope in turn applies a force to (pulls) its neighbor –
finally in turn, the force is applied to the block. We say “the force T is applied to
the block because of the tension in the rope.”
o Consider common usage of the word tension, where it is the tendency of the rope
to be pulled apart. The right side of the rope in the situation above is pulled by
force +F, while the left side of the rope is pulled the opposite direction (-F) by the
block attached to the rope (Newton’s 3rd law)
o We have and will continue to use the concept of a “massless” rope – no net force
is required to accelerate the rope, so the force applied to the rope is applied
undiminished to the attached object
The Restoring force of Springs (6.2 – Springs)
o Hooke’s law: FR = -kx
 x = displacement of end of spring away from equilibrium position
 k = spring constant (also called the force constant) with SI units of N/m
 the minus sign means the force points in the opposite direction of the
displacement, x
 assume the spring itself has no mass (an ideal spring)
Translational Equilibrium (6.3)
o F = 0
 Fx = 0 and Fy = 0
o An object in equilibrium is not accelerating, so there is no net force.
o The net force on an object in equilibrium is zero, so the object is not accelerating.
o Remember, an object with a constant velocity (is moving, but not changing
direction), is NOT being accelerated, so the net force is zero.
Connected Objects (6.4)
o Two ways to approach problem
 Method #1: Treat each object separately, and
 draw free body diagrams for each object separately
 adding up forces for each object separately
 remember that the acceleration is the same for all of the connected
objects (or the connecting string would stretch, break or bunch up)
 the magnitude of the tension is the same for all objects connected
by the same string
 Use algebra to combine separate equations and solve for the
unknowns.
 Method #2: Treat all the objects as a system
 Mass of system = sum of individual masses
 Only consider forces external (outside) the system,
o so the tension in a string connecting objects in a system
doesn’t count
 remember that the acceleration is the same for all of the connected
objects (or the connecting string would stretch, break or bunch up)
Circular Motion (6.5)
o Motion of an object traveling at a constant (uniform)speed on a circular path
o Direction is constantly changing, so the velocity is NOT constant
o Period, T – time required to travel once around the circle (one complete rotation)
2r
v
T
o
(speed)
o Centripetal Acceleration
v2
ac 
r
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(magnitude)
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4 2 r
ac  2
T
 Substituting for v, we get
(magnitude)
 Direction – always points toward center of circle.
o Centripetal Force
v2
 Fc  mac  m
r
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(magnitude)
 Not a new force, just the name we give to the net force or component of
force that points toward the center of the circle
o Banked Curves
 Component of the normal force that points toward the center of the circle.
(Occasionally, other forces or components of forces might contribute to
the centripetal force in addition to the component of the normal force.)
 Knowing the speed wanted and the radius of the (circular) curve, you can
solve for the angle of the incline (bank). It is independent of the mass of
the vehicle going around the curve. Example shown in class.
Work Done by a Constant Force (7.1)
o W = (Fcos)d
 F = magnitude of force applied
 d= magnitude of displacement
  = angle between the direction of the force applied and the direction of
the displacement (values between 0 and 180 degrees)
 Work is a scalar quantity, but can be either positive or negative
 SI units = Nm = J = Joule
The Work-Energy Theorem and Kinetic Energy (7.2)
o Wnet = KE = KEf - KE0 = ½ mvf2 + ½ mv02
o Kinetic Energy:
KE = ½ mv2
 SI units = J = Joule
Work done by a variable force (7.3)
o Work = area under the curve in a graph of Force versus Position (y vs x)
o Work to stretch or compress a spring a distance x from equilibrium
 WSp = ½ kx2
 always positive
o Work done BY A SPRING
 WSp = ½ kx2
 May be POSITIVE or NEGATIVE depending on the situation
 Look at direction of force from the spring and the direction of the
displacement to see if the work is positive or negative (see also textbook,
figure 7-12, Physics, by James S. Walker, 4th edition)
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Power (7.4)
W
t
o
o For constant speed, P = Fv
Conservative forces (8.1)
o Definition 1: “a force that does zero total work on any closed path.”
o Definition 2: “If the work done by a force in going from arbitrary point A to an
arbitrary point B is independent of the path from A to B, the force is
conservative.”
 Gravitational force (Weight)
 Elastic Spring force
 Electric force (Phy 123)
Non-Conservative forces (8.1)
 Static and kinetic friction forces
 Air resistance
 Normal force
 Tension force
Potential Energy and Work Done by Conservative Forces (8.2)
o Potential Energy, PE
o Work done by conservative forces change the potential energy
 WC = PE0-PEf = -(PEf-PE0) = -ΔPE
o Gravity
 PE = mg(h0 – hf) = mgh
o Springs
 PE = ½ kx2
 PE of spring is always positive or zero
The Conservation of Mechanical Energy (8.3)
o Mechanical Energy, ME = PE + KE
o If net work done by nonconservative forces is zero then mechanical energy is
conserved
 If WNC = 0, then Ef = E0
o This equation can also be written as
½ mvf2 + mghf + ½ kxf2 = ½ mv02 + mgh0 + ½ kx02
Work Done by Nonconservative forces (8.4)
o If nonconservative forces do net work, then the mechanical energy changes
 WNC = ME
 WNC = KE + PE
 WNC + KE0 +PE0 = KEf + PEf
Pave 
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How to do problems with work-energy
1. Draw a picture
2. Identify external conservative and non-conservative forces that act on an object.
3. If the net work done by non-conservative forces is zero (WNC = 0), then Conservation
of Mechanical Energy applies (Ef = E0). Otherwise, the work done by nonconservative forces is equal to the change in mechanical energy (WNC = KE + PE).
4. Choose the location where PEgravity = 0 (where all heights in the problem are relative
to). Don’t change this during the course of solving the problem. Usually PEspring = 0
at equilibrium position.
5. Now solve, using WNC = 0  Ef = E0, OR WNC = KE + PE, depending on what
you found regarding the work done by external non-conservative forces.
The following is arranged for a different textbook, so it may seem out of order compared to
Physics, 4th ed., by James S. Walker. – Dr. Nazareth, 2/17/2010.
 The Impulse-Momentum Theorem (9.3)
o Impulse = Ft = p
o Linear Momentum = p = mv
(9.1)
 The Principle of Conservation of Linear Momentum (9.4)
o If net external force is zero (Fext=0) then total momentum is conserved
p0 = pf
 Collisions (9.5 and 9.6)
o Elastic collision: KE0 = KEf
o Inelastic collision: KE0  KEf (energy before collision does NOT equal energy
after)
o Completely inelastic collision: Objects collide and “stick together”, moving as
one object after the collision with a combined mass.
 It is vital that you understand whether a collision is ELASTIC or INELASTIC!
 Center of Mass (9.7)
o X-coordinate of the center of mass
 XCM = ∑(mx)/M
o Y-coordinate of the center of mass
 YCM = ∑(my)/M
o Velocity of the center of mass
 VCM = ∑(mv)/M
o Acceleration of the center of mass
 ACM = ∑(ma)/M
o Newton’s 2nd law for a System of particles
 MACM = ∑Fext = Fnet, external
 Systems with Changing Mass – Rocket Propulsion
o NOT REQUIRED, only for one or two extra credit homework problems in ch. 9
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Angular Position (10.1)
o θ (radians) = angle measured from reference line
 counterclockwise rotation = +
 clockwise rotation =  2π radians = 360 degrees = 1 revolution
 1 rad. = 360 degrees/2π ≈ 57.3 degrees
 units for angles (radians or degrees or revolutions) = dimensionless – we
just write rad or radians to remind us which units we are using for angles
Angular Displacement (10.1)
o Δθ (in radians) = θ –θ0 = s/r
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Note: I am often a bit lazy and assume θ0 = 0, so that Δθ = θ, and just use
θ for the angular displacement (not the position). A bad habit, I know …
but so rarely is θ0 anything other than 0. – Dr. Nazareth
Angular Velocity (10.1)

 ave 
t
o Average angular velocity
 Units = radians/second = rad./s = s-1
 remember angles are dimensionless (see angular position)
 counterclockwise rotation = +
 clockwise rotation = o Instantaneous angular velocity, ω
 angular velocity at particular instant in time (in limit where Δt → 0)
 Units = radians/second = rad./s = s-1
 usually just say angular velocity
 counterclockwise rotation = +
 clockwise rotation = o Angular speed = magnitude of angular velocity
Angular Acceleration (10.1)

 ave 
t
o Average angular acceleration
2
2
-2
 Units = radians/second = rad./s = s
 remember angles are dimensionless (see angular position)
 if Δω > 0 then α = +
 if Δω < 0 then α =  if ω and α have same sign then rotation is increasing
 if ω and α have opposite sign then rotation is decreasing
o Instantaneous angular acceleration, α
 angular velocity at particular instant in time (in limit where Δt → 0)
 Units = radians/second2 = rad./s2 = s-2
 usually just say angular acceleration
 See average angular acceleration for sign convention
The Equations of Rotational Kinematics (10.2)
o ω = ω0 + t
o  = 0 + ½ (ω + ω0)t
o  = 0 + ω0t + ½ t2
o ω2 = ω02 + 2( - 0)
o Apply angular equations in the same way you did for the linear equations in
chapter 2, just the physical interpretation is different.
Connection between Linear and Rotational Variables (10.3)
o Tangential velocity vT = rω
(vT in m/s, r in m and ω in rad/s)
2
v
ac  T  r 2
r
o Centripetal acceleration
(ac in m/s2, r in m, ω in rad/s)
 Due to changing direction of motion
o Tangential acceleration
aT = r
(aT in m/s2, r in m and  in rad/s2)
 Due to changing angular speed
 If tangential speed is changing, then nonuniform circular motion.
o When both ac and aT are present
o
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a  ac2  aT2
(magnitude of the acceleration)
a 
  tan  T 
 ac 
o
(direction of the acceleration)
Rolling Motion (10.4)
o Example of car tire rolling.
o Linear speed of car = tangential speed of point on outer edge of the tire.
 v = vT = r
( in rad/s)
o Linear acceleration of car = tangential acceleration of point on outer edge of the
tire.
 a = aT = r
( in rad/s2)
Rotational Kinetic Energy and the Moment of Inertia (10.5)
o I = moment of inertia (Table 10.1 lists for common objects)
 Measure of the object’s resistance to a change in angular motion
 Depends on the mass of an object and how the mass is distributed about
the axis of rotation in question.
 For a system of particles
I = ∑miri2
 SI units = kgm2
o Rotational Kinetic Energy: KER = ½ I2
( in rad/s)
Conservation of Energy (10.6)
o Total Kinetic Energy (rolling motion):
KET = ½ mv2 + ½ I2
o Total Mechanical Energy:
ME = ½ mv2 + ½ I2 + mgh + ½ kx2
 ME = KEtranslational + KErotational + PEgravitational + PEspring
o Work-Energy Theorem:
 WNC + ½mv02 + ½I02 +mgh0 + ½kx02 = ½ mvf2 + ½ If2 + mghf + ½ kx2
 If WNC = 0, then Ef = E0
Torque (11.1)
o  = lF
 Where: F = applied force (N), and l = lever arm (m).
 Lever arm = moment arm
o =lxF
(cross product of l and F; General definition of torque)
o  = l(F sinφ)
(force is at angle φ relative to radial line)
o  = Fperpendicular d
(d = distance force is applied from axis of rotation)
o Convention: If force produces counterclockwise rotation = +.
If produces clockwise rotation = -.
Torque and Angular Acceleration (11.2)
o  = I
(I = moment of inertia;  = angular acceleration in rad/s2)
o A net torque causes an angular acceleration

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Magnitude depends on distribution of mass about axis of rotation/pivot
point
 Analogous to Newton’s 2nd law for translational motion
Zero Torque and Static Equilibrium (11.3)
o Fx = 0
and
Fy = 0
and
 = 0
Center of Mass and Balance (11.4)
o Center of Gravity = where object’s weight can be considered to act when
calculating the torque due to a weight = center of mass
 Geometric center of regular shaped object with uniform mass distribution
 If object supported directly under center of gravity (center of mass), object
balances
Angular Momentum (11.6)
o L = I
(SI units = kg m2/s)
 I = moment of inertia
  in rad/s)
o Angular momentum for a point particle
 L = rpsinφ
 φ = angle between p and radial direction
o Newton’s 2nd law for rotational motion
  = I = ΔL/Δt
Conservation of Angular Momentum (11.7)
o If ext = 0
then Lf = L0.
o Spinning figure skater example
 Arms out = slower rotation because moment of inertia is larger (more
mass farther from axis of rotation)
 Arms in = faster rotation because moment of inertia is smaller (mass is
closer to axis of rotation)
Rotational Work and Power (11.8)
o Rotational Work:
WR =  Δ 
( in rad)
2
o Rotational Kinetic Energy: KER = ½ I
( in rad/s)
2
2
o Total Kinetic Energy:
KET = ½ mv + ½ I
o Power produced by a torque:
P = Work/Δt =  Δ/Δt = ave =  if  = 0
Newton’s Law of Universal Gravitation (12.1)
Gm1m2
F
r2
o
(magnitude of force)
o m1 and m2 are mathematical points
o G = 6.6710-11 Nm2/kg2 = universal gravitation constant – same value for all pairs
of particles everywhere in the universe
o It is an attractive force
o The force that each particle exerts on the other is directed along the line joining
the particles
o (12.2) Also true for spherical bodies (like planets or stars), but distance is to
center of the sphere (planet)
o Weight = gravitational force that the earth (or another astronomical body like the
moon) exerts on an object
M Em
 mg
r2

(Weight = mg is the approximation)
 Weight always acts downward, toward the center of the earth
 An object has weight regardless of whether or not it is resting on the
surface of the earth
 Weight depends on G, the mass of the earth, ME, and distance, r, while the
mass of an object does not change
Kelpler’s Laws of Orbital Motion (12.3)
o First law: Planets follow elliptical orbits with the Sun at one focus of the ellipse.
o Second law: As a planet moves in its orbit, it sweeps out an equal amount of area
in an equal amount of time (See Figure 12-9 in textbook)
o Third law: The period, T, of a planet increases as its mean distance form the Sun,
r, raised to the 3/2 power. That is, T = (constant)r3/2
 For planet orbiting the Sun (circular orbit), T = (2π/sqrt(GMsun))r3/2
 Same formula works for orbiting the Earth or another planet – just replace
the mass of the Sun with the mass of the planet being orbitted
 Note the period depends on the mass of the object being orbited, not the
mass of the orbiting object
Satellites in Circular Orbits
o Satellite is kept on circular path by gravitational force.
GM
mv 2 GMm
v
Fc 
 2
r
r
r
o
(magnitude) →
o If a satellite is to remain at radius, r, it must have speed, v, precisely.
m4 2 r GMm
Fc 

T2
r 2 (magnitude)
o
Gravitational Potential Energy (12.4)
o PEg = -GmM/r
o PEg approaches zero as r approaches infinity
o Convenient to choose infinity as reference point for PEg = 0 when have distances
on astronomical scale
Energy Conservation (12.5)
o Mechanical energy of an object of mass, m, at distance, r, from the Earth
 ME = PE + KE = ½ mv2 – GmME/r
o Speed of asteroid striking earth, starting at infinite distance with v0 = 0
 vf = sqrt(2GME/RE) = 11,200 m/s ≈ 25000 miles per hour
o Escape speed of object just barely leaving Earth, with vf = 0 when reaches infinite
distance from Earth.
 This is the initial velocity it must have at the surface to escape Earth’s
gravity
 v0 = sqrt(2GME/RE) = 11,200 m/s ≈ 25000 miles per hour
W G
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