Fall 2008 Honors Physics Review Notes - Tom Strongtomstrong.org/physics/fall08h.pdfHonors Physics Review Notes Fall 2008 Tom Strong Science Department Mt Lebanon High School [email protected]

Honors Physics Review Notes

Fall 2008

Tom StrongScience Department

Mt Lebanon High [email protected]

The most recent version of this can be found at http://www.tomstrong.org/physics/

Chapter 1 — The Science of Physics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1Chapter 2 — Motion in One Dimension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3Chapter 3 — Two-Dimensional Motion and Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4Chapter 4 — Forces and the Laws of Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5Chapter 5 — Work and Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7Chapter 6 — Momentum and Collisions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8Chapter 7 — Rotational Motion and the Law of Gravity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9Chapter 8 — Rotational Equilibrium and Dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10Variables and Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12Mathematics Review for Physics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15Physics Using Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20Data Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22Midterm Exam Description . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23Semester Exam Equation Sheet . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24Mixed Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25Mixed Review Answers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

ii

These notes are meant to be a summary of important points covered in the Honors Physics class at Mt. Lebanon HighSchool. They are not meant to be a replacement for your own notes that you take in class, nor are they a replacement foryour textbook. Much of the material in here is taken from the textbook without specifically acknowledging each case, inparticular the organization and overall structure exactly match the 2002 edition of Holt Physics by Serway and Faughnand many of the expressions of the ideas come from there as well.

The mixed review exercises were taken from the supplementary materials provided with the textbook. They are arepresentative sampling of the type of mathematical problems you may see on the final exam for the course. There willalso be conceptual questions on the exam that may not be covered by the exercises included here. These exercises areprovided to help you to review material that has not been seen in some time, they are not meant to be your only resourcefor studying. Exercises are included from chapters 1–8, 12–17, and 19–22.

The answers at the end of the review are taken from the textbook, often without verifying that they are correct. Usethem to help you to solve the problems but do not accept them as correct without verifying them yourself.

This is a work in progress and will be changing and expanding over time. I have attempted to verify the correctnessof the information presented here, but the final responsibility there is yours. Before relying on the information in thesenotes please verify it against other sources.

Honors Physics 2008–2009 Mr. Strong

1

Chapter 1 — The Science of Physics

1.1 What is Physics?Some major areas of Physics:Mechanics — motion and its causes — falling objects,

friction, weightThermodynamics — heat and temperature — melting

and freezing processes, engines, refrigeratorsVibrations and Waves — specific types of repeating

motions — springs, pendulums, soundOptics — light — mirrors, lenses, colorElectromagnetism — electricity, magnetism, and light

— electrical charge, circuitry, magnetsRelativity — particles moving at very high speeds — par-

ticle accelerators, particle collisions, nuclear energyQuantum Mechanics — behavior of sub-microscopic

particles — the atom and its parts

The steps of the Scientific Method1. Make observations and collect data that lead to a

question2. Formulate and objectively test hypotheses by experi-

ments (sometimes listed as 2 steps)3. Interpret results and revise the hypotheses if neces-

sary4. State conclusions in a form that can be evaluated by

others

1.2 Measurements in ExperimentsMeasurementsThere are 7 basic dimensions in SI (Systeme International),the 3 we will use most often are:

• Length — meter (m) — was 1/10,000,000 of the dis-tance from the equator to the North Pole — now thedistance traveled by light in 3.3× 10−9 s

• Mass — kilogram (kg) — was the mass of 0.001 cubicmeters of water, now the mass of a specific platinum-iridium cylinder

• Time — second (s) — was a fraction of a mean so-lar day, now 9,162,631,700 times the period of a radiowave emitted by a Cesium-133 atom

Common SI Prefixes

Prefix Multiple Abbrev.

nano- 10−9 n deci- 10−1 dmicro- 10−6 µ kilo- 103 kmilli- 10−3 m mega- 106 Mcenti- 10−2 c giga- 109 G

Accuracy vs. Precision

• Accuracy describes how close a measured value is tothe true value of the quantity being measured

Problems with accuracy are due to error. To avoiderror:

– Take repeated measurements to be certain thatthey are consistent (avoid human error)

– Take each measurement in the same way (avoidmethod error)

– Be sure to use measuring equipment in goodworking order (avoid instrument error)

• Precision refers to the degree of exactness with whicha measurement is made and stated.

– 1.325 m is more precise than 1.3 m

– lack of precision is usually a result of the limi-tations of the measuring instrument, not humanerror or lack of calibration

– You can estimate where divisions would fall be-tween the marked divisions to increase the pre-cision of the measurement

1.3 The Language of Physics

There are many symbols that will be used in this class, someof the more common will be:

Symbol Meaning

∆x Change in xxi, xf Initial, final values of x∑

F Sum of all F

Dimensional analysis provides a way of checking tosee if an equation has been set up correctly. If the units re-sulting from the calculation are not those that are expectedthen it’s very unlikely that the numbers will be correct ei-ther. Order of magnitude estimates provide a quickway to evaluate the appropriateness of an answer — if theestimate doesn’t match the answer then there’s an errorsomewhere.


2

Counting Significant Figures in a Number

Rule Example

All counted numbers have an infinite numberof significant figures

10 items, 3 measurements

All mathematical constants have an infinitenumber of significant figures

1/2, π, e

All nonzero digits are significant 42 has two significant figures; 5.236 has four

Always count zeros between nonzero digits 20.08 has four significant figures; 0.00100409 has six

Never count leading zeros 042 and 0.042 both have two significant figures

Only count trailing zeros if the number con-tains a decimal point

4200 and 420000 both have two significant figures; 420.has three; 420.00 has five

For numbers in scientific notation apply theabove rules to the mantissa (ignore the expo-nent)

4.2010× 1028 has five significant figures

Counting Significant Figures in a Calculation

Rule Example

When adding or subtracting numbers, find the number which is knownto the fewest decimal places, then round the result to that decimalplace.

21.398 + 405 − 2.9 = 423 (3significant figures, roundedto the ones position)

When multiplying or dividing numbers, find the number with the fewestsignificant figures, then round the result to that many significant fig-ures.

0.049623 × 32.0/478.8 =0.00332 (3 significant figures)

When raising a number to some power count the number’s significantfigures, then round the result to that many significant figures.

5.82 = 34 (2 significantfigures)

Mathematical constants do not influence the precision of any compu-tation.

2× π × 4.00 = 25.1 (3 signif-icant figures)

In order to avoid introducing errors during multi-step calculations, keepextra significant figures for intermediate results then round properlywhen you reach the final result.

Rules for Rounding

Rule Example

If the hundredths digit is 0 through 4 drop it and all following digits. 1.334 becomes 1.3

If the hundredths digit is 6 though 9 round the tenths digit up to thenext higher value.

1.374 becomes 1.4

If the hundredths digit is a 5 followed by other non-zero digits thenround the tenths digit up to the next higher value.

1.351 becomes 1.4

If the hundredths digit is a 5 not followed by any non-zero digits thenif the tenths digit is even round down, if it is odd then round up.

1.350 becomes 1.4, 1.250 be-comes 1.2

(assume that the result was to be rounded to the nearest 0.1, for other precisions adjust accordingly)


3

Chapter 2 — Motion in One Dimension

2.1 Displacement and Velocity

The displacement of an object is the straight line (vector)drawn from the object’s initial position to its new position.Displacement is independent of the path taken and is notnecessarily the same as the distance traveled. Mathemati-cally, displacement is:

∆x = xf − xi

The average velocity, equal to the constant velocitynecessary to cover the given displacement in a certain timeinterval, is the displacement divided by the time intervalduring which the displacement occurred, measured in m

s

vavg =∆x∆t

=xf − xi

tf − ti

The instantaneous velocity of an object is equivalentto the slope of a tangent line to a graph of x vs. t at thetime of interest.

The area under a graph of instantaneous velocity vs.time (v vs. t) is the displacement of the object during thattime interval.

2.2 Acceleration

The average acceleration of an object is the rate ofchange of its velocity, measured in m

s2 . Mathematically, itis:

aavg =∆v∆t

=vf − vi

tf − ti

Like velocity, acceleration has both magnitude and di-rection. The speed of an object can increase or decreasewith either positive or negative acceleration, depending onthe direction of the velocity — negative acceleration doesnot always mean decrease in speed.

The instantaneous acceleration of an object is equiv-alent to the slope of a tangent line to the v vs. t graph atthe time of interest, while the area under a graph of instan-taneous acceleration vs. time (a vs. t) is the velocity ofthe object. In this class acceleration will almost always beconstant in any problem.

Displacement and velocity with constant uniform accel-eration can be expressed mathematically as any of:

vf = vi + a∆t

∆x = vi∆t+12a∆t2

∆x =12(vi + vf )∆t

v2f = v2

i + 2a∆x

2.3 Falling ObjectsIn the absence of air resistance all objects dropped near thesurface of a planet fall with effectively the same constantacceleration — called free fall. That acceleration is alwaysdirected downward, so in the customary frame of referenceit is negative, so:

ag = g = −9.81ms2


4

Chapter 3 — Two-Dimensional Motion and Vectors

3.1 Introduction to VectorsVectors can be added graphically.

Vectors can be added in any order:

~V1 + ~V2 = ~V2 + ~V1

To subtract a vector you add its opposite:

~V1 − ~V2 = ~V1 + (−~V2)

Multiplying a vector by a scalar results in a vector insame direction as the original vector with a magnitude equalto the original magnitude multiplied by the scalar.

3.2 Vector OperationsTo add two perpendicular vectors use the Pythagorean The-orem to find the resultant magnitude and the inverse of thetangent function to find the direction: Vr =

√V 2

x + V 2y and

the direction of Vr, θr, is tan−1 Vy

Vx

Just as 2 perpendicular vectors can be added, any vectorcan be broken into two perpendicular component vectors:~V = ~Vx + ~Vy where ~Vx = V cos θı and ~Vy = V sin θ;

Two vectors with the same direction can be added byadding their magnitudes, the resultant vector will have thesame direction as the vectors that were added.

Any two (or more) vectors can be added by first decom-posing them into component vectors, adding all of the x andy components together, and then adding the two remainingperpendicular vectors as described above.

3.3 Projectile MotionThe equations of motion introduced in chapter 2 are ac-tually vector equations. Once the new symbol for displace-ment, ~d = ∆xı+∆y has been introduced the most commonones can be rewritten as

~d = ~vi∆t+12~a∆t2

~vf = ~vi + ~a∆t

Since motion in each direction is independent of motion inthe other, objects moving in two dimensions are easier toanalyze if each dimension is considered separately.

~dx = ~vxi∆t+12~ax∆t2

~dy = ~vyi∆t+12~ay∆t2

In the specific case of projectile motion there is noacceleration in the horizontal (ı) direction (~ax = 0) andthe acceleration in the vertical () direction is constant(~ay = ~ag = ~g = −9.81 m

s2 ). While ~vy will change over time,~vx remains constant. Neglecting air resistance, the path fol-lowed by an object in projectile motion is a parabola. Thefollowing equations use these simplifications to describe themotion of a projectile launched with speed ~vi and directionθ up from horizontal:

~vx = vi cos θı = constant

~vyi = vi sin θ

~dx = vi cos θ∆tı

~dy = vi sin θ∆t+12~g∆t2

To solve a projectile motion problem use one part of theproblem to find ∆t, then once you know ∆t you can usethat to solve the rest of the problem. In almost every casethis will involve using the vertical part of the problem tofind ∆t which will then let you solve the horizontal partbut there may be some problems where the opposite ap-proach will be necessary. For the special case of a projectilelaunched on a level surface the range can be found with

dx =v2 sin 2θag


5

Chapter 4 — Forces and the Laws of Motion

4.1 Changes in MotionForce causes change in velocity. It can cause a station-ary object to move or a moving object to stop or otherwisechange its motion.

The unit of force is the newton (N), equivalent to kg·ms2 ,

which is defined as the amount of force that, when actingon a 1 kg mass, produces an acceleration of 1m

s2 .Contact forces act between any objects that are in

physical contact with each other, while field forces actover a distance.

A force diagram is a diagram showing all of the forcesacting on the objects in a system.

A free-body diagram is a diagram showing all of theforces acting on a single object isolated from its surround-ings.

4.2 Newton’s First LawNewton’s first law states that

An object at rest remains at rest, and an ob-ject in motion continues in motion with constantvelocity (that is, constant speed in a straightline) unless the object experiences a net exter-nal force.

This tendency of an object to not accelerate is calledinertia. Another way of stating the first law is that if thenet external force on an object is zero, then the accelerationof that object is also zero. Mathematically this is∑

~F = 0 −→ ~a = 0

An object experiencing no net external force is said tobe in equilibrium, if it is also at rest then it is in staticequilibrium

4.3 Newton’s Second and Third LawsNewton’s second law states that

The acceleration of an object is directly pro-portional to the net external force acting on theobject and inversely proportional to the object’smass.

Mathematically this can be stated as:∑~F = m~a

Which in the case of no net external force (∑ ~F = 0) also

illustrates the first law:∑~F = 0 −→ m~a = 0 −→ ~a = 0

Newton’s third law states that

If two objects interact, the magnitude of theforce exerted on object 1 by object 2 is equal tothe magnitude of the force simultaneously ex-erted on object 2 by object 1, and these twoforces are opposite in direction.

Mathematically, this can be stated as:

~F1,2 = −~F2,1

The two equal but opposite forces form an action-reactionpair.

4.4 Everyday Forces

The weight of an object (~Fg) is the gravitational force ex-erted on the object by the Earth. Mathematically:

~Fg = m~g where ~g = −9.81ms2j

The normal force (~Fn) is the force exerted on an objectby the surface upon which the object rests. This force isalways perpendicular to the surface at the point of contact.

The force of static friction (~Fs) is the force that op-poses motion before an object begins to move, it will pre-vent motion so long as it has a magnitude greater than theapplied force in the direction of motion. The maximummagnitude of the force of static friction is the product ofthe magnitude of the normal force times the coefficientof static friction (µs) and its direction is opposite thedirection of motion:

µs =Fs,max

Fn−→ Fs,max = µsFn

The force of kinetic friction (~Fk) is the force that op-poses the motion of an object that is sliding against a sur-face. The magnitude of the force of kinetic friction is theproduct of the magnitude of the normal force times thecoefficient of kinetic friction (µk) and its direction isopposite the direction of motion:

µk =Fk

Fn−→ Fk = µkFn

The force of friction between two solid objects dependsonly on the normal force and the coefficient of friction, itis independent of the the surface area in contact betweenthem.

For an object at rest with a force applied to it the fric-tional force will vary as the applied force is increased.


6

Fapp

Ffriction

µkFn

µsFn

µsFnstatic kinetic

Fs = Fapp

Fk = µkFn

Air resistance (~FR) is the force that opposes the mo-tion of an object though a fluid. For small speeds FR ∝ vwhile for large speeds FR ∝ v2. (Exactly what is small orlarge depends on things that are outside the scope of thisclass.) Air resistance is what will cause falling objects toeventually reach a terminal speed where FR = Fg.

Solving Friction Problems

When solving friction problems start by drawing a diagramof the system being sure to include the gravitational force(Fg), normal force (Fn), frictional force (Fk, Fs, or Fs,max)and any applied force(s).

Determine an appropriate frame of reference, rotatingthe x and y coordinate axes if that is convenient for theproblem (such as an object on an inclined plane). Resolveany vectors not lying along the coordinate axes into com-ponents.

Use Newton’s second law (∑ ~F = m~a) to find the rela-

tionship between the acceleration of the object (often zero)and the applied forces, setting up equations in both the xand y directions (

∑ ~Fx = m~ax and∑ ~Fy = m~ay) to solve

for any unknown quantities.


7

Chapter 5 — Work and Energy

5.1 WorkA force that causes a displacement of an object does work(W ) on the object. The work is equal to the product of thedistance that the object is displaced times the component ofthe force in the direction of the displacement, or if θ is theangle between the displacement vector and the (constant)net applied force vector, then:

W = Fnetd cos θ

The units of work are Joules (J) which are equivalentto N ·m or kg·m2

s2 . The sign of the work being done is sig-nificant, it is possible to do a negative amount of work.

5.2 EnergyWork done to change the speed of an object will accumu-late as the kinetic energy (KE , or sometimes K) of theobject. Kinetic energy is:

KE =12mv2

If work is being done on an object, the work-kineticenergy theorem shows that:

Wnet = ∆KE

In addition to kinetic energy, there is also potentialenergy (PE , or sometimes U) which is the energy storedin an object because of its position. If the energy is storedby lifting the object to some height h, then the equation forgravitational potential energy is:

PE g = mgh

Potential energy can also be stored in a compressedor stretched spring, if x is the distance that a spring isstretched from its rest position and k is the spring con-stant measuring the stiffness of the spring (in newtons permeter) then the elastic potential energy that is storedis:

PE elastic =12kx2

The units for all types of energy are the same as thosefor work, Joules.

5.3 Conservation of EnergyThe sum an objects kinetic energy and potential energy isthe object’s mechanical energy (ME ). In the absence offriction, the total mechanical energy of a system will remainthe same. Mathematically, this can be expressed as:

ME i = ME f

In the case of a single object in motion, this becomes:

12mv2

i +mghi =12mv2

f +mghf

5.4 PowerThe rate at which energy is transferred is called power (P ).The mathematical expression for power is:

P =W

∆t=Fd

∆t= F

d

∆t= Fv

The units for power are Watts (W) which are equivalentto J

s or kg·m2

s3


8

Chapter 6 — Momentum and Collisions

6.1 Momentum and ImpulseMomentum is a vector quantity described by the productof an object’s mass times its velocity:

~p = m~v

If an object’s momentum is known its kinetic energy canbe found as follows:

KE =p2

2m

The change in the momentum of an object is equal tothe impulse delivered to the object. The impulse is equalto the constant net external force acting on the object timesthe time over which the force acts:

∆~p = ~F∆t

Any force acting on on object will cause an impulse,including frictional and gravitational forces.

In one dimension the slope of a graph of the momen-tum of an object vs. time is the net external force actingon the object. The area under a graph of the net externalforce acting on an object vs. time is the total change inmomentum of that object.

Momentum and impulse are measured in kg·ms — there

is no special name for that unit.

6.2 Conservation of MomentumMomemtum is always conserved in any closed system:

Σ~pi = Σ~pf

For two objects, this becomes:

m1~v1i +m2~v2i = m1~v1f +m2~v2f

Any time two objects interact, the change in momen-tum of one object is equal in magnitude and opposite indirection to the change in momentum of the other object:

∆~p1 = −∆~p2

6.3 Elastic and Inelastic CollisionsThere are three types of collisions:

• Elastic — momentum and kinetic energy are con-served. Both objects return to their original shapeand move away separately. Generally:

Σ~pi = Σ~pf

ΣKEi = ΣKEf

For two objects:

m1~v1i +m2~v2i = m1~v1f +m2~v2f

12m1v

21i +

12m2v

22i =

12m1v

21f +

12m2v

22f

• Inelastic — momentum is conserved, kinetic energyis lost. One or more of the objects is deformed in thecollision. Generally:

Σ~pi = Σ~pf

ΣKEi > ΣKEf

For two objects:

m1~v1i +m2~v2i = m1~v1f +m2~v2f

12m1v

21i +

12m2v

22i >

12m1v

21f +

12m2v

22f

• Perfectly inelastic — momentum is conserved, ki-netic energy is lost. One or more objects may bedeformed and the objects stick together after the col-lision. Generally:

Σ~pi = Σ~pf

ΣKEi > ΣKEf

For two objects:

m1~v1i +m2~v2i = (m1 +m2)~vf

12m1v

21i +

12m2v

22i >

12(m1 +m2)v2

f

True elastic or perfectly inelastic collisions are very rarein the real world. If we ignore friction and other small en-ergy losses many collisions may be modeled by them.

Newton’s Laws in terms of Momentum1. Inertia:

Σ~F = 0 −→ ~p = constant

2. ~F = m~a:∆~p = ~F∆t

3. Every action has an equal and opposite reaction:

∆~p1 = −∆~p2


9

Chapter 7 — Rotational Motion and the Law of Gravity

7.1 Measuring Ratational Motion

For an object moving in a circle with radius r through anarc length of s, the angle θ (in radians) swept by the objectis:

θ =s

r

The conversion between radians and degrees is:

θ(rad) =π

180oθ(deg)

The angular displacement (∆θ) through which anobject moves from θi to θf is, in rad:

∆θ = θf − θi =sf − si

r=

∆sr

The average angular speed (ω) of an object is, in rads ,

the ratio between the angular displacement and the timeinterval required for that displacement:

ωavg =θf − θi

∆t=

∆θ∆t

The average angular acceleration (α) is, in rads2 :

αavg =ωf − ωi

∆t=

∆ω∆t

For each quantity or relationship in linear motion thereis a corresponding quantity or relationship in angular mo-tion:

Linear Angular

x θv ωa α

vf = vi + a∆t ωf = ωi + α∆t

∆x = vi∆t+ 12a∆t

2 ∆θ = ωi∆t+ 12α∆t2

v2f = v2

i + 2a∆x ω2f = ω2

i + 2α∆θ

∆x = 12 (vi + vf )∆t ∆θ = 1

2 (ωi + ωf )∆t

7.2 Tangential and Centripetal AccelerationIn addition to the angular speed of an object moving withcircular motion it is also possible to measure the object’stangential speed (vt) or instantaneous linear speed whichis measured in m

s as follows:

vt = rω

An object’s tangential acceleration (at) can also bemeasured, in m

s2 , as follows:

at = rα

The centripetal acceleration (ac) of an object is di-rected toward the center of the object’s rotation and hasthe following magnitude:

ac =v2

t

r= rω2

7.3 Causes of Circular MotionThe centripetal force (Fc) causing a centripetal accel-eration is also directed toward the center of the object’srotation, and has the following magnitude:

Fc = mac =mv2

t

r= mrω2

The centripetal force keeping planets in orbit is a grav-itational force (Fg) and it is found with Newton’s Univer-sal Law of Gravitation:

Fg = Gm1m2

r2

where G is the constant of universal gravitation whichhas been determined experimentally to be

G = 6.673× 10−11 N ·m2

kg2

An object in a circular orbit around the Earth will sat-isfy the equation

v =

√GME

r

where ME is the mass of the Earth.


10

Chapter 8 — Rotational Equilibrium and Dynamics

8.1 Torque

Just as a net external force acting on an object causes linearacceleration, a net torque (τ) causes angular acceleration.The torque caused by a force acting on an object is:

τ = rF sin θ

where F is the force causing the torque, r is the distancefrom the center to the point where the force acts on theobject, and θ is the angle between the force and a radialline from the object’s center through the point where theforce is acting. Torque is measured in N ·m.

The convention used by the book is that torque in acounterclockwise direction is positive and torque in a clock-wise direction is negative (this corresponds to the right-hand rule). If more that one force is acting on an objectthe torques from each force can be added to find the nettorque:

τnet =∑

τ

8.2 Rotation and Inertia

The center of mass of an object is the point at which allthe mass of the object can be said to be concentrated. Ifthe object rotates freely it will rotate about the center ofmass.

The center of gravity of an object is the point throughwhich a gravitational force acts on the object. For mostobjects the center of mass and center of gravity will be thesame point.

The moment of inertia (I) of an object is the object’sresistance to changes in rotational motion about some axis.Moment of inertia in rotational motion is analogous to massin translational motion.

Some moments of inertia for various common shapes are:

Shape I

Point mass at a distance r from the axis mr2

Solid disk or cylinder of radius r aboutthe axis

12mr

2

Solid sphere of radius r about its diameter 25mr

2

Thin spherical shell of radius r about itsdiameter

23mr

2

Thin hoop of radius r about the axis mr2

Thin hoop of radius r about the diameter 12mr

2

Thin rod of length l about its center 112ml

2

Thin rod of length l about its end 13ml

2

An object is said to be in rotational equilibrium whenthere is no net torque acting on the object. If there is also

no net force acting on the object (translational equilib-rium) then the object is in equilibrium (without any qual-ifying terms).

Type Equation Meaning

TranslationalEquilibrium

∑ ~F = 0 The net force on the ob-ject is zero

Rotational Equi-librium

∑τ = 0 The net torque on the

object is zero

8.3 Rotational DynamicsNewton’s second law can be restated for angular motion as:

τnet = Iα

This is parallel to the equation for translational motionas follows:

Type of Motion Equation

Translational ~F = m~aRotational τ = Iα

Just as moment of inertia was analogous to mass, theangular momentum (L) in rotational motion is analogousto the momentum of an object in translational motion.

This is parallel to the equation for translational motionas follows:

Type of Motion Equation

Translational ~p = m~vRotational L = Iω

The angular momemtum of an object is conserved in theabsence of an external force or torque.

Rotating objects have rotational kinetic energy ac-cording to the following equation:

KE r =12Iω2 =

L2

2I

Just as other types of mechanical energy may be con-served, rotational kinetic energy is also conserved in theabsence of friction.

8.4 Simple MachinesThere are six fundamental types of machines, called sim-ple machines, with which any other machine can be con-structed. They are levers, inclined planes, wheels, wedges,pulleys, and screws.

The main purpose of a machine is to magnify the outputforce of the machine compared to the input force, the ratioof these forces is called the mechanical advantage (MA)


11

of the machine. It is a unitless number according to thefollowing equation:

MA =output forceinput force

=Fout

Fin

When frictional forces are accounted for, some of theoutput force is lost, causing less work to be done by themachine than by the original force. The ratio of work doneby the machine to work put in to the machine is called theefficiency (eff ) of the machine:

eff =Wout

Win

If a machine is perfectly efficient (eff = 1) then the idealmechanical advantage (IMA) can be found by comparingthe input and output distances:

IMA =din

dout

This leads to another way to find the efficiency of themachine as well:

eff =MAIMA


12

Variables and Notation

SI Prefixes

Prefix Mult. Abb. Prefix Mult. Abb.

yocto- 10−24 y yotta- 1024 Yzepto- 10−21 z zetta- 1021 Zatto- 10−18 a exa- 1018 Efemto- 10−15 f peta- 1015 Ppico- 10−12 p tera- 1012 Tnano- 10−9 n giga- 109 Gmicro- 10−6 µ mega- 106 Mmilli- 10−3 m kilo- 103 kcenti- 10−2 c hecto- 102 hdeci- 10−1 d deka- 101 da

Units

Symbol Unit Quantity Composition

kg kilogram Mass SI base unitm meter Length SI base units second Time SI base unitA ampere Electric current SI base unitcd candela Luminous intensity SI base unitK kelvin Temperature SI base unit

mol mole Amount SI base unit

Ω ohm Resistance VA or m2·kg

s3·A2

C coulomb Charge A · sF farad Capacitance C

V or s4·A2

m2·kg

H henry Inductance V·sA or m2·kg

A2·s2

Hz hertz Frequency s−1

J joule Energy N ·m or kg·m2

s2

N newton Force kg·ms2

rad radian Angle mm or 1

T tesla Magnetic field NA·m

V volt Electric potential JC or m2·kg

s3·A

W watt Power Js or kg·m2

s3

Wb weber Magnetic flux V · s or kg·ms2·A

Notation

Notation Description

~x Vectorx Scalar, or the magnitude of ~x|~x| The absolute value or magnitude of ~x∆x Change in x∑x Sum of all x

Πx Product of all xxi Initial value of xxf Final value of xx Unit vector in the direction of x

A −→ B A implies BA ∝ B A is proportional to BA B A is much larger than B

Greek Alphabet

Name Maj. Min. Name Maj. Min.

Alpha A α Nu N νBeta B β Xi Ξ ξ

Gamma Γ γ Omicron O oDelta ∆ δ Pi Π π or $

Epsilon E ε or ε Rho P ρ or %Zeta Z ζ Sigma Σ σ or ςEta H η Tau T τ

Theta Θ θ or ϑ Upsilon Υ υIota I ι Phi Φ φ or ϕ

Kappa K κ Chi X χLambda Λ λ Psi Ψ ψ

Mu M µ Omega Ω ω


13

Variables

Variable Description Units

α Angular acceleration rads2

θ Angular position radθc Critical angle o (degrees)θi Incident angle o (degrees)θr Refracted angle o (degrees)θ′ Reflected angle o (degrees)∆θ Angular displacement radτ Torque N ·mω Angular speed rad

sµ Coefficient of friction (unitless)µk Coefficient of kinetic friction (unitless)µs Coefficient of static friction (unitless)~a Acceleration m

s2

~ac Centripetal acceleration ms2

~ag Gravitational acceleration ms2

~at Tangential acceleration ms2

~ax Acceleration in the x direction ms2

~ay Acceleration in the y direction ms2

A Area m2

B Magnetic field strength TC Capacitance F~d Displacement m

d sin θ lever arm m~dx or ∆x Displacement in the x direction m~dy or ∆y Displacement in the y direction m

E Electric field strength NC

f Focal length m~F Force N~Fc Centripetal force N

~Felectric Electrical force N~Fg Gravitational force N~Fk Kinetic frictional force N

~Fmagnetic Magnetic force N~Fn Normal force N~Fs Static frictional force N~F∆t Impulse N · s or kg·m

s~g Gravitational acceleration m

s2

Variable Description Units

h Object height mh′ Image height mI Current AI Moment of inertia kg ·m2

KE or K Kinetic energy JKE rot Rotational kinetic energy JL Angular Momentum kg·m2

sL Self-inductance Hm Mass kgM Magnification (unitless)M Mutual inductance HMA Mechanical Advantage (unitless)ME Mechanical Energy Jn Index of refraction (unitless)p Object distance m~p Momentum kg·m

s

P Power WPE or U Potential Energy JPE elastic Elastic potential energy JPE electric Electrical potential energy J

PE g Gravitational potential energy Jq Image distance m

q or Q Charge CQ Heat, Entropy JR Radius of curvature mR Resistance Ωs Arc length mt Time s

∆t Time interval s~v Velocity m

s

vt Tangential speed ms

~vx Velocity in the x direction ms

~vy Velocity in the y direction ms

V Electric potential V∆V Electric potential difference VV Volume m3

W Work Jx or y Position m


14

Constants

Symbol Name Established Value Value Used

ε0 Permittivity of a vacuum 8.854 187 817× 10−12 C2

N·m2 8.85× 10−12 C2

N·m2

φ Golden ratio 1.618 033 988 749 894 848 20

π Archimedes’ constant 3.141 592 653 589 793 238 46

g, ag Gravitational acceleration constant 9.79171 ms2 (varies by location) 9.81 m

s2

c Speed of light in a vacuum 2.997 924 58× 108 ms (exact) 3.00× 108 m

s

e Natural logarithmic base 2.718 281 828 459 045 235 36

e− Elementary charge 1.602 177 33× 1019 C 1.60× 1019 C

G Gravitational constant 6.672 59× 10−11 N·m2

kg2 6.67× 10−11 N·m2

kg2

kC Coulomb’s constant 8.987 551 788× 109 N·m2

C2 8.99× 109 N·m2

C2

NA Avogadro’s constant 6.022 141 5× 1023 mol−1

Astronomical Data

Symbol Object Mean Radius Mass Mean Orbit Radius Orbital Period

Á Moon 1.74× 106 m 7.36× 1022 kg 3.84× 108 m 2.36× 106 s

À Sun 6.96× 108 m 1.99× 1030 kg — —

Â Mercury 2.43× 106 m 3.18× 1023 kg 5.79× 1010 m 7.60× 106 s

Ã Venus 6.06× 106 m 4.88× 1024 kg 1.08× 1011 m 1.94× 107 s

Ê Earth 6.37× 106 m 5.98× 1024 kg 1.496× 1011 m 3.156× 107 s

Ä Mars 3.37× 106 m 6.42× 1023 kg 2.28× 1011 m 5.94× 107 s

Ceres1 4.71× 105 m 9.5× 1020 kg 4.14× 1011 m 1.45× 108 s

Å Jupiter 6.99× 107 m 1.90× 1027 kg 7.78× 1011 m 3.74× 108 s

Æ Saturn 5.85× 107 m 5.68× 1026 kg 1.43× 1012 m 9.35× 108 s

Ç Uranus 2.33× 107 m 8.68× 1025 kg 2.87× 1012 m 2.64× 109 s

È Neptune 2.21× 107 m 1.03× 1026 kg 4.50× 1012 m 5.22× 109 s

É Pluto1 1.15× 106 m 1.31× 1022 kg 5.91× 1012 m 7.82× 109 s

Eris21 2.4× 106 m 1.5× 1022 kg 1.01× 1013 m 1.75× 1010 s

1Ceres, Pluto, and Eris are classified as “Dwarf Planets” by the IAU2Eris was formerly known as 2003 UB313


15

Mathematics Review for Physics

This is a summary of the most important parts of math-ematics as we will use them in a physics class. There arenumerous parts that are completely omitted, others aregreatly abridged. Do not assume that this is a completecoverage of any of these topics.

AlgebraFundamental properties of algebra

a+ b = b+ a Commutative law for ad-dition

(a+ b) + c = a+ (b+ c) Associative law for addi-tion

a+ 0 = 0 + a = a Identity law for additiona+ (−a) = (−a) + a = 0 Inverse law for addition

ab = ba Commutative law formultiplication

(ab)c = a(bc) Associative law for mul-tiplication

(a)(1) = (1)(a) = a Identity law for multipli-cation

a 1a = 1

aa = 1 Inverse law for multipli-cation

a(b+ c) = ab+ ac Distributive law

Exponents(ab)n = anbn (a/b)n = an/bn

anam = an+m 0n = 0an/am = an−m a0 = 1(an)m = a(mn) 00 = 1 (by definition)

Logarithms

x = ay −→ y = loga x

loga(xy) = loga x+ loga y

loga

(x

y

)= loga x− loga y

loga(xn) = n loga x

loga

(1x

)= − loga x

loga x =logb x

logb a= (logb x)(loga b)

Binomial Expansions

(a+ b)2 = a2 + 2ab+ b2

(a+ b)3 = a3 + 3a2b+ 3ab2 + b3

(a+ b)n =n∑

i=0

n!i!(n− i)!

aibn−i

Quadratic formulaFor equations of the form ax2 +bx+c = 0 the solutions are:

x =−b±

√b2 − 4ac

2a

Geometry

Shape Area Volume

Triangle A = 12bh —

Rectangle A = lw —

Circle A = πr2 —

Rectangularprism

A = 2(lw + lh+ hw) V = lwh

Sphere A = 4πr2 V = 43πr

3

Cylinder A = 2πrh+ 2πr2 V = πr2h

Cone A = πr√r2 + h2 + πr2 V = 1

3πr2h

TrigonometryIn physics only a small subset of what is covered in atrigonometry class is likely to be used, in particular sine, co-sine, and tangent are useful, as are their inverse functions.As a reminder, the relationships between those functionsand the sides of a right triangle are summarized as follows:

θ

opp

adj

hypsin θ =

opp

hypcos θ =

adj

hyp

tan θ =opp

adj=

sin θcos θ

The inverse functions are only defined over a limited range.The tan−1 x function will yield a value in the range −90 <θ < 90, sin−1 x will be in −90 ≤ θ ≤ 90, and cos−1 xwill yield one in 0 ≤ θ ≤ 180. Care must be taken toensure that the result given by a calculator is in the correctquadrant, if it is not then an appropriate correction mustbe made.

Degrees 0o 30o 45o 60o 90o 120o 135o 150o 180o

Radians 0 π6

π4

π3

π2

2π3

3π4

5π6 π

sin 0 12

√2

2

√3

2 1√

32

√2

212 0

cos 1√

32

√2

212 0 − 1

2 −√

22 −

√3

2 −1

tan 0√

33 1

√3 ∞ −

√3 −1 −

√3

3 0

sin2 θ + cos2 θ = 1 2 sin θ cos θ = sin(2θ)


16

Trigonometric functions in terms of each other

sin θ = sin θ√

1− cos2 θ tan θ√1+tan2 θ

cos θ =√

1− sin2 θ cos θ 1√1+tan2 θ

tan θ = sin θ√1−sin2 θ

√1−cos2 θcos θ tan θ

csc θ = 1sin θ

1√1−cos2 θ

√1+tan2 θtan θ

sec θ = 1√1−sin2 θ

1cos θ

√1 + tan2 θ

cot θ =√

1−sin2 θ

sin θcos θ√

1−cos2 θ1

tan θ

sin θ = 1csc θ

√sec2 θ−1sec θ

1√1+cot2 θ

cos θ =√

csc2 θ−1csc θ

1sec θ

cot θ√1+cot2 θ

tan θ = 1√csc2 θ−1

√sec2 θ − 1 1

cot θ

csc θ = csc θ sec θ√sec2 θ−1

√1 + cot2 θ

sec θ = csc θ√csc2 θ−1

sec θ√

1+cot2 θcot θ

cot θ =√

csc2 θ − 1 1√sec2 θ−1

cot θ

Law of sines, law of cosines, area of a triangle

HH

HHHH

A

B

C

a

b

c

a2 = b2 + c2 − 2bc cosAb2 = a2 + c2 − 2ac cosBc2 = a2 + b2 − 2ab cosC

a

sinA=

b

sinB=

c

sinC

let s = 12 (a+ b+ c)

Area=√s(s− a)(s− b)(s− c)

Area= 12bc sinA = 1

2ac sinB = 12ab sinC

VectorsA vector is a quantity with both magnitude and direction,such as displacement or velocity. Your textbook indicatesa vector in bold-face type as V and in class we have beenusing ~V . Both notations are equivalent.

A scalar is a quantity with only magnitude. This caneither be a quantity that is directionless such as time ormass, or it can be the magnitude of a vector quantity suchas speed or distance traveled. Your textbook indicates ascalar in italic type as V , in class we have not done any-thing to distinguish a scalar quantity. The magnitude of ~Vis written as V or |~V |.

A unit vector is a vector with magnitude 1 (a dimen-sionless constant) pointing in some significant direction. Aunit vector pointing in the direction of the vector ~V is in-dicated as V and would commonly be called V -hat. Anyvector can be normalized into a unit vector by dividing it byits magnitude, giving V = ~V

V . Three special unit vectors, ı,, and k are introduced with chapter 3. They point in thedirections of the positive x, y, and z axes, respectively (asshown below).

6

?

-

y+

−

x+−

z+

−6

-

ı

k

Vectors can be added to other vectors of the same di-mension (i.e. a velocity vector can be added to anothervelocity vector, but not to a force vector). The sum of allvectors to be added is called the resultant and is equivalentto all of the vectors combined.

Multiplying VectorsAny vector can be multiplied by any scalar, this has theeffect of changing the magnitude of the vector but not itsdirection (with the exception that multiplying a vector bya negative scalar will reverse the direction of the vector).As an example, multiplying a vector ~V by several scalarswould give:

- -

-

~V 2~V

.5~V −1~V

In addition to scalar multiplication there are also twoways to multiply vectors by other vectors. They will not bedirectly used in class but being familiar with them may helpto understand how some physics equations are derived. Thefirst, the dot product of vectors ~V1 and ~V2, represented as~V1 · ~V2 measures the tendency of the two vectors to point inthe same direction. If the angle between the two vectors isθ the dot product yields a scalar value as

~V1 · ~V2 = V1V2 cos θ

The second method of multiplying two vectors, thecross product, (represented as ~V1× ~V2) measures the ten-dency of vectors to be perpendicular to each other. It yieldsa third vector perpendicular to the two original vectors withmagnitude

|~V1 × ~V2| = V1V2 sin θ

The direction of the cross product is perpendicular to thetwo vectors being crossed and is found with the right-handrule — point the fingers of your right hand in the directionof the first vector, curl them toward the second vector, andthe cross product will be in the direction of your thumb.

Adding Vectors GraphicallyThe sum of any number of vectors can be found by drawingthem head-to-tail to scale and in proper orientation thendrawing the resultant vector from the tail of the first vector


17

to the point of the last one. If the vectors were drawn ac-curately then the magnitude and direction of the resultantcan be measured with a ruler and protractor. In the ex-ample below the vectors ~V1, ~V2, and ~V3 are added to yield~Vr

-

AAAAAAU

~V1

~V2 ~V3

~V1, ~V2, ~V3

-AAAAAAU

~V1

~V2

~V3

~V1 + ~V2 + ~V3

-AAAAAAU

~V1

~V2

~V3

-~Vr

~Vr = ~V1 + ~V2 + ~V3

Adding Parallel VectorsAny number of parallel vectors can be directly added byadding their magnitudes if one direction is chosen as pos-itive and vectors in the opposite direction are assigned anegative magnitude for the purposes of adding them. Thesum of the magnitudes will be the magnitude of the resul-tant vector in the positive direction, if the sum is negativethen the resultant will point in the negative direction.

Adding Perpendicular VectorsPerpendicular vectors can be added by drawing them as aright triangle and then finding the magnitude and directionof the hypotenuse (the resultant) through trigonometry andthe Pythagorean theorem. If ~Vr = ~Vx + ~Vy and ~Vx ⊥ ~Vy

then it works as follows:

-~Vx

6

~Vy

-~Vx

6

~Vy

~Vr

θ

Since the two vectors to be added and the resultant forma right triangle with the resultant as the hypotenuse thePythagorean theorem applies giving

Vr = |~Vr| =√V 2

x + V 2y

The angle θ can be found by taking the inverse tangent ofthe ratio between the magnitudes of the vertical and hori-zontal vectors, thus

θ = tan−1 Vy

Vx

As was mentioned above, care must be taken to ensure thatthe angle given by the calculator is in the appropriate quad-rant for the problem, this can be checked by looking at the

diagram drawn to solve the problem and verifying that theanswer points in the direction expected, if not then makean appropriate correction.

Resolving a Vector Into ComponentsJust two perpendicular vectors can be added to find asingle resultant, any single vector ~V can be resolved intotwo perpendicular component vectors ~Vx and ~Vy so that~V = ~Vx + ~Vy.

~V

θ ~Vx

~Vy

~V

θ

-

6

As the vector and its components can be drawn as aright triangle the ratios of the sides can be found withtrigonometry. Since sin θ = Vy

V and cos θ = Vx

V it followsthat Vx = V cos θ and Vy = V sin θ or in a vector form,~Vx = V cos θı and ~Vy = V sin θ. (This is actually an appli-cation of the dot product, ~Vx = (~V · ı)ı and ~Vy = (~V · ),but it is not necessary to know that for this class)

Adding Any Two Vectors AlgebraicallyOnly vectors with the same direction can be directly added,so if vectors pointing in multiple directions must be addedthey must first be broken down into their components, thenthe components are added and resolved into a single resul-tant vector — if in two dimensions ~Vr = ~V1 + ~V2 then

*

~V1

~V2

~Vr

~Vr = ~V1 + ~V2

*

-

-6

6

~V1

~V2

~Vr

~V1x

~V2x

~V1y

~V2y

~V1 = ~V1x + ~V1y

~V2 = ~V2x + ~V2y

~Vr = (~V1x + ~V1y) + (~V2x + ~V2y)

*

- -

6

6

~V1

~V2

~Vr

~V1x~V2x

~V1y

~V2y

~Vr = (~V1x + ~V2x) + (~V1y + ~V2y)

Once the sums of the component vectors in each direc-tion have been found the resultant can be found from themjust as an other perpendicular vectors may be added. Since


18

from the last figure ~Vr = (~V1x+~V2x)+(~V1y +~V2y) and it waspreviously established that ~Vx = V cos θı and ~Vy = V sin θit follows that

~Vr = (V1 cos θ1 + V2 cos θ2) ı+ (V1 sin θ1 + V2 sin θ2)

and

Vr =√

(V1x + V2x)2 + (V1y + V2y)2

=√

(V1 cos θ1 + V2 cos θ2)2 + (V1 sin θ1 + V2 sin θ2)

2

with the direction of the resultant vector ~Vr, θr, being foundwith

θr = tan−1 V1y + V2y

V1x + V2x= tan−1 V1 sin θ1 + V2 sin θ2

V1 cos θ1 + V2 cos θ2

Adding Any Number of Vectors Algebraically

For a total of n vectors ~Vi being added with magnitudes Vi

and directions θi the magnitude and direction are:

*

- - -

6

6

6

~V1

~V2

~V3

~Vr

~V1x~V2x

~V3x

~V1y

~V2y

~V3y

~Vr =

(n∑

i=1

Vi cos θi

)ı+

(n∑

i=1

Vi sin θi

)

Vr =

√√√√( n∑i=1

Vi cos θi

)2

+

(n∑

i=1

Vi sin θi

)2

θr = tan−1

n∑i=1

Vi sin θi

n∑i=1

Vi cos θi

(The figure shows only three vectors but this method willwork with any number of them so long as proper care istaken to ensure that all angles are measured the same wayand that the resultant direction is in the proper quadrant.)

CalculusAlthough this course is based on algebra and not calculus,it is sometimes useful to know some of the properties ofderivatives and integrals. If you have not yet learned cal-culus it is safe to skip this section.

DerivativesThe derivative of a function f(t) with respect to t is afunction equal to the slope of a graph of f(t) vs. t at everypoint, assuming that slope exists. There are several waysto indicate that derivative, including:

ddtf(t)

f ′(t)

f(t)

The first one from that list is unambiguous as to the in-dependent variable, the others assume that there is only onevariable, or in the case of the third one that the derivativeis taken with respect to time. Some common derivativesare:

ddttn = ntn−1

ddt

sin t = cos t

ddt

cos t = − sin t

ddtet = et

If the function is a compound function then there are afew useful rules to find its derivative:

ddtcf(t) = c

ddtf(t) c constant for all t

ddt

[f(t) + g(t)] =ddtf(t) +

ddtg(t)

ddtf(u) =

ddtu

dduf(u)

IntegralsThe integral, or antiderivative of a function f(t) withrespect to t is a function equal to the the area under agraph of f(t) vs. t at every point plus a constant, assumingthat f(t) is continuous. Integrals can be either indefinite ordefinite. An indefinite integral is indicated as:∫

f(t) dt

while a definite integral would be indicated as∫ b

a

f(t) dt

to show that it is evaluated from t = a to t = b, as in:∫ b

a

f(t) dt =∫f(t) dt|t=b

t=a =∫f(t) dt|t=b −

∫f(t) dt|t=a

Some common integrals are:∫tn dt =

tn+1

n+ 1+ C n 6= −1


19

∫t−1 dt = ln t+ C∫

sin tdt = − cos t+ C∫cos tdt = sin t+ C∫et dt = et + C

There are rules to reduce integrals of some compoundfunctions to simpler forms (there is no general rule to reducethe integral of the product of two functions):∫

cf(t) dt = c

∫f(t) dt c constant for all t∫

[f(t) + g(t)] dt =∫f(t) dt+

∫g(t) dt

Series Expansions

Some functions are difficult to work with in their normalforms, but once converted to their series expansion can bemanipulated easily. Some common series expansions are:

sinx = x− x3

3!+x5

5!− x7

7!+ · · · =

∞∑i=0

(−1)i x2i+1

(2i+ 1)!

cosx = 1− x2

2!+x4

4!− x6

6!+ · · · =

∞∑i=0

(−1)i x2i

(2i)!

ex = 1 + x+x2

2!+x3

3!+x4

4!+ · · · =

∞∑i=0

xi

(i)!


20

Physics Using Calculus

In our treatment of mechanics the majority of the equa-tions that have been covered in the class have been eitherapproximations or special cases where the acceleration orforce applied have been held constant. This is requiredbecause the class is based on algebra and to do otherwisewould require the use of calculus.

None of what is presented here is a required part of theclass, it is here to show how to handle cases outside theusual approximations and simplifications. Feel free to skipthis section, none of it will appear as a required part of anyassignment or test in this class.

Notation

Because the letter ‘d’ is used as a part of the notation ofcalculus the variable ‘r’ is often used to represent the posi-tion vector of the object being studied. (Other authors use‘s’ for the spatial position or generalize ‘x’ to two or moredimensions.) In a vector form that would become ~r to giveboth the magnitude and direction of the position. It is as-sumed that the position, velocity, acceleration, etc. can beexpressed as a function of time as ~r(t), ~v(t), and ~a(t) butthe dependency on time is usually implied and not shownexplicitly.

Translational Motion

Many situations will require an acceleration that is not con-stant. Everything learned about translational motion sofar used the simplification that the acceleration remainedconstant, with calculus we can let the acceleration be anyfunction of time.

Since velocity is the slope of a graph of position vs. timeat any point, and acceleration is the slope of velocity vs.time these can be expressed mathematically as derivatives:

~v =ddt~r

~a =ddt~v =

d2

dt2~r

The reverse of this relationship is that the displacementof an object is equal to the area under a velocity vs. timegraph and velocity is equal to the area under an accelera-tion vs. time graph. Mathematically this can be expressedas integrals:

~v =∫~adt

~r =∫~v dt =

∫∫~adt2

Using these relationships we can derive the main equa-tions of motion that were introduced by starting with the

integral of a constant velocity ~a(t) = ~a as∫ tf

ti

~adt = ~a∆t+ C

The constant of integration, C, can be shown to be the ini-tial velocity, ~vi, so the entire expression for the final velocitybecomes

~vf = ~vi + ~a∆t

or~vf (t) = ~vi + ~at

Similarly, integrating that expression with respect to timegives an expression for position:∫ tf

ti

(~vi + ~at) dt = ~vi∆t+12~a∆t2 + C

Once again the constant of integration, C, can be shown tobe the initial position, ~ri, yielding an expression for positionvs. time

~rf (t) = ~ri + ~vit+12~at2

While calculus can be used to derive the algebraic formsof the equations the reverse is not true. Algebra can han-dle a subset of physics and only at the expense of needingto remember separate equations for various special cases.With calculus the few definitions shown above will sufficeto predict any linear motion.

Work and PowerThe work done on an object is the integral of dot product ofthe force applied with the path the object takes, integratedas a contour integral over the path.

W =∫

C

~F · d~r

The power developed is the time rate of change of thework done, or the derivative of the work with respect totime.

P =ddtW

MomentumNewton’s second law of motion changes from the familiar~F = m~a to the statement that the force applied to an objectis the time derivative of the object’s momentum.

~F =ddt~p

Similarly, the impulse delivered to an object is the in-tegral of the force applied in the direction parallel to themotion with respect to time.

∆~p =∫

~F dt


21

Rotational MotionThe equations for rotational motion are very similar tothose for translational motion with appropriate variablesubstitutions. First, the angular velocity is the time deriva-tive of the angular position.

~ω =d~θdt

The angular acceleration is the time derivative of the an-gular velocity or the second time derivative of the angularposition.

~α =d~ωdt

=d2~θ

dt2

The angular velocity is also the integral of the angularacceleration with respect to time.

~ω =∫~α dt

The angular position is the integral of the angular veloc-ity with respect to time or the second integral of the angularacceleration with respect to time.

~θ =∫~ω dt =

∫∫~α dt2

The torque exerted on an object is the cross product ofthe radius vector from the axis of rotation to the point ofaction with the force applied.

~τ = ~r × ~F

The moment of inertia of any object is the integral ofthe square of the distance from the axis of rotation to eachelement of the mass over the entire mass of the object.

I =∫r2 dm

The torque applied to an object is the time derivativeof the object’s angular momentum.

~τ =ddt~L

The integral of the torque with respect to time is theangular momentum of the object.

~L =∫~τ dt


22

Data Analysis

Comparative Measures

Percent ErrorWhen experiments are conducted where there is a calcu-lated result (or other known value) against which the ob-served values will be compared the percent error betweenthe observed and calculated values can be found with

percent error =∣∣∣∣observed value − accepted value

accepted value

∣∣∣∣× 100%

Percent DifferenceWhen experiments involve a comparison between two ex-perimentally determined values where neither is regardedas correct then rather than the percent error the percentdifference can be calculated. Instead of dividing by the ac-cepted value the difference is divided by the mean of thevalues being compared. For any two values, a and b, thepercent difference is

percent difference =∣∣∣∣ a− b

12 (a+ b)

∣∣∣∣× 100%

Dimensional Analysis and Units on Con-stantsConsider a mass vibrating on the end of a spring. The equa-tion relating the mass to the frequency on graph might looksomething like

y =3.658x2

− 1.62

This isn’t what we’re looking for, a first pass would be tochange the x and y variables into f and m for frequencyand mass, this would then give

m =3.658f2

− 1.62

This is progress but there’s still a long way to go. Massis measured in kg and frequency is measured in s−1, sincethese variables represent the entire measurement (includ-ing the units) and not just the numeric portion they do not

need to be labeled, but the constants in the equation doneed to have the correct units applied.

To find the units it helps to first re-write the equationusing just the units, letting some variable such as u (or u1,u2, u3, etc.) stand for the unknown units. The exampleequation would then become

kg =u1

(s−1)2+ u2

Since any quantities being added or subtracted must havethe same units the equation can be split into two equations

kg =u1

(s−1)2and kg = u2

The next step is to solve for u1 and u2, in this case u2 = kgis immediately obvious, u1 will take a bit more effort. Afirst simplification yields

kg =u1

s−2

then multiplying both sides by s−2 gives

(s−2)(kg) =( u1

s−2

)(s−2)

which finally simplifies and solves to

u1 = kg · s−2

Inserting the units into the original equation finally yields

m =3.658 kg · s−2

f2− 1.62 kg

which is the final equation with all of the units in place.Checking by choosing a frequency and carrying out the cal-culations in the equation will show that it is dimensionallyconsistent and yields a result in the units for mass (kg) asexpected.


23

Midterm Exam Description

• 99 Multiple Choice Items (31 problems, 68 questions)

• Calculators will not be permitted

• Many test items contain numbers, but do not require calculations to solve. Before plugging values into an equation,look at the possible responses. The correct response may be obvious due to significant figures, order of magnitudeor simple common sense.

• Since all test items are worth one point, you should first complete the items that you are most sure of and then gothrough the exam a second time. You should make sure that you answer every question on the exam.

• Exam items are from the following categories:

– Chapter 1 — The Science of Physicsunits of measurementsignificant figures

– Chapter 2 — Motion in One Dimensiondisplacement and velocityaccelerationfalling objectsgraphs of motion

– Chapter 3 — Two-Dimensional Motion and Vectorsvectors and vector operationsprojectilesrelative motion

– Chapter 4 — Forces and the Laws of MotionNewton’s 1st LawNet forcesNewton’s 2nd LawNewton’s 3rd LawFriction

– Chapter 5 — Work and Energyworktypes of energyconservation of energypower

– Chapter 6 — Momentum and Collisionsmomentum and impulseconservation of momentumtypes of collisions

– Chapter 7 — Rotational Motion and the Law of Gravitymeasuring rotational motionrotational kinematics equationscentripetal force and accelerationgravitational forces

Chapter 8 will not be included on the first semester exam.


24

Semester Exam Equation Sheet

This is a slightly reduced copy of the equation sheet that you will receive with the semester exam.

Honors Physics – 1st Semester Final Constants and Equations

Kinematics Linear

avd v t= Δ

( )12 i fd v v t= + Δ

( )212id v t a t= Δ + Δ

f iv v a t= + Δ

2 2 2f iv v ad= +

x xd v t= Δ ( )21

2y y yd v t a t= Δ + Δ

sinyv v θ=

cosxv v θ=

Kinematics Rotational

dr

θ =

vr

ω =

ar

α =

av tθ ω= Δ

( )212i t tθ ω α= Δ + Δ

( )12 i f tθ ω ω= + Δ

f i tω ω α= + Δ

2 2f i 2ω ω αθ= +

Dynamics Linear

F ma=∑

gF mg=

cosgF F θ⊥ =

singF F θ=

f NF Fμ=

tanμ θ=

springF kx= −

1 22grav

m mF Gr

=

Dynamics Circular

2 rvTπ

=

2 2

2

4c

v rar T

π= =

2 2

2

4c

v rF m mr T

π= =

1T f=

Dynamics Rotational

Iτ α=∑

2point massI mr=

Work, Power, Energy Linear

cosW Fd θ=

gPE mgh=

212elasticPE kx=

212KE mv=

0PE KE WΔ + Δ + =

WP Fvt

= =Δ

Work, Power, Energy Rotational

W τθ= 21

2KE Iω=

P τω=

Momentum Linear

P mv=

1 1 2 2 1 1 2 2i f f fm v m v m v m v+ = +

F t m vΔ = Δ

Rotational

L Iω=

Constants

210. msg =

2

211 N m

kg6.67 10G − ⋅= ×

8 ms3.00 10c = ×

2

29 N m

C9.00 10K ⋅= ×

Note that some equations may be slightly different than the ones we have used in class, the sheet that you receivewill have these forms of the equations, it will be up to you to know how to use them or to know other forms that youare able to use.


25

Mix

edRev

iew

Exe

rcises

HRW material copyrighted under notice appearing earlier in this book.

Hol

t Phy

sics

Sec

tion

Revi

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orks

heet

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3.W

ith

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ific

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ect

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ific

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26HRW material copyrighted under notice appearing earlier in this book.

Mot

ion

in O

ne D

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sion

Mix

ed R

evie

wH

OLT

PH

YS

ICS

Ch

apte

r

2

Chap

ter 2

9

1.D

uri

ng

a re

lay

race

alo

ng

a st

raig

ht

road

,th

e fi

rst

run

ner

on

a t

hre

e-

pers

on te

am r

un

s d 1

wit

h a

con

stan

t ve

loci

ty v

1.T

he

run

ner

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en h

ands

off

the

bato

n to

th

e se

con

d ru

nn

er,w

ho

run

s d 2

wit

h a

con

stan

t ve

loci

ty

v 2.T

he

bato

n is

th

en p

asse

d to

th

e th

ird

run

ner

,wh

o co

mpl

etes

th

e ra

ce

by t

rave

ling

d 3w

ith

a c

onst

ant

velo

city

v3.

a.In

term

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d th

e ti

me

it t

akes

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each

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er to

com

plet

e

a se

gmen

t of

the

race

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3 ren

nu

R2 re

nn

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1 ren

nu

R

b.W

hat

is t

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l dis

tan

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fth

e ra

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ours

e?

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hat

is t

he

tota

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e it

tak

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team

to c

ompl

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race

?

2.T

he e

quat

ion

s be

low

incl

ude

the

equ

atio

ns

for

stra

ight

-lin

e m

otio

n.

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each

of

the

follo

win

g pr

oble

ms,

indi

cate

whi

ch e

quat

ion

or

equ

atio

ns

you

wou

ld u

se to

sol

ve th

e pr

oble

m,b

ut d

o n

ot a

ctu

ally

per

form

the

calc

ula

tion

s.

a.D

uri

ng

take

off,

a pl

ane

acce

lera

tes

at 4

m/s

2an

d ta

kes

40 s

to r

each

take

off

spee

d.W

hat

is t

he

velo

city

of

the

plan

e at

tak

eoff

?

b.A

car

wit

h a

n in

itia

l spe

ed o

f31

.4 k

m/h

acc

eler

ates

at

a u

nif

orm

rat

e

of1.

2 m

/s2

for

1.3

s.W

hat

is t

he

fin

al s

peed

an

d di

spla

cem

ent

ofth

e

car

duri

ng

this

tim

e?

∆x=

1 2 (v i

+v f

)∆t

∆x=

1 2 (v f

)∆t

∆x=v i

(∆t)

+1 2 a

(∆t)

2∆x

=1 2 a

(∆t)

2

v f=v i

+a(

∆t)

v f=

a(∆t

)

v f2

=v i

2+

2a∆x

v f2

=2a

∆x


Hol

t Phy

sics

Sec

tion

Revi

ew W

orks

heet

s103.

Bel

ow is

th

e ve

loci

ty-t

ime

grap

h o

fan

obj

ect

mov

ing

alon

g a

stra

igh

t

path

.Use

th

e in

form

atio

n in

th

e gr

aph

to fi

ll in

th

e ta

ble

belo

w.

HO

LT P

HY

SIC

S

Mix

ed R

evie

w co

ntin

ued

Ch

apte

r

2

Tim

eM

otio

nv

ain

terv

alA B C D E

4.A

bal

l is

thro

wn

upw

ard

wit

h a

n in

itia

l vel

ocit

y of

9.8

m/s

from

th

e to

p

ofa

build

ing.

a.Fi

ll in

th

e ta

ble

belo

w s

how

ing

the

ball’

s po

siti

on,v

eloc

ity,

and

acce

l-

erat

ion

at

the

end

ofea

ch o

fth

e fi

rst

4 s

ofm

otio

n.

Tim

eP

osit

ion

Vel

ocit

yA

ccel

erat

ion

(s)

(m)

(m/s

)(m

/s2 )

1 2 3 4

b.In

wh

ich

sec

ond

does

th

e ba

ll re

ach

th

e to

p of

its

flig

ht?

c.In

wh

ich

sec

ond

does

th

e ba

ll re

ach

th

e le

vel o

fth

e ro

of,o

n t

he

way

dow

n?

For

each

oft

he le

tter

ed in

terv

als

belo

w,i

ndi

cate

the

mot

ion

oft

he o

bjec

t

(whe

ther

it is

spe

edin

g up

,slo

win

g do

wn

,or

at r

est)

,the

dir

ecti

on o

fthe

velo

city

(+,

−,or

0),

and

the

dire

ctio

n o

fthe

acc

eler

atio

n (

+,−,

or 0

).

Velocity (m/s)

time

(s)

15 10 5 070

100

AB

CD

E30

5020

6040


27

Two-

Dim

ensi

onal

Mot

ion

and

Vect

ors

Mix

ed R

evie

wH

OLT

PH

YS

ICS

Ch

apte

r

3

Chap

ter 3

15

1.T

he

diag

ram

bel

ow in

dica

tes

thre

e po

siti

ons

to

whi

ch a

wom

an tr

avel

s.Sh

e st

arts

at p

osit

ion

A,

trav

els

3.0

km to

the

wes

t to

poin

t B,t

hen

6.0

km

to

the

nor

th to

poi

nt C

.She

then

bac

ktra

cks,

and

trav

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2.0

km to

the

sou

th to

poi

nt D

.

a.In

th

e sp

ace

prov

ided

,dia

gram

th

e di

spla

cem

ent

vect

ors

for

each

seg

men

t of

the

wom

an’s

tri

p.

b.W

hat i

s th

e to

tal d

ispl

acem

ent o

fth

e w

oman

from

her

init

ial p

osit

ion

,A,t

o he

r fi

nal

pos

itio

n,D

?

c.W

hat i

s th

e to

tal d

ista

nce

trav

eled

by

the

wom

an

from

her

init

ial p

osit

ion,

A,t

o he

r fin

al p

osit

ion,

D?

2.Tw

o pr

ojec

tile

s ar

e la

un

ched

from

th

e gr

oun

d,an

d bo

th r

each

th

e sa

me

vert

ical

hei

ght.

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ever

,pro

ject

ile B

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tw

ice

the

hor

izon

tal d

ista

nce

as p

roje

ctile

A b

efor

e h

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ng

the

grou

nd.

a.H

ow la

rge

is t

he

vert

ical

com

pon

ent

ofth

e in

itia

l vel

ocit

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proj

ec-

tile

B c

ompa

red

wit

h t

he

vert

ical

com

pon

ent

ofth

e in

itia

l vel

ocit

y

ofpr

ojec

tile

A?

b.H

ow la

rge

is t

he

hor

izon

tal c

ompo

nen

t of

the

init

ial v

eloc

ity

ofpr

o-

ject

ile B

com

pare

d w

ith

th

e h

oriz

onta

l com

pon

ent

ofth

e in

itia

l

velo

city

of

proj

ecti

le A

?

c.Su

ppos

e pr

ojec

tile

A is

lau

nch

ed a

t an

an

gle

of45

°to

th

e h

oriz

onta

l.

Wh

at is

th

e ra

tio,v B

/vA

,of

the

spee

d of

proj

ecti

le B

,vB

,com

pare

d

wit

h t

he

spee

d of

proj

ecti

le A

,vA

?


C D BA


Hol

t Phy

sics

Sec

tion

Revi

ew W

orks

heet

s163.

A p

asse

nge

r at

an

air

port

ste

ps o

nto

a m

ovin

g si

dew

alk

that

is 1

00.0

m

lon

g an

d is

mov

ing

at a

spe

ed o

f1.

5 m

/s.T

he

pass

enge

r th

en s

tart

s w

alk-

ing

at a

spe

ed o

f1.

0 m

/s in

th

e sa

me

dire

ctio

n a

s th

e si

dew

alk

is m

ovin

g.

Wh

at is

th

e pa

ssen

ger’

s ve

loci

ty r

elat

ive

to t

he

follo

win

g ob

serv

ers?

a.A

per

son

sta

ndi

ng

stat

ion

ary

alon

gsid

e to

th

e m

ovin

g si

dew

alk.

b.A

per

son

sta

ndi

ng

stat

ion

ary

onth

e m

ovin

g si

dew

alk.

c.A

per

son

wal

kin

g al

ongs

ide

the

side

wal

k w

ith

a s

peed

of

2.0

m/s

an

d

in a

dir

ecti

on o

ppos

ite

the

mot

ion

of

the

side

wal

k.

d.A

per

son

rid

ing

in a

car

t alo

ngs

ide

the

side

wal

k w

ith

a sp

eed

of5.

0 m

/s

and

in th

e sa

me

dire

ctio

n in

whi

ch th

e si

dew

alk

is m

ovin

g.

e.A

per

son

rid

ing

in a

car

t w

ith

a s

peed

of

4.0

m/s

an

d in

a d

irec

tion

perp

endi

cula

r to

th

e di

rect

ion

in w

hic

h t

he

side

wal

k is

mov

ing.

4.U

se t

he

info

rmat

ion

giv

en in

item

3 to

an

swer

th

e fo

llow

ing

ques

tion

s:

a.H

ow lo

ng

does

it t

ake

for

the

pass

enge

r w

alki

ng

on t

he

side

wal

k to

get

from

on

e en

d of

the

side

wal

k to

th

e ot

her

en

d?

b.H

ow m

uch

tim

e do

es th

e pa

ssen

ger

save

by

taki

ng

the

mov

ing

side

-

wal

k in

stea

d of

wal

kin

g al

ongs

ide

it?

HO

LT P

HY

SIC

S

Mix

ed R

evie

w co

ntin

ued

Ch

apte

r

3


28

Forc

es a

nd th

e La

ws

ofM

otio

n

Mix

ed R

evie

wH

OLT

PH

YS

ICS

Ch

apte

r

4

Chap

ter 4

21

1.A

cra

te r

ests

on

the

hori

zont

al b

ed o

fa p

icku

p tr

uck.

For

each

sit

uati

on d

e-

scri

bed

belo

w,i

ndic

ate

the

mot

ion

ofth

e cr

ate

rela

tive

to th

e gr

ound

,the

mot

ion

ofth

e cr

ate

rela

tive

to th

e tr

uck,

and

whe

ther

the

crat

e w

ill h

it th

e

fron

t wal

l oft

he tr

uck

bed,

the

back

wal

l,or

nei

ther

.Dis

rega

rd fr

icti

on.

a.St

arti

ng

at r

est,

the

tru

ck a

ccel

erat

es to

th

e ri

ght.

b.T

he

crat

e is

at

rest

rel

ativ

e to

th

e tr

uck

wh

ile t

he

tru

ck m

oves

to t

he

righ

t w

ith

a c

onst

ant

velo

city

.

c.T

he

tru

ck in

item

b s

low

s do

wn

.

2.A

bal

l wit

h a

mas

s of

mis

th

row

n t

hro

ugh

th

e ai

r,as

sh

own

in t

he

figu

re.


a.W

hat

is t

he

grav

itat

ion

al fo

rce

exer

ted

on t

he

ball

by E

arth

?

b.W

hat

is t

he

forc

e ex

erte

d on

Ear

th b

y th

e ba

ll?

c.If

the

surr

oun

din

g ai

r ex

erts

a fo

rce

on t

he

ball

that

res

ists

its

mot

ion

,

is t

he

tota

lfor

ce o

n t

he

ball

the

sam

e as

th

e fo

rce

calc

ula

ted

in p

art

a?

d.If

the

surr

oun

din

g ai

r ex

erts

a fo

rce

on t

he

ball

that

res

ists

its

mot

ion

,

is t

he

grav

itat

iona

l for

ce o

n t

he

ball

the

sam

e as

th

e fo

rce

calc

ula

ted

in

part

a?


Hol

t Phy

sics

Sec

tion

Revi

ew W

orks

heet

s223.

Two

bloc

ks o

fm

asse

s m

1an

dm

2,r

espe

ctiv

ely,

are

plac

ed in

con

tact

wit

h

each

oth

er o

n a

sm

ooth

,hor

izon

tal s

urf

ace.

A c

onst

ant

hor

izon

tal f

orce

F

to t

he

righ

t is

app

lied

to m

1.A

nsw

er t

he

follo

win

g qu

esti

ons

in te

rms

of

F,m

1,an

dm

2.

a.W

hat

is t

he

acce

lera

tion

of

the

two

bloc

ks?

b.W

hat

are

th

e h

oriz

onta

l for

ces

acti

ng

on m

2?

c.W

hat

are

th

e h

oriz

onta

l for

ces

acti

ng

on m

1?

d.W

hat

is t

he

mag

nit

ude

of

the

con

tact

forc

e be

twee

n t

he

two

bloc

ks?

4.A

ssu

me

you

hav

e th

e sa

me

situ

atio

n a

s de

scri

bed

in it

em 3

,on

ly t

his

tim

e th

ere

is a

fric

tion

al fo

rce,

F k,b

etw

een

th

e bl

ocks

an

d th

e su

rfac

e.

An

swer

th

e fo

llow

ing

ques

tion

s in

term

s of

F,F k

,m1,

and

m2.

a.W

hat

is t

he

acce

lera

tion

of

the

two

bloc

ks?

b.W

hat

are

th

e h

oriz

onta

l for

ces

acti

ng

on m

2?

c.W

hat

are

th

e h

oriz

onta

l for

ces

acti

ng

on m

1?

d.W

hat

is t

he

mag

nit

ude

of

the

con

tact

forc

e be

twee

n t

he

two

bloc

ks?

HO

LT P

HY

SIC

S

Mix

ed R

evie

w co

ntin

ued

Ch

apte

r

4


29

Wor

k an

d En

ergy

Mix

ed R

evie

wH

OLT

PH

YS

ICS

Ch

apte

r

5

Chap

ter 5

27

1.A

bal

l has

a m

ass

of3

kg.W

hat

is t

he

wor

k do

ne

on t

his

bal

l by

the

grav

i-

tati

onal

forc

e ex

erte

d by

Ear

th if

the

ball

mov

es 2

m a

lon

g ea

ch o

fth

e

follo

win

g di

rect

ion

s?

a.do

wnw

ard

(alo

ng

the

forc

e)

b.u

pwar

d (o

ppos

ite

the

forc

e)

2.A

sto

ne

wit

h a

mas

s of

mis

th

row

n o

ffa

build

ing.

As

the

ston

e pa

sses

poin

tA

,it

has

a s

peed

ofv A

at a

n a

ngl

e of

qto

th

e h

oriz

onta

l.T

he

ston

e

then

tra

vels

a v

erti

cal d

ista

nce

hto

poi

nt

B,w

her

e it

has

a s

peed

vB

.


a.W

hat

is t

he

wor

k do

ne

on t

he

ston

e by

th

e gr

avit

atio

nal

forc

e du

e to

Ear

th w

hile

th

e st

one

mov

es fr

om A

to B

?

b.W

hat

is t

he

chan

ge in

th

e ki

net

ic e

ner

gy o

fth

e st

one

as it

mov

es fr

om

Ato

B?

c.W

hat

is t

he

spee

d v B

ofth

e st

one

in te

rms

ofv A

,g,a

nd

h?

d.D

oes

the

chan

ge in

th

e st

one’

s sp

eed

betw

een

Aan

dB

depe

nd

on t

he

mas

s of

the

ston

e?

e.D

oes

the

chan

ge in

th

e st

one’

s sp

eed

betw

een

Aan

dB

depe

nd

on t

he

angl

e q

?

h

A

Bθ


Hol

t Phy

sics

Sec

tion

Revi

ew W

orks

heet

s283.

An

em

pty

coff

ee m

ug

wit

h a

mas

s of

0.40

kg

gets

kn

ocke

d of

fa

tabl

etop

0.75

m a

bove

th

e fl

oor

onto

th

e se

at o

fa

chai

r 0.

45 m

abo

ve t

he

floo

r.

Ass

um

e th

at t

he

grav

itat

ion

al p

oten

tial

en

ergy

,PE

g,is

mea

sure

d u

sin

g

the

floo

r as

th

e ze

ro e

ner

gy le

vel.

a.W

hat

is t

he

init

ial g

ravi

tati

onal

pot

enti

al e

ner

gy a

ssoc

iate

d w

ith

th

e

mu

g’s

posi

tion

on

th

e ta

ble?

b.W

hat

is t

he

fin

al g

ravi

tati

onal

pot

enti

al e

ner

gy a

ssoc

iate

d w

ith

th

e

mu

g’s

posi

tion

on

th

e ch

air

seat

?

c.W

hat

was

th

e w

ork

don

e by

th

e gr

avit

atio

nal

forc

e as

it fe

ll fr

om t

he

tabl

e to

th

e ch

air?

d.Su

ppos

e th

at z

ero

leve

l for

th

e en

ergy

was

tak

en to

be

the

ceili

ng

of

the

room

rat

her

th

an t

he

floo

r .W

ould

th

e an

swer

s to

item

s a

to c

be

the

sam

e or

dif

fere

nt?

4.A

car

ton

of

shoe

s w

ith

a m

ass

ofm

slid

es w

ith

an

init

ial s

peed

ofv i

m/s

dow

n a

ram

p in

clin

ed a

t an

an

gle

of23

°to

the

hor

izon

tal.

Th

e ca

rton

’s

init

ial h

eigh

t is

hi,

and

its

fin

al h

eigh

t is

hf,

and

it t

rave

ls a

dis

tan

ce o

f

ddo

wn

th

e ra

mp.

Th

ere

is a

fric

tion

al fo

rce,

F k,b

etw

een

th

e ra

mp

and

the

cart

on.

a.W

hat i

s th

e in

itia

l mec

han

ical

en

ergy

,ME

i,of

the

cart

on?

(Hin

t:

App

ly th

e la

w o

fco

nse

rvat

ion

of

ener

gy.)

b.Ifm

is t

he

coef

fici

ent

offr

icti

on b

etw

een

th

e ra

mp

and

the

cart

on,

wh

at is

Fk?

c.E

xpre

ss t

he

fin

al s

peed

,vf,

ofth

e ca

rton

in te

rms

ofv i

,g,d

,an

dm.

HO

LT P

HY

SIC

S

Mix

ed R

evie

w co

ntin

ued

Ch

apte

r

5


30

Hol

t Phy

sics

Sec

tion

Revi

ew W

orks

heet

s32

Mom

entu

m a

nd C

ollis

ions

Mix

ed R

evie

wH

OLT

PH

YS

ICS

Ch

apte

r

6

1.A

pit

cher

th

row

s a

soft

ball

tow

ard

hom

e pl

ate.

Th

e ba

ll m

ay b

e h

it,s

end-

ing

it b

ack

tow

ard

the

pitc

her

,or

it m

ay b

e ca

ugh

t,br

ingi

ng

it to

a s

top

in

the

catc

her

’s m

itt.

a.C

ompa

re t

he

chan

ge in

mom

entu

m o

fth

e ba

ll in

th

ese

two

case

s.

b.D

iscu

ss t

he

mag

nit

ude

of

the

impu

lse

on t

he

ball

in t

hes

e tw

o ca

ses.

c.In

th

e sp

ace

belo

w,d

raw

a v

ecto

r di

agra

m fo

r ea

ch c

ase,

show

ing

the

init

ial m

omen

tum

of

the

ball,

the

impu

lse

exer

ted

on t

he

ball,

and

the

resu

ltin

g fi

nal

mom

entu

m o

fth

e ba

ll.

2.a.

Usi

ng

New

ton’

s th

ird

law

,exp

lain

why

th

e im

puls

e on

on

e ob

ject

in a

colli

sion

is e

qual

in m

agn

itu

de b

ut

oppo

site

in d

irec

tion

to t

he

im-

puls

e on

th

e se

con

d ob

ject

.

b.E

xten

d yo

ur

disc

uss

ion

of

impu

lse

and

New

ton’

s th

ird

law

to t

he

case

ofa

bow

ling

ball

stri

kin

g a

set

of10

bow

ling

pin

s.



Chap

ter 6

33

3.St

arti

ng

wit

h t

he

con

serv

atio

n o

fto

tal m

omen

tum

,pf

=p

i,sh

ow

that

th

e fi

nal

vel

ocit

y fo

r tw

o ob

ject

s in

an

inel

asti

c co

llisi

on is

v f=

m1m +1 m

2

v 1,i

+ m

1m +2 m2

v 2

,i.

4.Tw

o m

ovin

g bi

lliar

d ba

lls,e

ach

wit

h a

mas

s of

M,u

nde

rgo

an e

last

ic

colli

sion

.Im

med

iate

ly b

efor

e th

e co

llisi

on,b

all A

is m

ovin

g ea

st a

t

2 m

/s a

nd

ball

B is

mov

ing

east

at

4 m

/s.

a.In

term

s of

M,w

hat i

s th

e to

tal m

omen

tum

(m

agn

itu

de a

nd

dire

c-

tion

) im

med

iate

ly b

efor

e th

e co

llisi

on?

b.T

he

fin

al m

omen

tum

,M(v

A,f

+v B

,f),

mu

st e

qual

th

e in

itia

l mom

en-

tum

.If

the

fin

al v

eloc

ity

ofba

ll A

incr

ease

s to

4 m

/s e

ast

beca

use

of

the

colli

sion

,wh

at is

th

e fi

nal

mom

entu

m o

fba

ll B

?

c.Fo

r ea

ch b

all,

com

pare

th

e fi

nal

mom

entu

m o

fth

e ba

ll to

th

e in

itia

l

mom

entu

m o

fth

e ot

her

ball.

Th

ese

resu

lts

are

typi

cal o

fh

ead-

on

elas

tic

colli

sion

s.W

hat

gen

eral

izat

ion

abo

ut

hea

d-on

ela

stic

col

lisio

ns

can

you

mak

e?

HO

LT P

HY

SIC

S

Mix

ed R

evie

w co

ntin

ued

Ch

apte

r

6


31HRW material copyrighted under notice appearing earlier in this book.

Rota

tiona

l Mot

ion

and

the

Law

ofG

ravi

ty

Mix

ed R

evie

wH

OLT

PH

YS

ICS

Ch

apte

r

7

Chap

ter 7

37

1.C

ompl

ete

the

follo

win

g ta

ble.

s(m

)r(

m)

q(r

ad)

t(s)

w(r

ad/s

)v t

(m/s

)a c

(m/s

2 )

a.4.

51.

50.

50

b.5.8

5.805.0

c.85

02.02.3

d.

1250

2.0

17

e.68

0570573

2.D

escr

ibe

the

forc

e th

at m

ain

tain

s ci

rcu

lar

mot

ion

in t

he

follo

win

g ca

ses.

a.A

car

exi

ts a

free

way

an

d m

oves

aro

un

d a

circ

ula

r ra

mp

to r

each

th

e

stre

et b

elow

.

b.T

he

moo

n o

rbit

s E

arth

.

c.D

uri

ng

gym

cla

ss,a

stu

den

t h

its

a te

ther

bal

l on

a s

trin

g.

3.D

eter

min

e th

e ch

ange

in g

ravi

tati

onal

forc

e u

nde

r th

e fo

llow

ing

chan

ges.

a.on

e of

the

mas

ses

is d

oubl

ed

b.bo

th m

asse

s ar

e do

ubl

ed

c.th

e di

stan

ce b

etw

een

mas

ses

is d

oubl

ed

d.th

e di

stan

ce b

etw

een

mas

ses

is h

alve

d

e.th

e di

stan

ce b

etw

een

mas

ses

is t

ripl

ed


Hol

t Phy

sics

Sec

tion

Revi

ew W

orks

heet

s384.

Som

e pl

ans

for

a fu

ture

spa

ce s

tati

on m

ake

use

ofro

tati

onal

forc

e to

sim

u-

late

gra

vity

.In

ord

er to

be

effe

ctiv

e,th

e ce

ntr

ipet

al a

ccel

erat

ion

at t

he o

uter

rim

oft

he s

tati

on s

houl

d eq

ual a

bout

1 g

,or

9.81

m/s

2 .How

ever

,hum

ans

can

wit

hsta

nd

a di

ffer

ence

ofo

nly

1/1

00 g

bet

wee

n th

eir

head

an

d fe

et

befo

re th

ey b

ecom

e di

sori

ente

d.A

ssum

e th

e av

erag

e hu

man

hei

ght i

s

2.0

m,a

nd

calc

ulat

e th

e m

inim

um r

adiu

s fo

r a

safe

,eff

ecti

ve s

tati

on.

(Hin

t:T

he r

atio

oft

he c

entr

ipet

al a

ccel

erat

ion

ofa

stro

nau

t’s fe

et to

the

cen

trip

etal

acc

eler

atio

n o

fthe

ast

ron

aut’s

hea

d m

ust b

e at

leas

t 99/

100.

)

5.A

s an

ele

vato

r be

gin

s to

des

cen

d,yo

u fe

el m

omen

tari

ly li

ghte

r.A

s th

e

elev

ator

sto

ps,y

ou fe

el m

omen

tari

ly h

eavi

er.S

ketc

h t

he

situ

atio

n,a

nd

expl

ain

th

e se

nsa

tion

s u

sin

g th

e fo

rces

in y

our

sket

ch.

6.Tw

o ca

rs s

tart

on

opp

osit

e si

des

ofa

circ

ula

r tr

ack.

On

e ca

r h

as a

spe

ed

of0.

015

rad/

s;th

e ot

her

car

has

a s

peed

of

0.01

2 ra

d/s.

Ifth

e ca

rs s

tart

pra

dian

s ap

art,

calc

ula

te t

he

tim

e it

tak

es fo

r th

e fa

ster

car

to c

atch

up

wit

h t

he

slow

er c

ar.

HO

LT P

HY

SIC

S

Mix

ed R

evie

w co

ntin

ued

Ch

apte

r

7


32

Rota

tiona

l Equ

ilibr

ium

and

Dyn

amic

s

Mix

ed R

evie

wH

OLT

PH

YS

ICS

Ch

apte

r

8

Chap

ter 8

43

1.a.

On

som

e do

ors,

the

door

knob

is in

th

e ce

nte

r of

the

door

.Wh

at

wou

ld a

phy

sici

st s

ay a

bou

t th

e pr

acti

calit

y of

this

arr

ange

men

t? W

hy

wou

ld p

hysi

cist

s de

sign

doo

rs w

ith

kn

obs

fart

her

from

th

e h

inge

?

b.H

ow m

uch

mor

e fo

rce

wou

ld b

e re

quir

ed to

ope

n t

he

door

from

th

e

cen

ter

rath

er t

han

from

th

e ed

ge?

2.Fi

gure

ska

ters

com

mon

ly c

han

ge t

he

shap

e of

thei

r bo

dy in

ord

er to

ach

ieve

spi

ns

on t

he

ice.

Exp

lain

th

e ef

fect

s on

eac

h o

fth

e fo

llow

ing

quan

titi

es w

hen

a fi

gure

ska

ter

pulls

in h

is o

r h

er a

rms.

a.m

omen

t of

iner

tia

b.an

gula

r m

omen

tum

c.an

gula

r sp

eed

3.Fo

r th

e fo

llow

ing

item

s,as

sum

e th

e ob

ject

s sh

own

are

in r

otat

ion

al e

quili

briu

m.

a.W

hat

is t

he

mas

s of

the

sph

ere

to t

he

righ

t?

b.W

hat

is t

he

mas

s of

the

port

ion

of

the

met

er-

stic

k to

th

e le

ft o

fth

e pi

vot?

(H

int:

20%

of

the

mas

s of

the

met

erst

ick

is o

n t

he

left

.How

mu

ch m

ust

be

on t

he

righ

t?)


1.0

kg

0 cm

50 c

m25

cm

75 c

m10

0 cm

1.0

kg0 cm

50 c

m25

cm

75 c

m10

0 cm


Hol

t Phy

sics

Sec

tion

Revi

ew W

orks

heet

s444.

A fo

rce

of25

N is

app

lied

to t

he

end

ofa

un

ifor

m r

od t

hat

is 0

.50

m lo

ng

and

has

a m

ass

of0.

75 k

g.

a.Fi

nd

the

torq

ue,

mom

ent o

fin

erti

a,an

d an

gula

r ac

cele

rati

on if

the

rod

is a

llow

ed to

piv

ot a

rou

nd

its

cen

ter

ofm

ass.

b.Fi

nd

the

torq

ue,

mom

ent o

fin

erti

a,an

d an

gula

r ac

cele

rati

on if

the

rod

is a

llow

ed to

piv

ot a

rou

nd

the

end,

away

from

the

appl

ied

forc

e.

5.A

sat

ellit

e in

orb

it a

rou

nd

Ear

th is

init

ially

at

a co

nst

ant

angu

lar

spee

d of

7.27

×10

–5ra

d/s.

Th

e m

ass

ofth

e sa

telli

te is

45

kg,a

nd

it h

as a

n o

rbit

al

radi

us

of4.

23 ×

107

m.

a.Fi

nd

the

mom

ent

ofin

erti

a of

the

sate

llite

in o

rbit

aro

un

d E

arth

.

b.Fi

nd

the

angu

lar

mom

entu

m o

fth

e sa

telli

te.

c.Fi

nd

the

rota

tion

al k

inet

ic e

ner

gy o

fth

e sa

telli

te a

rou

nd

Ear

th.

d.Fi

nd

the

tan

gen

tial

spe

ed o

fth

e sa

telli

te.

e.Fi

nd

the

tran

slat

ion

al k

inet

ic e

ner

gy o

fth

e sa

telli

te.

6.A

ser

ies

oftw

o si

mpl

e m

ach

ines

is u

sed

to li

ft a

133

00 N

car

to a

hei

ght

of3.

0 m

.Bot

h m

ach

ines

hav

e an

eff

icie

ncy

of

0.90

(90

per

cen

t).M

ach

ine

A m

oves

th

e ca

r,an

d th

e ou

tpu

t of

mac

hin

e B

is t

he

inpu

t to

mac

hin

e A

.

a.H

ow m

uch

wor

k is

don

e on

th

e ca

r?

b.H

ow m

uch

wor

k m

ust

be

don

e on

mac

hin

e A

in o

rder

to a

chie

ve t

he

amou

nt

ofw

ork

don

e on

th

e ca

r?

c.H

ow m

uch

wor

k m

ust

be

don

e on

mac

hin

e B

in o

rder

to a

chie

ve t

he

amou

nt

ofw

ork

from

mac

hin

e A

?

d.W

hat

is t

he

over

all e

ffic

ien

cy o

fth

is p

roce

ss?

HO

LT P

HY

SIC

S

Mix

ed R

evie

w co

ntin

ued

Ch

apte

r

8


33

Mix

edRev

iew

Answ

ers

Sect

ion

Thre

e—Se

ctio

n Re

view

Wor

kshe

et A

nsw

ers

III–1

1.a.

2.2

×10

5s

b.3.

5×

107

mm

c.4.

3×

10−4

km

d.2.

2×

10−5

kg

e.6.

71×

1011m

g

f.8.

76×

10−5

GW

g.1.

753

×10

−1ps

2.a.

3

b.4

c.10

d.3

e.2

f.4

3.a.

4

b.5

c.3

4.a.

1.00

54;−

0.99

52;5

.080

×10

−3;

5.07

6×

10−3

b.4.

597

×10

7 ;3.8

66 ×

107 ;

1.54

6×

1014

;11.

58

5.15

.9 m

2

6.T

he g

raph

sho

uld

be a

str

aigh

t lin

e.

Chap

ter 1

Mix

ed R

evie

w

Hol

t Phy

sics

Sol

utio

n M

anua

lIII

–2

III

2.a.

v f=

a(∆t

)

b.v f

=v i

+a(

∆t);

∆x=

1 2 (v i

+v f

)∆t

or∆x

=v i

(∆t)

+1 2 a

(∆t)

2

1.a.

t 1=

d 1/v

1;t 2

=d 2

/v2;t

3=

d 3/v

3

b.to

tal d

ista

nce

=d 1

+d 2

+d 3

c.to

tal t

ime

=t 1

+t 2

+t 3

3.

Tim

e in

terv

alTy

pe o

fmot

ion

v(m

/s)

a(m

/s2 )

Asp

eedi

ng

up

++

Bsp

eedi

ng

up

++

Cco

nst

ant

velo

city

+0

Dsl

owin

g do

wn

+−

Esl

owin

g do

wn

+−

.b

.a

.4

1 s

c.2

sTi

me

(s)

Posi

tion

(m)

v(m

/s)

a(m

/s2 )

14.

90

−9.8

1

20

−9.8

−9.8

1

3−1

4.7

−19.

6−9

.81

4−3

9.2

−29.

4−9

.81

Chap

ter 2

Mix

ed R

evie

w

Hol

t Phy

sics

Sol

utio

n M

anua

lIII

–4

1.a.

Th

e di

agra

m s

hou

ld in

dica

te t

he

rela

tive

dis

tan

ces

and

dire

ctio

ns

for

each

seg

men

t of

the

path

.

b.5.

0 km

,slig

htl

y n

orth

of

nor

thw

est

c.11

.0 k

m

2.a.

Th

e sa

me

b.Tw

ice

as la

rge

c.1.

58

3.a.

2.5

m/s

,in

th

e di

rect

ion

of

the

side

wal

k’s

mot

ion

b.1.

0 m

/s,i

n t

he

dire

ctio

n o

fth

e si

dew

alk’

s m

otio

n

c.4.

5 m

/s,i

n t

he

dire

ctio

n o

fth

e si

dew

alk’

s m

otio

n

d.2.

5 m

/s,i

n t

he

dire

ctio

n o

ppos

ite

to t

he

side

wal

k’s

mot

ion

e.4.

7 m

/s,q

=32

°

4.a.

4.0

×10

1se

con

ds

b.6.

0×

101

seco

nds

Chap

ter 3

Mix

ed R

evie

w

Sect

ion

Thre

e—Se

ctio

n Re

view

Wor

kshe

et A

nsw

ers

III–5

1.a.

at r

est,

mov

es to

th

e le

ft,h

its

back

wal

l

b.m

oves

to t

he

righ

t (w

ith

vel

ocit

y v)

,at

rest

,nei

ther

c.m

oves

to t

he

righ

t,m

oves

to t

he

righ

t,h

its

fron

t w

all

2.a.

mg,

dow

n

b.m

g,u

p

c.n

o

d.ye

s

3.a.

a=

m1

+F

m2

b.m

2a

c.F

−m

2a

=m

1a

d. m

1m +1 m2

F

4.a.

a= mF 1− +F mk

2

b.m

2a

−F k

c.F

−m

2a

−F k

=m

1a−

F k

d. m

1m +1 m2

(F

−F k

)

Chap

ter 4

Mix

ed R

evie

w

Hol

t Phy

sics

Sol

utio

n M

anua

lIII

–6

1.a.

60 J

b.−6

0 J

2.a.

mgh

b.m

gh

c.v B

=v A

2+

2gh

d.n

o

e.n

o

3.a.

2.9

J

b.1.

8 J

c.1.

2 J

d.a,

b:di

ffer

ent;

c:sa

me

4.a.

1 2 mv i

2+

mgh

i=

1 2 mv f

2+

mgh

f+

F kd

b.F k

=mm

g(co

s 23

°)

c.v f

= mv i

2+

2g(d

sin

23°−

mco

s23

°)

Chap

ter 5

Mix

ed R

evie

w

Sect

ion

Thre

e—Se

ctio

n Re

view

Wor

kshe

et A

nsw

ers

III–7

III

Copyright ©by Holt, Rinehart and Winston.All rights reserved.

1.a.

Th

e ch

ange

du

e to

th

e ba

t is

gre

ater

th

an t

he

chan

gedu

e to

th

e m

itt.

b.T

he

impu

lse

due

to t

he

bat

is g

reat

er t

han

th

e im

-pu

lse

due

to t

he

mit

t.

c.C

hec

k st

ude

nt

diag

ram

s.B

at:v

ecto

r sh

owin

g in

itia

lm

omen

tum

an

d a

larg

er v

ecto

r in

th

e op

posi

te d

i-re

ctio

n s

how

ing

impu

lse

ofba

t,re

sult

is t

he

sum

of

the

vect

ors.

Mit

t:ve

ctor

sh

owin

g in

itia

l mom

entu

man

d an

equ

al le

ngt

h v

ecto

r sh

owin

g im

puls

e of

mit

t,re

sult

is t

he

sum

,wh

ich

is e

qual

to z

ero.

2.a.

Th

e im

puls

es a

re e

qual

,bu

t op

posi

te fo

rces

,occ

ur-

rin

g du

rin

g th

e sa

me

tim

e in

terv

al.

b.T

he

tota

l for

ce o

n t

he

bow

ling

ball

is t

he

sum

of

forc

es o

n p

ins.

Th

e fo

rce

on t

he

pin

s is

equ

al b

ut

op-

posi

te o

fto

tal f

orce

on

bal

l.

3.m

1v1i

+m

2v2i

=(m

1+

m2)v f

;

m1v

1i/(

m1

+m

2)+

m2v

2i/(

m1

+m

2)=v f

4.a.

M(6

m/s

)

b.2

m/s

c.ob

ject

s tr

ade

mom

entu

m;i

fm

asse

s ar

e eq

ual

,ob-

ject

s tr

ade

velo

citi

es

Chap

ter 6

Mix

ed R

evie

w

Hol

t Phy

sics

Sol

utio

n M

anua

lIII

–8

1.a.

3.0,

3.0,

9.0,

27

b.4.

3,1.

0,4.

3,37

c.16

,0.2

8,11

,6.0

×10

2

d.63

0,0.

11,7

4,8.

7

e.5.

0,44

,0.1

1,9.

9

2.a.

fric

tion

b.gr

avit

atio

nal

forc

e

c.te

nsi

on in

str

ing

3.a.

dou

bled

b.qu

adru

pled

c.re

duce

d to

1 4

d.qu

adru

pled

e.re

duce

d to

1 9

4.19

0 m

5.St

uden

t dia

gram

s sh

ould

sho

w v

ecto

rs fo

r w

eigh

t an

dn

orm

al fo

rce

from

ele

vato

r;de

scen

t sho

uld

show

nor

mal

forc

e le

ss th

an w

eigh

t;st

oppi

ng

shou

ld s

how

nor

mal

forc

e gr

eate

r th

an w

eigh

t;“w

eigh

tles

snes

s”fe

elin

g is

due

to a

ccel

erat

ion

.

6.10

50 s

(17

.5 m

in)

Chap

ter 7

Mix

ed R

evie

w


34

Sect

ion

Thre

e—Se

ctio

n Re

view

Wor

kshe

et A

nsw

ers

III–9

1.a.

Ifth

e kn

ob is

fart

her

from

th

eh

inge

,tor

que

is in

crea

sed

torq

ue

for

a gi

ven

forc

e.

b.tw

ice

as m

uch

2.a.

Rot

atio

nal

iner

tia

is r

edu

ced.

b.A

ngu

lar

mom

entu

m r

emai

ns

the

sam

e.

c.A

ngu

lar

spee

d in

crea

ses.

3.a.

2.0

kg

b.0.

67 k

g

4.a.

6.2

N• m

,0.0

16 k

g• m

2 ,39

0 ra

d/s2

b.12

N• m

,0.0

62 k

g•m

2 ,190

rad

/s2

5.a.

8.1

×10

16kg

• m2

b.5.

9×

1012

kg• m

2 /s

c.2.

1×

108

J

d.3.

1×

103

m/s

e.2.

2×

108

J

6.a.

4.0

×10

4J

b.4.

4×

104

J

c.4.

9×

104

J

d.0.

81

Chap

ter 8

Mix

ed R

evie

w


Documents

Fall 2008 Honors Physics Review Notes - Tom Strongtomstrong.org/physics/fall08h.pdfHonors Physics Review Notes Fall 2008 Tom Strong Science Department Mt Lebanon High School [email protected]