Welcome to the archival Web page for U.C. Berkeley's Physics H7ABC, Honors Physics for Scientistsand Engineers, Fall 1998, Spring 1999, and Fall 1999.Instructor:(Prof.) Mark Strovink. I have a research web page, a standardized U.C. Berkeley web page, and astatement of research interests.

Physics H7A (Mechanics and Vibrations)Problem set solutions initially composed by E.A. ("Ted") BaltzGraduate Student Instructors: David Bacon and Elizabeth Wu

Physics H7B (Electromagnetism and Thermal Physics):Most problem set solutions composed by Peter PeblerGraduate Student Instructor: Robin Blume-Kohout

Physics H7C (Physical Optics and Modern Physics):Problem set solutions composed by Graduate Student Instructor: Derek Kimball

Most documents linked on this page are in PDF format. They are typeset except where indicated. Thedocuments are intended to be displayed by Adobe Acrobat [Reader], version 4 or later (version 3 mayalso work). (You may optimize Acrobat's rendering of equations by unchecking "Use Greek TextBelow:" on File-Preferences-General.)You may right-click to download a single .pdf file (19.5 MB) that includes every displayable image onthis Web page.Physics H7A (Mechanics and Vibrations)Texts: Kleppner/Kolenkow, An Introduction to Mechanics; French, Vibrations and Waves.

General Information including schedules and rooms.Course Outline including all reading assignments.Questionnaire that was filled out by the students.Typos in Kleppner & Kolenkow

Problem Set 1 Solution Set 1Problem Set 2 Solution Set 2Problem Set 3 Solution Set 3Problem Set 4 Solution Set 4Problem Set 5 Solution Set 5Problem Set 6 Solution Set 6Problem Set 7 Solution Set 7Problem Set 8 Solution Set 8Problem Set 9 Solution Set 9Problem Set 10 Solution Set 10Problem Set 11 Solution Set 11

Physics H7ABC

http://d0lbln.lbl.gov/h7abc-web.htm (1 of 3) [1/5/2000 9:50:55 AM]

Practice Exam 1 Solution to Practice Exam 1Exam 1 Solution to Exam 1Exam 2 Solution to Exam 2Final Exam Solution to Final ExamYou may right-click to download a single .tar.gz archive (1.0 MB) that includes the source files requiredto build every H7A file linked above.Physics H7B (Electromagnetism and Thermal Physics)Text: Purcell, Electricity and Magnetism.

General Information including schedules and rooms.Course Outline including all reading assignments.Special Relativity Notes* also used in H7C.

Problem Set 1 Solution Set 1Problem Set 2 Solution Set 2Problem Set 3 Solution Set 3Problem Set 4 Solution Set 4Problem Set 5 Solution Set 5Problem Set 6 Solution Set 6Problem Set 7 Solution Set 7Problem Set 8 Solution Set 8Problem Set 9 Solution Set 9Problem Set 10 Solution Set 10Problem Set 11 Solution Set 11Problem Set 12 Solution Set 12Problem Set 13 Solution Set 13

Midterm Solution to MidtermFinal Exam Solution to Final Exam** handwrittenYou may right-click to download a single .tar.gz archive (20 MB) containing the source files required tobuild every typeset H7B file linked above.Physics H7C (Physical Optics and Modern Physics)Texts: Fowles, Introduction to Modern Optics; Rohlf, Modern Physics from alpha to Z0.General Information including schedules and rooms.Course Outline including all reading assignments.Note on H7C texts, required and supplementary.Errata for Fowles, Introduction to Modern Optics, second edition.Special Relativity Notes* also used in H7B.

Physics H7ABC

http://d0lbln.lbl.gov/h7abc-web.htm (2 of 3) [1/5/2000 9:50:55 AM]

Problem Set 1 Solution Set 1Problem Set 2 Solution Set 2Problem Set 3 Solution Set 3Problem Set 4 Solution Set 4Problem Set 5 Solution Set 5Problem Set 6 Solution Set 6Problem Set 7 Solution Set 7Problem Set 8 Solution Set 8Problem Set 9 Solution Set 9Problem Set 10 Solution Set 10Problem Set 11 Solution Set 11Problem Set 12 Solution Set 12

Midterm 1 Solution to Midterm 1Midterm 2 Solution to Midterm 2Final Exam Solution to Final Exam* handwrittenYou may right-click to download a single .tar.gz archive (0.5 MB) containing the source files required tobuild every typeset H7C file linked above.

Physics H7ABC

http://d0lbln.lbl.gov/h7abc-web.htm (3 of 3) [1/5/2000 9:50:55 AM]

University of California, BerkeleyPhysics H7A, Fall 1998 (Strovink)

General Information (1 Sep 98)

Instructors: Prof. Mark Strovink, 437 LeConte; (LBL) 486-7087; (home, before 10) 486-8079; (UC) 642-9685.Email: strovink@lbl.gov . Web: http://d0lbln.lbl.gov/ . Office hours: M 3:15-4:15, Tu 10-11,Th 10-11. Mr. David Bacon, 214 LeConte; (UC) 642-5430; (home, before 1 AM) 666-9867. Email:dabacon@wco.com . Office hours: W 12-2. Ms. Elizabeth Wu, 214 LeConte; (UC) 642-5430. Email:ecmwu@euler.berkeley.edu . Office hours in 262 LeConte: M 10-11. You may also get help in the 7ACourse Center, 262 LeConte.

Lectures: Tu-Th 11:10-12:30, 2 LeConte. Lecture attendance is essential, since not all of the course content can befound in either of the course texts.

Labs: In the second week, in 270 LeConte, please enroll in one of only 3 special H7A lab sections [(A) #134, M 4-6; (B) #241, Th 4-6; or (C) #312, F 8-10]. Section 134 is taught by Ms. Wu and Sections 241 and 312 are taught byMr. Bacon. If you can make more than one of these lab (and section, see below) slots, please attempt to enroll in theearliest of these lab slots. Depending on crowding, you may be asked to move to a later lab. During "off" weeks notrequiring lab apparatus, your lab section will still meet, in (#134) 336 LeConte, (#241) 395 LeConte, or (#312) 335LeConte.

Discussion Sections: Beginning in the second week, please enroll in the only one of the 1 hr H7A discussionsections corresponding to your H7A lab section: (A) #134, W 1-2, 343 LeConte; (B) #241, Tu 4-5, 395 LeConte;(C) #312, W 8-9, 385 LeConte. You are especially encouraged to attend discussion section regularly. There youwill learn techniques of problem solving, with particular application to the assigned exercises.

Texts (both required): Kleppner/Kolenkow, An Introduction to Mechanics, (McGraw-Hill, 1973). A.P. French,Vibrations and Waves, Paper Edition (Norton, 1971).

Problem Sets: Twelve problem sets are assigned and graded, with solutions provided as part of the Syllabus. Theyare due on Wednesday at 5 PM on weeks (including Thanksgiving) in which there is no exam, beginning in week 2.Deposit problem sets in the box labeled "H7A" outside 201 LeConte. You are encouraged to attempt all theproblems. Students who do not do them find it almost impossible to learn the material and to succeed on theexaminations. Work independently. Credit for collective effort, which is easy to identify, will be divided among thecollectivists. Late papers will not be graded. Your lowest problem set score will be dropped, in lieu of due dateextensions for any reason.

Syllabus: H7A has two syllabus cards. The first card is mandatory; it will be collected at the time of the first in-class examination. This card pays for the experiment descriptions and instructions that you will receive from yourGSI at the beginning of each laboratory. The second syllabus card is optional. It entitles you to pick up printedsolutions to the problem set assignments from Copy Central. These solutions will also be made available on theWeb. Both cards will be available for purchase at Copy Central beginning in the second week of class.

Exams: There will be two 80-minute in-class examinations and one 3-hour final examination. Before confirmingyour enrollment in this class, please check that its final Exam Group 9 does not conflict with the Exam Group for anyother class in which you are enrolled. Please verify that you will be available for both in-class examinations (Th 24Sep and Th 5 Nov, 11:10-12:30), and for the final examination, F 11 Dec, 5-8 PM. Except for unforeseeableemergencies, it will not be possible for these in-class or final exams to be rescheduled. Passing H7A requirespassing the final exam.

Grading: 35% in-class examinations; 20% problem sets; 40% final exam; 5% lab. Grading is not "curved" -- itdoes not depend on your performance relative to that of your H7A classmates. Rather it is based on comparing yourwork to that of a generation of earlier lower division Berkeley physics students, with due allowance for educationaltrends.

COURSE OUTLINE

Week Week Lecture Topic Problem Due LabNo. of... chapter (K&K = Kleppner/Kolenkow, Intro. to Mechanics ) Set No. 5 PM on...

(French = Vibrations & Waves)(Feynman = Lectures on Physics Vol. II)

K&K1 24-Aug 1.1-8 Introduction; vectors, kinematics. none

(do experiment in lab=expt)(have discussion in lab=disc)

2 31-Aug 1.9-2.3 Motion in polar coordinates; exptNewton's laws; units. 1 Wed 2 Sep

3 7-Sep 2.4-2.5 LABOR DAY HOLIDAY discApplication of Newton's laws; forces. 2 9-Sep

4 14-Sep 3.1-Note3.1 Momentum; center of mass. expt3 16-Sep

5 21-Sep 4.1-6 Work; kinetic energy. disc24-Sep EXAM 1 (covers PS 1-3)

6 28-Sep 4.7-14 Potential energy; nonconservative forces; discenergy conservation; power; collisions. 4 30-Sep

7 5-Oct 6.1-7 Angular momentum; fixed axis rotation; discrotation with translation. 5 7-Oct

8 12-Oct 7.1-5 Vector angular momentum; conservation thereof. expt8.1-8.4 Noninertial systems; fictitious forces. 6 14-Oct

9 19-Oct 8.5-Note8.2 Rotating coordinate systems; equivalence principle. expt9.1-9.5 Central forces. 7 21-Oct

10 26-Oct 9.6-9.7 Planetary motion. discFrench 8 28-Oct10-15,43-45,77-89 Damped forced harmonic oscillator.

11 2-Nov 62-70,92-96 Transient response. disc5-Nov EXAM 2 (covers PS 1-8)

12 9-Nov 19-27,119-129 Coupled oscillator; beats. disc161-170,189-196 Fourier expansion in normal modes. 9 11-Nov

13 16-Nov 201-209,213-215, Waves: travelling, sinusoidal, modulated; expt228-234 phase and group velocity. 10 18-Nov

14 23-Nov 45-62,209-212 Longitudinal waves; sound. disc170-178,274-279 Boundary reflections of waves; Doppler effect. 11 25-Nov

26,27-Nov THANKSGIVING HOLIDAY15 30-Nov Feynman disc

II.2-1,2,3,4,5;II.40-1,2,3 Fluid statics and nonviscous dynamics. 12 Fri 4 DecLAST LECTURE (review)

16 7-Dec11-Dec 5-8 PM FINAL EXAM (Group 9) (covers PS 1-12)

Physics H7A Fall 1998 (Strovink)

University of California, Berkeley

Physics H7A Fall 1998 (Strovink)

STUDENT QUESTIONNAIRE

Please complete this questionnaire if you are considering enrolling inPhysics H7A. Among other purposes, it will be used to make up the initialclass roll.

Are you qualified for this course? Your interest in Honors Physics isalready a good indication that you are. It is unlikely that the issue ofyour eligibility will come into question. If it does, you will beinterviewed promptly by the instructor.

Last name__________________________First_______________________Initial_________

Registration No.___________________ Year (Freshman, etc.)_____________

SAT scores:

SAT I Quantitative______ SAT II Math 1______

SAT I Verbal______ SAT II Math 1C______

SAT II Math 2______

SAT II Physics______ SAT II Math 2C______

AP Calculus Exam: Grade______ AB BC (circle one)

AP Physics Exam: Grade______ B C (circle one)

Year of last math course_______ Where taken?________________________________

Course title and no._______________________________________________________

Course text_______________________________________ Grade received_______

Year of last physics course____ Where taken?________________________________

Course title and no._______________________________________________________

Course text_______________________________________ Grade received_______

List majors that you are considering at Berkeley_______________________________

___________________________________________________________________________

Why are you (might you be) interested in Honors in place of standard lower-

division physics at Berkeley?__________________________________________________

_______________________________________________________________________________

_______________________________________________________________________________

_______________________________________________________________________________

_______________________________________________________________________________

_______________________________________________________________________________

University of California, BerkeleyPhysics H7A Fall 1998 (Strovink)

TYPOS IN KLEPPNER & KOLENKOW

This is a list of typos in Kleppner & Kolenkowthat have been brought to the attention of H7AF98 sta by students in the course. Thanks tothem for pointing these out.

Page 205, Example 5.2:

In the equation below the dierential of f is,replace the second dy with dx.

Page 276, Note 6.2:

In the equation following E = K + U , the rstterm 12 l

22 should be multiplied by m.

Page 392, Equation 9.22:

The rst instance of x should be x2 instead.

University of California, BerkeleyPhysics H7A Fall 1998 (Strovink)

PROBLEM SET 1

1. Specify the properties of two vectors a and bsuch that

(a.) a+ b = c and |a|+ |b| = |c|.(b.) a+ b = a b.(c.) a+ b = c and |a|2 + |b|2 = |c|2.(d.) |a+ b| = |a b|.(e.) |a+ b| = |a| = |b|.2. K&K problem 1.2 Find the cosine ofthe angle between A = (3i + j + k) and B =(2i 3j k).3. The relation between Cartesian (x, y, z) andspherical polar (r, , ) coordinates is:

x = r sin cosy = r sin sinz = r cos .

Consider two points on a sphere of radius R:(R, 1, 1) and (R, 2, 2). Use the dot productto nd the cosine of the angle 12 between thetwo vectors which point to the origin from thesetwo points. You should obtain:

cos 12 = cos 1 cos 2+sin 1 sin 2 cos (1 2).

4. New York has North Latitude (= 90 )= 41 and West Longitude (= 360 ) = 74.Sydney has South Latitude (= 90) = 34and East Longitude (= ) = 151. Take theearth to be a sphere of radius 6370 km; use theresult of Problem 3.

(a.) Find the length in km of an imaginarystraight tunnel bored between New Yorkand Sydney.

(b.) Find the distance of the shortest possiblelow-altitude ight between the two cities.(Hint: The great circle distance along thesurface of a sphere is just R12, where 12 isthe angle between the two points, measuredin radians.)

5. K&K problem 1.6 Prove the law of sines us-ing the cross product. It should only take a cou-ple of lines. (Hint: Consider the area of a triangleformed by A, B, C, where A + B + C = 0.)

6. K&K problem 1.11 Let A be an arbitraryvector and let n be a unit vector in some xed di-rection. Show thatA = (A n) n+(nA)n.7. If the air velocity (velocity with respect tothe air) of an airplane is u, and the wind ve-locity with respect to the ground is w, then theground velocity v of the airplane is

v = u+w.

An airplane les a straight course (with respectto the ground) from P to Q and then back to P ,with air speed |u| which is always equal to a con-stant U0, regardless of the wind. Find the timerequired for one round trip, under the followingconditions:

(a.) No wind.

(b.) Wind of speed W0 blowing from P to Q.

(c.) Wind of speed W0 blowing perpendicular toa line connecting P and Q.

(d.) Wind of speed W0 blowing at an angle from a line connecting P and Q.

(e.) Show that the round trip ying time is al-ways least for part (a.).

(f.) What happens to the answers to (b.)-(d.)when W0 > U0? Interpret this limitingcondition physically.

8. A particle moves along the curve y = Ax2

such that its x position is given by x = Bt (t =time).

(a.) Express the vector position r(t) of the par-ticle in the form

r(t) = if(t) + jg(t) [or xf(t) + yg(t)]

where i and j [or x and y] are unit vectors,and f(t) and g(t) are functions of t.

(b.) Find the (vector) velocity v(t) as a functionof t.

(c.) Find the (vector) acceleration a(t) as a func-tion of t.

(d.) Find the (scalar) speed |v(t)| as a functionof t.

(e.) Find the (vector) average velocity v(t0)between t = 0 and t = t0 where t0 is anypositive time.

9. Below are some measurements taken ona stroboscopic photograph of a particle under-going accelerated motion. The distance s ismeasured from a xed point, but the zero oftime is set to coincide with the rst strobe ash:

time (sec) distance (m)0 0.561 0.842 1.173 1.574 2.005 2.536 3.087 3.718 4.39

Plot a straight-line graph, based on these data,to show that they are tted by the equation

s = a(t t0)2/2,

where a and t0 are constants, and extrapolatethe line to evaluate t0.

1University of California, BerkeleyPhysics H7A Fall 1998 (Strovink)

SOLUTION TO PROBLEM SET 1Composed and formatted by E.A. Baltz and M. Strovink; proofed by D. Bacon

1. You may remember the law of cosines fromtrigonometry. It will be useful for several partsof this problem, so we will state it here. If thelengths of the sides of a triangle are a, b, and c,and the angle opposite the side c is , then

c2 = a2 + b2 2ab cos

(a.) When two vectors add up to a third vector,the three vectors form a triangle. If the anglebetween a and b is , then the angle oppositethe side formed by c is 180 .

The law of cosines then tells us that

|c|2 = |a|2 + |b|2 2|a||b| cos (180 )

From trigonometry, remember that

cos (180 ) = cos

which gives

|c|2 = |a|2 + |b|2 + 2|a||b| cos

We know that |a| + |b| = |c|. Squaring thisequation, we get

|c|2 = |a|2 + |b|2 + 2|a||b|

If we compare this with the equation above, wecan see that cos has to be equal to one. Thisonly happens when = 0. What this means isthat the two vectors are parallel to each other,and they point in the same direction. If the an-gle between them were 180, then they would beparallel but point in opposite directions.

(b.) This part is simple. Just subtract the vec-tor a from both sides to see that b = b. Theonly way that this can happen is if b = 0, thezero vector.

(c.) This part can also be done by the law ofcosines. Like part (a.), we have the followingtwo equations

|c|2 = |a|2 + |b|2 + 2|a||b| cos This is just the law of cosines again, where is the angle between |a| and b|. The problemstates that

|c|2 = |a|2 + |b|2Comparing this with the equation above, we ndthat cos = 0. This happens at = 90. Thismeans that the vectors must be perpendicular toeach other.

(d.) Yet again, we can use the law of cosines. Ifthe angle between a and b is , then the anglebetween a and b is 180 . The lengths ofthe sum and dierence are

|a+ b|2 = |a|2 + |b|2 + 2|a||b| cos |a b|2 = |a|2 + |b|2 2|a||b| cos

For these to be equal, we need cos = 0, whichhappens when = 90. Again, this means thatthe vectors are perpendicular.

(e.) Guess what? Yup, law of cosines. We knowthat |a| = |b| = |a+ b|. Adding a to b is goingto look like two vectors stuck together to formtwo sides of a triangle. If the angle between thevectors is , the law of cosines gives

|a+ b|2 = 2|a|2 + 2|a|2 cos where we have used the fact that a and b havethe same length. We also know that |a+b| = |a|.Using this we get

|a|2 = 2|a|2 + 2|a|2 cos Dividing by |a|2, we nd a condition on the angle

cos = 12

= 120

22. K&K problem 1.2

We can use the dot product, also known asthe inner product, of two vectors here. Remem-ber that

A B = |A||B| cos where is the angle between the vectors. We canuse the formula for computing the dot productfrom the vector components

A B = AxBx +AyBy +AzBz

The vectors are given as follows: A = 3i+ j+ kand B = 2i 3j k. Multiplying, we nd thatA B = 10. We need the lengths of A and B.Remember that |A|2 = A A. This tells us that|A|2 = 11 and |B|2 = 14. Dividing, we nd that

cos =1011 14 = 0.805

3. Using the formulas on the problem set, wecan convert the points on the surface of thesphere to Cartesian coordinates.

x1 = R sin 1 cos1y1 = R sin 1 sin1z1 = R cos 1x2 = R sin 2 cos2y2 = R sin 2 sin2z2 = R cos 2

As in problem 2, we need to know the length ofthese vectors in order to calculate the angle be-tween them from the dot product. It is fairlyobvious that the lengths of the vectors are justR, because that is the radius of the sphere; wewill show this explicitly.

|(R, , )|2 = R2(sin2 cos2 + sin2 sin2 + cos2 )

Using the fact that sin2 + cos2 = 1, we get

|(R, , )|2 = R2 (sin2 + cos2 )

We can just repeat the previous step for nowand get

|(R, , )| = R

as we expected in the rst place. Now we calcu-late the dot product. Let x1 be the vector to therst point and x2 be the vector to the secondpoint. We nd that

x1x2 = R2(sin 1 sin 2 cos1 cos2+ sin 1 sin 2 sin1 sin2 + cos 1 cos 2)

This can be simplied if we remember the for-mula for the cosine of a sum of two angles.

cos( ) = cos cos sin sin

Using this formula, we get the result

x1x2 = R2(sin 1 sin 2 cos(1 2)+ cos 1 cos 2)

To get the angle we just divide by the lengthsof each vector, which are both R. This gives thenal result.

cos 12 = cos 1 cos 2+ sin 1 sin 2 cos(1 2)

4. This problem is an application of the resultsof problem 3.

(a.) A straight tunnel between Sydney and NewYork can be represented by the dierence of thevectors pointing to their locations. To say it an-other way, the distance between the ends of twovectors is the length of the dierence of the vec-tors. Adjusting for the fact that latitude andlongitude are not quite the same as the coordi-nates and , we nd the polar coordinates ofthe cities.

XNY = (6370 km, 49, 286)XSydney = (6370 km, 124, 151)

Converting to Cartesian coordinates (x, y, z) us-ing the formulas from problem 3 we get

XNY = (1325 km,4621 km, 4179 km)XSydney = (4619 km, 2560 km,3562 km)

3The distance between New York and Sydneythrough the earth is just |XNY XSydney|. Theresult of the calculation is

Distance = 12, 117 km

(b.) Using the result from problem 3 to calcu-late the angle between Sydney and New York,we nd that cos 12 = 0.809, thus 12 = 144.0.To calculate the distance along the earths sur-face we need to express this angle in radians.The conversion formula is

(radians) =

180(degrees)

Thus 12 = 2.513 radians. Multiplying this bythe radius of the earth, we get the great circledistance between New York and Sydney:

Distance = 16, 010 km

5. K&K problem 1.6

This question asks you to prove the law ofsines using the cross product. Let A,B, and Cbe the lengths of the vectors making the threesides of the triangle. Let a, b and c be the an-gle opposite each of those sides. The law of sinesstates that

sin aA

=sin bB

=sin cC

Remember that the length of the cross productof two vectors is equal to the area of the paral-lelogram dened by them. Remember also thatthe the length of the cross product is equal tothe product of the lengths times the sine of theangle between them: |AB| = |A||B| sin . Wehave three vectors to play with in this problem,and using the cross product we can compute thearea of the triangle from any two of them. Wend that

Area = AB sin c = BC sin a = AC sin b

We just divide the whole thing by ABC and werecover the law of sines.

6. K&K problem 1.11

Let A be an arbitrary vector and let n be aunit vector in some xed direction. Show that

A = (A n) n+ (nA) nForm a triangle from the three vectors in thisequation. Let B = (A n) n and let C =(nA) n. Let the angle between A andn be . What this formula does is to break upthe vectorA into a piece parallel to n and a pieceperpendicular to n. B gives the parallel piece.Its length is just |B| = |A| cos . The length ofthe perpendicular piece must then be |A| sin .

Examining the vector C, we see that inside theparentheses is a vector whose length is |A| sin and is perpendicular to n. This vector is thencrossed into n. Since it is perpendicular to n,the length of the nal vector is |A| sin , whichis what we want. Now we are just concernedwith the direction. The rst cross product isperpendicular to the plane containing n and A.The second cross product is perpendicular to therst, thus it is coplanar with n and A. It isalso perpendicular to n. Thus it represents thecomponent of A that is perpendicular to n. Becareful about the sign here.

A useful vector identity that you will be see-ing again is the so-called BAC-CAB rule. It isan identity for the triple cross product.

A(BC) = B(A C)C(A B)Its fairly obvious why this is called the BAC-CAB rule. Using this rule, we see that

n(nA) = n(n A)A(n n)This immediately gives

A = (A n) n+ (nA)n

4Of course we havent derived the BAC-CAB rulehere. Its a mess.

7. The idea in all of the parts of this problemis that the plane must oppose any perpendicularwind speed to maintain its straight path. If thewind is blowing with a speed v perpendicular tothe path, the planes airspeed must be v per-pendicular to the path. The airspeed is u, thewind speed relative to the ground is w, and theground speed is v = u +w. |u| = U0. Let thetotal distance traveled be D.

(a.) No wind, w = 0 so u = v. T = D/U0.

(b.) Wind of speed W0 blowing parallel to thepath. When the wind is going with the plane,v = W0 + U0, when it opposes the plane, theground speed is v = U0 W0. The time for therst leg is T1 = D/2(U0 +W0). The time forthe second leg is T2 = D/2(U0 W0). The totaltime is the sum

T =D

2

(1

U0 +W0+

1U0 W0

)

This can be simplied, and we get the nal an-swer, which agrees with part (a.) when W0 = 0.

T =DU0

U20 W 20

(c.) Wind of speed W0 blowing perpendicular tothe path. This part is a little harder. The planewill not be pointed straight along the path be-cause it has to oppose the wind trying to blow ito course. The airspeed of the plane in the per-pendicular direction will be W0, and we knowwhat the total airspeed is, so we can calculatethe airspeed along the path.

U20 = U2 + U

2 U =

U20 W 20

The wind has no component along the path ofmotion, so the airspeed in the parallel directionis the same as the ground speed in the paralleldirection. The ground speed is furthermore thesame on both legs of the trip. The nal answeragain agrees with part (a.) when W0 = 0

T =D

U20 W 20

(d.) Wind of speed W0 blowing at an angle to the direction of travel. The plane again needsto cancel the component of the wind blowing inthe perpendicular direction. The perpendicularcomponent of the wind speed is W0 =W0 sin .As in part (c.) the airspeed in the parallel direc-tion can be computed

U20 = U2 + U

2 U =

U20 W 20 sin2

In this case, the wind has a component alongthe direction of travel. This parallel componentis W0 = W0 cos . On one leg of the trip, thisadds to the ground velocity. On the other leg, itsubtracts. This gives us the following formula:

T =D

2

(1

U20 W 20 sin2 +W0 cos

+1

U20 W 20 sin2 W0 cos

)

This can be simplied considerably:

T =DU20 W 20 sin2 U20 W 20

If you look at this carefully, you will realize thatit reduces to the correct answer for parts (a.),(b.), and (c.) with the proper values for W0 and. If = 90, the wind blows perpendicular tothe path and we get the result from part (c.). If = 0, the wind blows parallel to the directionof travel and we recover the result from part (b.).

(e.) (f.) This part requires some calculus. Weneed to do a minimization of a function. Whatthis part asks is to study the travel time as afunction of wind speed for an arbitrary angle .We need to consider the result from part (d.) asa function of the wind speed:

T (W0) =DU20 W 20 sin2 U20 W 20

5For now we are going to ignore the fact that italso depends on U0 and . Remember that func-tions have maxima and minima at places wherethe derivative vanishes, so we need to take thederivative of T with respect to W0:

d

dW0T (W0) = D

(2W0U20 W 20 sin2

(U20 W 20 )2

W0 sin2

U20 W 20 sin2 (U20 W 20 )

)

The derivative is clearly zero when W0 = 0. Inthis case the travel time T = D/U0 as in part(a.). There is another case we have to worryabout though. We divide out what we can toget an equation for another value where thederivative vanishes

U20 sin2 W 20 sin2 = 2U20 2W 20 sin2

This gives us the other point where the deriva-tive is zero

W 20 = U20

2 sin2 sin2

Notice that this point always occurs when thewind speed is greater than the air speed. Noprogress can be made against the wind if thisis the case, so the trip cannot occur. The -nal possibility to consider is the case where thewind speed is the same as the air speed. Look-ing at the formula, the time taken is innite.The only possibility is that the minimum is atW0 = 0. The nal piece of this problem is toobserve what happens to the time taken whenW0 > U0. For one thing, it becomes negative.In some circumstances it can even become imag-inary. There is really no interpretation of thisother than ask a stupid question, get a stupidanswer. The answer doesnt make sense be-cause the question didnt make sense. The tripcannot occur when W0 > U0, so it is meaninglessto ask how long it would take.

8. A particle moves along the curve y = Ax2

and its x position is given by x = Bt.

(a.) We can just plug the x equation into the yequation to get the y position as a function of

time, y = AB2t2. In vector form, the position isthen

r(t) = xBt+ yAB2t2

(b.) The vector velocity is obtained from the vec-tor position by dierentiating with respect to t

v(t) =d

dtr(t) = xB + y2AB2t

(c.) The vector acceleration is obtained fromthe vector velocity by again dierentiating withrespect to t

a(t) =d

dtv(t) = y2AB2

(d.) The scalar speed is just the length of thevelocity vector. Remember that |A| = A A.

|v(t)| =v(t)v(t) =

B2 + 4A2B4t2

(e.) The vector average velocity is the integralof the velocity vector over a time interval, di-vided by the time interval. In general, the (time)average of a quantity A is given by

A = 1(t2 t1)

t2t1

A(t)dt

Applying this formula, we see the integral thatneeds to be evaluated:

v(t0) = 1t0

t00

v(t)dt

=1t0

t00

(xB + y2AB2t)dt

We could also use the fact that the integral of thevelocity is the position to get a simpler lookingformula for the average velocity

v(t0) = 1t0(r(t0) r(0))

Evaluating this integral, we get an answer thatis not surprising

v(t0) = xB + yAB2t0

6This is just (r(t0) r(0))/t0! The average veloc-ity is just the distance traveled divided by thetime it took.

9. The idea behind this problem is to make agraph of position vs. time data and show thatthey t the equation s = a(t t0)2/2. In addi-tion you are supposed to nd t0. The way todo this is to plot the square root of the distancevs. time, which will give a straight line graph:s =

(a/2)(t t0). The slope of this graph is

approximately 0.168, so we can use that to ex-trapolate back to zero. We nd that the graphreaches zero at about t = 4.45, so this meansthat t0 = 4.45 to make the distance traveledequal zero at t = 4.45.

University of California, BerkeleyPhysics H7A Fall 1998 (Strovink)

PROBLEM SET 2

1. Calculate the following centripetal accel-erations as fractions or multiples of g (= 9.8m/sec2):

(a.) The acceleration toward the earths axis of aperson standing on the earth at 45 latitude.

(b.) The acceleration of the moon toward theearth.

(c.) The acceleration of an electron movingaround a proton at a speed of 2106 m/secin a circular orbit of radius 0.5 Angstroms(1 Angstrom = 1010 m).

(d.) The acceleration of a point on the rim of abicycle wheel of 26 in diameter, traveling ata constant speed of 25 mph.

2. K&K problem 1.17 A particle moves in aplane....

3. K&K problem 1.20 A particle moves out-ward along....

4. At t=0 an object is released from rest at thetop of a tall building. At the time t0 a secondobject is dropped from the same point.

(a.) Ignoring air resistance, show that the timeat which the objects have a vertical separa-tion l is given by

t =l

gt0+t02.

How do you interpret this result for l t0, the separationbetween them is l = d1 d2. Thus we get

l =12gt2 1

2g(t2 2tt0 + t20) = gtt0

12gt20

Solving for t as a function of l, we get

t =l

gt0+t02

When l < gt20/2 this time is negative. This doeshave a sensible interpretation, unlike the nega-tive time in the airplane problem of the previousproblem set. Think of the problem as if wewanted to throw both objects at the same time,but still have the initial conditions given. At t =0, when we drop the rst object, from where dowe have to throw the other object? The answeris we want to throw upwards from below in sucha way that at time t = t0, the ball has reachedthe peak of its path and is momentarily at restat the point where the rst ball was dropped.During the time between t = 0 and t = t0, theseparation between the objects can be negative,meaning that the second one is below the rst.

(b.) In this part we want to calculate the op-timal value of t0 so that the separation reachessome value l0 at the earliest time possible. Inother words, we want to minimize the functiont(t0, l0). First we take the derivative and set itto zero to nd local extrema.

dt

dt0=

12 l0gt20

= 0 t0 =

2l0g

Notice that this is just the time that it takes therst object to fall a distance l0. The endpointshere are t0 = 0, where the separation remainsat l = 0 forever and t0 = , where the time toreach a separation of l0 is also innity. Thusthis best time to drop it is at t0 =

2l0/g. This

means that the best thing to do is to drop the

3second object when the rst object has alreadyfallen a distance l0.

5. K&K problem 1.21

This is another maximization problem. Wewant to know the optimal angle to throw a balldown a hill with slope angle . Splitting thisinto the x and y directions is the easiest wayto do the problem. First put the origin at thetop of the hill. If the ball is thrown up at anangle with speed v, the initial velocities arevx = v cos and vy = v sin .

Taking into account the acceleration of gravity,the positions are given by

x = vt cos

y = vt sin 12gt2

We need to know where the ground is in thesecoordinates. At a position x on the ground, they coordinate is given by yground = x tan. Wecan now nd the time at which the ball hits theground. Plugging into the equation for distancetraveled in y, we get

yground = x tan = vt sin 12gt2

v sin 12gt = v cos tan

This gives the time at which the ball hits theground:

t =2vg(sin + cos tan)

We now plug this time into the equation fordistance traveled in the x direction, giving thedistance that the ball traveled:

x =2v2

g(cos sin + cos2 tan)

=2v2

g(12sin 2 + cos2 tan)

Treating this as a function of , we can maxi-mize the range by dierentiating with respect to. Note that the endpoints in this problem arenot interesting. Throwing the ball straight up( = 90) and throwing it at an angle bothresult in the ball traveling no distance in the xdirection.

dx

d=

2v2

g(cos (2) 2 cos sin tan)

=2v2

g(cos (2) sin (2) tan) = 0

Solving for , we get

cos(2) = sin(2) tan cot(2) = tan

Remembering that cot = tan(/2 ), we seethe nal result:

tan(2 2

)= tan =

4

2

Note that on a level surface, when = 0, theoptimal angle is 45, as you might already know.

6. K&K problem 2.1

This is the rst problem where you are askedto consider the forces causing acceleration. Theforce on a 5 kg mass is given by F = 4t2x 3tyNewtons. Apply Newtons second law of mo-tion, namely F = ma, to get the acceleration,a = (4t2/5)x (3t/5)ym/sec.(a.) We can get velocity from acceleration byintegrating

v(t) v(t0) = tt0

a(t)dt

Plugging the acceleration we just determinedinto this integral, and knowing that the velocityat t = 0 is zero, we get the velocity as a functionof time:

v(t) =415t3x 3

10t2y m/sec

4(b.) We get the position by integrating again:

r(t) r(t0) = tt0

v(t)dt

Applying this to the result of part (a.), and re-membering that at t = 0 the mass is at the originso r(0) = 0, we get the position as a function oftime:

r(t) =115t4x 1

10t3y m

(c.) Now all that is left is to take the cross prod-uct of the position with the velocity. We ndthat

r v =( 3150

t6 +4150

t6)z =

t6

150z

7. This problem asks you to consider two blockssliding on a table together. The larger block,with mass M , has ve forces acting on it. Theyare F x, the applied force, a contact force thatI will call CM x, the force of gravity gy, thenormal force N y, and the force of friction. Be-cause there is no acceleration in the y direction,we can easily nd that N = Mg, so that thereis no net force in the y direction. From the nor-mal force we can determine the force of friction.An object that is sliding with friction along asurface is acted upon by a force opposing the di-rection of motion with magnitude N(= Mg),where is the coecient of sliding friction. Wenow can write an expression for the accelerationof the large block in the x direction

ax =FxM

=F

M CM

M g

There are two unknowns here, ax and CM . Weneed another equation. Luckily, there is anotherblock that we can write equations about. Thesmall block, having mass m, is aected by fourforces. They are the contact force, the force ofgravity, the normal force, and the force of fric-tion. The contact force is exactly opposite tothe contact force on the rst block. This is dueto Newtons third law of motion, which statesthat every force has an equal and opposite force.Thus Cm = CM and the contact force is CM x.The force of gravity is just mgy, the normalforce exactly opposes gravity as before, +mgy,and the force of friction is again mg. Weget the equation of motion for the second block,noticing that acceleration of the second block isthe same as for the rst block because they aremoving together:

ax =CMm

g

We are only concerned with CM here, so we cansimplify the solution of these equations. Subtractthe second equation from the rst to get

F

M CM

(1M

+1m

)= 0

We can now solve for CM in terms of F :

CM = F(

m

M +m

)

If we follow the same procedure when the forceis acting on the second block, we get a very simi-lar answer, but the mass in the numerator is thelarger mass

CM = F(

M

M +m

)

This is a factor of M/m larger, which is whatwe expect. The only force pushing on the sec-ond block in each case is the contact force,and the acceleration doesnt depend on whichside we push the combined system from. The

5force of friction acts in proportion to the massin this case, so it does not aect this argu-ment. It can be thought of as a force thatacts on the combined system, not on the indi-vidual blocks, because it is proportional to themass. Notice that neither of these expressionsdepend on the value of , which would indicatethat the friction was aecting the contact force.

8. K&K problem 2.5

This is the rst pulley problem, and it wontbe the last. The pulley is massless and friction-less, and supports two masses M and m by amassless rope connecting them. The rst thingto notice about this problem is that the ten-sion in the rope must be the same on both sidesof the pulley. If the dierent sides had dier-ent tensions, there would be a tendency to causean angular acceleration in the pulley. Since itis massless, this acceleration would be innite,which is unphysical, so the tensions must beequal. You will see this in more detail when youdo rigid body motion later this term.

The second thing to notice is that the moremassive block will fall and the less massive blockwill rise, and their accelerations will be the same,but in opposite directions. This just means thatthe rope isnt stretching. For this problem I willset down to be the positive direction. Theequation for the larger mass is

Ma =Mg T T =M(g a)Using the expression for tension derived in theabove equation, the equation for the smallermass is the following:

ma = mg + T m(a+ g) =M(g a)

Solving for the acceleration,

a = g(M mM +m

)

The larger mass accelerates with magnitude adownward. The tension is found by plugging theacceleration into either of the starting equations

T =Mg Mg(M mM +m

)= g

(2MmM +m

)

9. Nope, this isnt the last pulley problem either.Again, the pulley and cords are massless and thepulley is frictionless. A force F is applied up-ward, and various things will happen dependingon what F is. The rst thing to notice is that thepulley is massless. This means that the tensionson the two ropes must be equal, otherwise a niteangular force would be applied to a massless ob-ject, which again is unphysical. The second thingto notice is that the upward force must exactlybalance the sum of the tensions. If this werentthe case, there would be a net force applied to amassless object. This cant happen, so to balancethe forces we just need F = 2T . With these twopoints in mind, we can do the rest of the problem.

(a.) The boundary between regimes (i) and (ii)is where the lighter block lifts o the ground.Consider the forces on this block. They are grav-ity mg, the tension T , and the normal force N .The equation of motion is ma = T +N mg. Atthe boundary between regimes (i) and (ii) whenthe block just barely can be lifted upward, thenormal force N is zero, but so is the accelera-tion. This gives us T = mg. We know also thatT = F/2, so the minimum force to lift the lighterblock is F = 2mg. This is just twice the weightof the lighter block, which we expect because theforce applied gets divided in half by the pulley.The boundary between regimes (ii) and (iii) isfound in a similar way. The equation of motionis T +N Mg = Ma. Again both a and N arezero at the transition point, so T = Mg, whichgives the force F = 2Mg. The nal results are

regime(i) {F < 2mg}regime(ii) {2mg < F < 2Mg}regime(iii) {2Mg < F}

6(b.) Now we want to nd the accelerationsfor regimes (ii) and (iii). This is easy be-cause we have already determined the equationsof motion. For regime (ii), only the lighterblock accelerates. The equation of motion isF/2mg = ma. This gives the result

regime (ii) aM = 0

am =F

2m g

In regime (iii), the equation of motion of therst block is the same, so we get the same resultfor the acceleration. The equation of motion ofthe larger block is F/2Mg = Ma. These givethe results

regime (iii) aM =F

2M g

am =F

2m g

10. K&K problem 2.6

The cement mixers drum has a radius R.We want to know how fast it can rotate so thatthe material will not stick to the walls all of thetime. We just need to gure out at what speedthe drum can oppose gravity all of the time. Fora glob of material of mass m, the worst case isat the top. To remain in contact with the drum,at that point the glob must feel a downwardforce from the drum that is positive. In additionit feels the downward force mg due to gravity.So the total downward force on it must be atleast mg. Now, what acceleration accompaniesthis force? We dont want the glob to leave thedrum, so its radial velocity must remain equalto zero. The only remaining possible downwardacceleration is the centripetal acceleration dueto the circular motion. This is 2R. Equatingthe mass m times this acceleration to the totaldownward force, we conclude that m2R mg,or g/R. For the material to not alwaysstick, we need the nal result

. We now calculatethe total distance traveled in the second part ofthe trip. We take the nal time to be t =.

x() = (M +m)pkM

(1 exp

( kM +m

))

This is exactly the opposite of the distancetraveled in the rst part. Thus the boat willeventually return to its starting point.

(b.) Here is a quick, elegant way to prove theresult of part (b.). It deserves full grading credit.We do not mention only this method because,as seen above, the problem is amenable to solu-tion by systematic calculation as well as brilliantinsight.

Consider the impulse applied by the forceFext of the water on the boat. To specify theimpulse, which is the time integral of Fext, we

must specify the time interval. We choose theinterval from t = 0, just before any motionstarts, to t =, at which time all motion musthave stopped due to eects of viscosity. Atboth of those times the total momentum of theboat+man system, whose rate of change is con-trolled by Fext, is zero. Therefore the impulse inquestion, which is equal to the dierence P ()P (0) of the boat+man system, must vanish.

The same impulse can also be written as

0 = 0

Fextdt

= k 0

dx

dtdt = k

0

dx

= k(x() x(0)) ,

where x is the position of the boat. This provesthat the boat returns to its original position.

(c.) The result of part (b.) says that any viscousforce, no matter how small, results in the boatreturning to its original location. The result ofpart (a.) says that when there is no viscous force,the boat moves some distance. Mathematically,the dierence between the two results is due tothe order in which the limits are taken. In part(a.), the rst thing done is to take the limitas k 0, no viscous force. Then the limit ast is taken. If we look at the result of part(b.), we rst take the limit as t , then weconsider what happens when there is no viscousforce. This is an instance in which we cannotreverse the order of taking limits. Denoting theresults from part (a.) and (b.) by capital letters,we see that

limt limk0

A = limk0

limtB

So much for the reason why, mathematically, theresults of (a.) and (b.) are not the same. Phys-ically, they are not in conict. As the coecientk approaches zero in part (b.), the speed withwhich the boat ultimately migrates back to itsoriginal position approaches zero also. This can-not be distinguished by physical measurementfrom the limiting case (a).

56. The Great Pyramid at Gizeh is h=150 mhigh and has a square base of side s = 230 m. Ithas a density = 2.5 g/cc.

(a.) If all the stone is initially at ground level,it must be raised to its position in the pyramid.The work required to this is

W =Mghcm =gz dV

The volume element is the area of the square at aheight z times dz, the dierential of height. Thesquare has side s at z = 0 and side 0 at z = h.The width of the square decreases linearly withheight, so the width and area at height z is givenby

w(z) = s(1 z

h

)A(z) = s2

(1 z

h

)2

The volume element dV is given by dV =A(z) dz. We can now perform the integral.Expanding the polynomial in z

W = gs2 h0

(z 2z

2

h+z3

h2

)dz

This is a simple integral to perform:

W = gs2h2(12 23+14

)=

112gs2h2

Plugging in the values for these constants, we getthe amount of work required to erect the pyramid

W = 2.43 1012 Joules

(b.) The slaves employed in building this pyra-mid consumed 1500 Calories per day, which is6.3 106 joules per day. With 100,000 slavesworking for 20 years, this is 730 million slave-days of work to build the pyramid. The totalenergy the slaves spent is thus 4.6 1015 joules.The eciency thus implied is low, " = 5.3104.This does not necessarily reect a low intrinsiceciency, since the slaves undoubtedly expendedmost of their energy on activities other than lift-ing the stone blocks to their nal position.

7. A force f(t) has magnitude F at t = 0, mag-nitude 0 at t = T , and it decreases linearly with

time. The direction remains the same. Themagnitude of the force is thus

f(t) = F(1 t

T

)

The force acts on a particle of mass m initiallyat rest. The kinetic energy at t = T is just theintegral

K = T0

F

(1 t

T

)dx =

T0

F

(1 t

T

)v dt

We can nd v by applying Newtons second law,but once we have it, we dont need to do theintegral because we know that K = mv2/2

f(t) = mdv

dt= F

(1 t

T

)

We can just directly integrate both sides withrespect to t, with limits t = 0 and t = T

v(T ) =F

2mT

We now have the answer

K =18F 2T 2

m

8. Instantaneously after the collision of the bul-let and block, after the bullet has come to restbut before the frictional force on the block hashad time to slow it down more than an inn-tesimal amount, we can apply momentum con-servation to the bullet-block collision. At thattime the total momentum of the block+bulletsystem is (M +m)v0, where v

0 is the velocity of

the block+bullet system immediately after thecollision. Momentum conservation requires thatmomentum to be equal to the initial momentummv of the bullet. Thus

v0 =mv

M +m.

After the collision, the normal force on theblock+bullet system from the table is (M+m)g,giving rise to a frictional force

N = (M +m)g

6on the sliding block+bullet system. This causesa constant acceleration g of that system oppo-site to its velocity.

Take t = 0 at the time of collision. After-ward, the block+bullet systems velocity in thehorizontal direction will be v(t) = v0 gt. Itwill continue sliding until v(t) = 0, at whichpoint the frictional force will disappear and itwill remain at rest. Solving, the time at whichthe block-bullet system stops is

t = v0/(g) .

The distance traveled in that time is

x = v0t12gt2 =

12v0t =

(v0)2

2g.

Plugging in the already deduced value for v0,this distance is

x =( mM +m

)2 v22g

.

University of California, BerkeleyPhysics H7A Fall 1998 (Strovink)

PROBLEM SET 5

1. Mass M rests on a table, and mass m issupported above it by a massless spring whichconnects the two masses.

(a.) Find the minimum downward force thatmust be exerted on m such that the entireassembly will barely leave the table whenthis force is suddenly removed.

(b.) Consider this problem in the time-reversedsituation: Let the assembly be supportedabove the table by supports attached to m.Lower the system until M barely touchesthe table and then release the supports.How far will m drop before coming to astop? Does knowledge of this distance helpyou solve the original problem?

(c.) Now that you have the answer, check itagainst your intuition by (1) letting M bezero and (2) letting m be zero. Especiallyin the second case, does the theoretical re-sult agree with your common sense? If not,discuss possible sources of error.

2. It has been claimed that a rocket would riseto a greater height if, instead of being ignited atground level (P ), it were ignited at a lower level(Q) after it had been allowed to slide from restalong a frictionless chute see the gure. To an-alyze this claim, consider a simplied model inwhich the body of the rocket is represented bya mass M , the fuel is represented by a mass m,and the chemical energy released in the burn-ing of the fuel is represented by a compressedspring between M and m which stores a deniteamount of potential energy, U , sucient to ejectm. (This corresponds to instantaneous burningand ejection of all the fuel i.e. an explosion.)Then proceed as follows:

(a.) Assuming a value of g independent of height,calculate how high the rocket would rise ifred directly upward from rest at (P ).

(b.) Let (Q) be a distance h vertically lower than(P ), and suppose that the rocket is red at(Q) after sliding down the frictionless chute.What is the velocity of the rocket at (Q)just before the spring is released? Just afterthe spring is released?

(c.) To what height above (P ) will the rocketrise now? Is this higher than the earliercase? By how much?

(d.) Remembering energy conservation, can youanswer a skeptic who claims that someonehas been cheated of some energy?

3. K&K problem 4.7 A ring of mass Mhangs....

4. Assume the moon to be a sphere of uniformdensity with radius 1740 km and mass 7.31022kg. Imagine that a straight smooth tunnel isbored though the moon so as to connect any twopoints on its surface. The gravitational force onan object by a uniform sphere is equal to theforce that would be exerted by the fraction of thespheres mass which lies at smaller radius thanthe object, as if that fraction were concentratedat the center of the sphere.

(a.) Show that the motion of objects along thistunnel under the action of gravity would besimple harmonic (neglect friction with thewalls of the tunnel).

(b.) Calculate the period of oscillation.

(c.) Compare this with the period of a satel-lite travelling around the moon in a circularorbit at the moons surface.

5. K&K problem 4.23 A small ball of mass mis placed on top....

6. In an historic piece of research, James Chad-wick in 1932 obtained a value for the mass of theneutron by studying elastic collisions of fast neu-trons with nuclei of hydrogen and nitrogen. Hefound that the maximum recoil velocity of hy-drogen nuclei (initially stationary) was 3.3107m/sec, and that the maximum recoil velocity ofnitrogen-14 nuclei was 4.7106 m/sec with anuncertainty on the latter of 10%. What doesthis tell you about:

(a.) The mass of the neutron (in amu)?

(b.) The initial velocity of the neutrons used?

(Take the uncertainty of the nitrogen measure-ment into account. Take the mass of a hydrogennucleus as 1 amu and the mass of a nitrogen-14nucleus as 14 amu.)

7. K&K problem 4.13 A commonly used po-tential energy function....

1University of California, BerkeleyPhysics H7A Fall 1998 (Strovink)

SOLUTION TO PROBLEM SET 5Composed and formatted by E.A. Baltz and M. Strovink; proofed by D. Bacon

1. A mass M rests on a table, and a mass mis supported on top of it by a massless springconnecting it to M .

(a.) We want to nd the force F needed topush down on the spring so that the whole sys-tem will barely leave the table. We solve thisby conservation of energy. The energy stored ina spring is given by kx2/2 where x is the dis-placement from equilibrium of the spring. Wecan measure the gravitational potential of thesmall mass relative to the equilibrium point ofthe spring. Initially then, the spring is com-pressed with a force F + mg which is just theweight of the small mass plus the added force.Hookes law tells us that x = (F +mg)/k. Wecan now write the initial energy of the system

Ei = mgk(F +mg) +

(F +mg)2

2k

The rst term is the gravitational energy relativeto the equilibrium point of the spring, and thesecond term is the energy stored in the spring.We want the spring to be able to lift the massM o the table. To do this it must apply aforce equal to Mg, its weight. When the springis released, it will oscillate. At the top of theoscillation, there will be no kinetic energy. Thedisplacement y of the spring must barely providethe force to lift the lower block: ky = Mg. Theenergy here is the following

Ef =Mmg2

k+M2g2

2k

Conservation of energy tells us that these arethe same, so now we can solve for F .

mgk(F +mg) +

(F +mg)2

2k=Mmg2

k+M2g2

2k

Cancelling k and using the quadratic formula tosolve for F +mg,

F +mg =(mg

m2g2 + 2Mmg2 +M2g2

)

The expression under the square root is just(M +m)g, so the expression simplies a lot:

F = (M +m)g

We obviously want the plus sign.

(b.) This is a similar situation. The mass M isdangling and barely touching the table. The dis-placement of the spring just supports the weightof the block so kx = Mg. At the other end,the small mass is momentarily at rest at somedistance y from equilibrium. The energies are

Ei =Mmg2

k+M2g2

2k

Ef = mgy + ky2

2

We equate these energies and solve for y.

mgy + ky2

2=Mmg2

k+M2g2

2k

Using the quadratic formula again

ky = mg m2g2 + 2Mmg2 +M2g2

Again the discriminant is a perfect square, andwe want the positive value of y, so

y =(2m+M)g

k

The total distance that the mass falls is x+y = d.

d = 2(M +m)g

k

This is just twice the displacement caused by theforce in part (a.), which makes sense, becausethe displacement upwards should be the same asthe displacement downwards.

2(c.) (1) When M is zero, the necessary appliedforce is mg. This is just the weight of the smallmass. The spring will bounce back with thesame force, so this is what is needed to lift thewhole assembly. The distance fallen makes sensealso, because the spring starts at its equilibriumlength. The mass wants to sit at mg/k belowthis, to just support the weight. Thus it willoscillate down to 2mg/k below this point.

(2) When m is zero, the force needed tomove the assembly is Mg; again this is the totalweight. The distance traveled by the end of thespring in the second case is just 2Mg/k. Theend of the spring isMg/k away from equilibriumwhen it begins, so the total distance traveled bythe end is 2Mg/k. While this seems to work out,it does not necessarily agree with common sense:a massless spring would not seem to be able topull a massive block o the table by virtue ofits own motion. However, we realize that, as thespring mass approaches zero in this idealization,its maximum velocity approaches innity. Thisexplains why the spring is still able to pull theblock o the table, defying our intuition.

2.

(a.) The rocket is red directly upward from theground. The initial energy is just U , the energyin the fuel. After the fuel is spent, the fuel massm is moving down at speed u and the remainingrocket mass M is moving upwards at speed v.Because momentum is conserved over this veryshort time, mu = Mv. The energy of the sys-tem is given by conservation of energy, and atlaunch, all of the energy is kinetic:

U =12Mv2 +

12mu2 =

Mv2

2

(1 +

M

m

)

Now we want to consider the motion of the massM alone. Its kinetic energy is Mv2/2, which wecan nd from the previous equation.

KM =U

1 +M/m

Energy for the mass M is now conserved, so wecan just set KM = Mgd, where d is the maxi-mum height achieved by the rocket. This gives

the answer to part (a.):

d =U

Mg

11 +M/m

(b.) This is a little more involved. The rockethas gone around part of an oval track and isnow a distance h below where it started. Thegravitational energy (M +m)gh gets convertedto kinetic energy, so we get the velocity v0 of therocket before the fuel is used:

(M +m)gh =12(M +m)v20 v0 =

2gh

Ignoring for the moment the gravitational en-ergy, the energy of the rocket at this point is

Ei = U +12(M +m)v20 = U + (M +m)gh

The spring (fuel) imparts a change in veloc-ity v to M and u to m. As in part (a.),instantaneous conservation of momentum givesMv = mu. After the spring is released, theenergy corresponding to Ei is

Ef =12M(v0 +v)2 +

12m(v0 u)2

We know by conservation of energy that Ei =Ef , and we have Mv = mu, so we can ndv:

v =

2U

M(1 +M/m)

The total velocity of M is now v = v0 +v:

v =2gh+

2U

M(1 +M/m)

(c.) We can easily nd the kinetic energy ofthe remaining rocket, and, using energy conser-vation, the maximum height H to which it risesabove its current position:

K =12M(v0 +v)2 =MgH

Using v0 =2gh, we can solve for H:

H =2gh+v

2gh+v2

2g

3Plugging in the result for v, we arrive at thenal answer

H = h+U

Mg(1 +M/m)+

hU

Mg(1 +M/m)

This is an interesting result. The rst termjust gets the rocket back to the height where itstarted in the rst place. The second term getsit to the maximum height of the rocket in part(a.). The fact that the third term is positivemeans that the rocket actually ies higher in thiscase. The gain in height is just the third term

H =

hU

Mg(1 +M/m)

(d.) This result does not conict with energyconservation, which says only that the total en-ergy of the system is conserved. We have beenneglecting what happens to the mass m, whichwill take away a smaller amount of energy in thesecond case. If we looked at the total energy ofboth pieces, it would be conserved.

3. K&K problem 4.7. This problem is one inwhich both force and energy need to be consid-ered. The forces on the ring are gravity, thetension in the thread T , and the normal forcesdue to the beads. The forces on the ring in thevertical direction are

Fring = T Mg 2N() cos where is the angle of the beads position fromthe top, and N() is taken to be positive out-ward. The two beads will move symmetrically.We now need to nd the normal force N().First we determine the velocity of the bead fromconservation of energy. It yields the following:

E/2 = mgL(cos 1) + 12mv2() = 0

This gives the velocity, and thus the centripetalacceleration ac, as a function of :

v2() = 2gL(1 cos ) ac = 2g(1 cos )The centripetal acceleration is provided by grav-ity and the normal force. Since a positive normal

force is outward, at the top the normal force willbe negative. The radial equation of motion willbe

mac = N +mg cos N() = mg(2 3 cos )

Now we can go back to the force equation for thering and use this result. The total force on thering will be zero, but the ring will just start tomove upwards when the thread is slack, which iswhen T = 0. Using both of these facts, we getan equation for

Mg = 2mg cos (3 cos 2)

This is just a quadratic equation in cos . Multi-plying it out, we can apply the quadratic formulato get the answer

cos =13 13

1 3M

2m

There is a small problem here in that the dis-criminant can be negative, making the cosine ofthe angle complex. This of course is unphysical.The problem is that for suciently small m, themotion of the small masses is never importantenough to cause the tension in the rope to vanish,so our calculation is wrong from the start. In-sisting that cos be real, we obtain the condition

m >32M

Taking the positive root, the nal answer is

= cos1(13+13

1 3M

2m

)

4. We assume that the moon is a uniform sphereof mass M = 7.3 1022 kg and radius R = 1740km. A straight, frictionless tunnel connects twopoints on the surface. Given the mass andradius, the density is just = 3M/4R3. Weneed to know the acceleration due to gravity ata distance r from the center of the moon. This isalso straightforward. Recall that a spherical shellof mass exerts no force on objects inside it, so ata radius r, the only force we need to consider is

4due to the mass in the moon interior to radius r.This is just the density times the volume interiorto r, or M(r) = Mr3/R3. The accelerationdue to gravity is then just g(r) = GM(r)/r2 =GMr/R3. Thus the acceleration due to gravityincreases linearly as one moves away from thecenter of a uniform solid sphere.

(a.) In a spherical polar coordinate system withits z axis at the moons north pole, assumethat the tunnel lies in a straight line between(r, 0, ) = (R, 0, 0) and (R, 0, ), i.e. betweentwo points at the same north latitude /2 0having the largest possible dierence in longi-tude. This means that the distance along a greatcircle between the ends of the tunnel is 20, whilethe distance from the center of the moon to thecenter of the tunnel is z0 = R cos 0. Now assumethat the mass makes an angle (2 < < 2 )with a line connecting the center of the moonand the center of the tunnel, i.e. with the z axis.The distance of the mass from the center of thetunnel is then x = z0 tan, while its distancefrom the center of the moon is r = z0/ cos. Wenow need to know the component Fx of the grav-itational force GMmr/R3 which lies in the (x)direction of the tunnel, which makes an angle/2 with the radial direction. This is

Fx = GMmrR3

cos (/2 )

= GMmR3

z0cos

sin

= GMmR3

x

This is like the force from a Hookes law springwith eective spring constant ke = GMm/R3,yielding simple harmonic oscillation with reso-nant angular frequency

0 =

kem

=

GM

R3

(b.) Plugging in values of M and R for themoon, and using T = 2/0, we get for theperiod of oscillation

T = 6536 seconds = 109 minutes

(c.) A satellite traveling in a circular orbitmust have centripetal acceleration provided bygravity, which means that

v2

R=GM

R2= 2R

From the last equality we see that the angu-lar frequency of a circular orbit of radius Raround the moon is the same as 0 above. Ofcourse the period is the same as well.

5. K&K problem 4.23. Two balls of masses Mand m are dropped from height h and collideelastically. The small ball is on top of the largerball. Conservation of energy for the system givesits speed v right before the balls hit the ground:

(M +m)gh =12(M +m)v2 v =

2gh

The ball M collides with the ground rst. Inorder to conserve energy, it must still havespeed v instantaneously after it bounces fromthe ground. Now it immediately collides withthe small ball. (Think of this problem as if therewere a very small gap between the two balls sothat the rst ball to hit the ground has a chanceto bounce before the second one hits it.) We con-sider the elastic collision between the two balls,each moving at speed v towards the other.

The easiest frame in which to study thiscollision is a comoving (inertial) frame that isinstantaneously at rest with respect to the largeball M immediately after it has rebounded withvelocity v =

2gh from its elastic collision with

the ground. In this frame, M is instantaneouslyat rest, and m has (upward) velocity 2v. Whenthe collision occurs, if m M as stated in theproblem, M seems to m like a brick wall fromwhich it bounces back elastically with the samespeed. Thus, in the comoving frame immediatelyafter the collision, m has velocity +2v. Finally,transforming back to the lab frame, m acquiresan extra velocity increment v, for a total of 3v.Since the height that m reaches is proportionalto the square of its velocity, this means thatm reaches nine times the height from which itoriginally was dropped.

5A less elegant approach considers the colli-sion between M and m in the lab frame. Hereis it essential not to apply the approximationm M until near the end, since cancellationsoccur which may make nonleading terms moreimportant than would initially be suspected.

In the lab frame, conservation of momentumgives

Mv mv =MVM +mVmWe also have conservation of energy through thiscollision. This condition gives

12(M +m)v2 =

12MV 2M +

12mV 2m

These are two equations in the two unknownsVm and VM , since we already know v =

2gh.

We are interested in Vm, which yields the de-sired nal height V 2m/2g of m, but we are notinterested in VM . So we plan to eliminate VM bysolving for it using the rst equation and thensubstituting for it in the second.

Before proceeding with this algebra, it isconvenient to substitute

$ = m/Mu = Vm/vU = VM/v

so that all terms are dimensionless. The twoequations above become

1 $ = U + $u1 + $ = U2 + $u2

Solving the rst equation for U ,

U = 1 $ $u

Substituting this value for U in the second equa-tion,

1 + $ = 1 2$+ $2 2$u+ 2$2u+ $2u2 + $u20 = $(1 + $)u2 2$(1 $)u $(3 $)0 = u2 21 $

1 + $u 3 $

1 + $

Neglecting $ with respect to 3 or 1 in bothquotients, the polynomial is

u2 2u 3 = (u 3)(u+ 1)

with the physical solution

u = 3

Vm = 3v =18gh

h =V 2m2g

= 9h

as before.

6. This is a collision problem that has dierentunknown quantities than those to which you areaccustomed, but it is still solvable. We have twocollisions to study, and the unknowns are theneutron mass and the initial and nal speeds ofthe neutrons. The initial speeds are the same, sothere are four unknowns in total. We have twocollisions, each of which yields two equations(one for momentum conservation, one for en-ergy conservation since the collisions are elastic).Therefore the system can be solved uniquely.The directions of the scattered neutrons rela-tive to the incident directions do not representadditional unknowns, since the maximum recoilvelocities of the target nucleii will occur whenthe collisions take place head-on, with the incom-ing neutrons bouncing straight back. Thus wecan take this to be a one dimensional problem.

The equations are the following (the energyequations have been multiplied by 2):

mnv = mnv +mHvH mnv2 = mnv2 +mHv2H

mnv = mnv +mNvN mnv2 = mnv2 +mNv2N

Solving these equations for mn and v requirescareful algebra. We square the rst momentumequation to get a relation between v, mn, and v

v2 =(mnv mHvH)2

m2n

Now we plug this into the rst energy equation

mnv2 =

(mnv mHvH)2mn

+mHv2H

6Expanding,

m2Hv2H 2mnmHvvH +mnmHv2H = 0

Writing this as an equation for v, we get

v =12

(1 +

mHmn

)vH

This is fairly simple result. If we perform thesame manipulations on the nitrogen equations,we will get an analogous result

v =12

(1 +

mNmn

)vN

We can now use these to solve for mn and v.Equating the right hand sides, we get a singleequation for the mass.

mnvH +mHvH = mnvN +mNvN

mn =mNvN mHvH

vH vN .

We can now use this to nd the initial velocityof the neutrons:

v =vH2

(1 +

mH(vH vN )mNvN mHvH

)=vH2

(mNvN mHvNmNvN mHvH

)=vHvN2

( mN mHmNvN mHvH

).

We want to know the mass of the neutron inamu, so we plug in mH = 1 and mN = 14(greater accuracy is unnecessary, since the recoilvelocities are measured only to 10%). We alsolook at both boundaries of the nitrogen veloc-ity, calling these results m and v. Pluggingin numbers, the values of mn are

mn = 1.159 amum+ = 1.415 amum = 0.911 amu .

Chadwicks experimental work is seen to be re-liable; todays accepted value for the neutronmass is 1.008665 amu, or 938.27231 0.00028

MeV/c2, well within his experimental range.The range of initial neutron velocity is given by

v = 3.07 107 m/secv+ = 2.82 107 m/secv = 4.13 107 m/sec .

7. K&K problem 4.13. The Lennard-Jonespotential is given by

U = $[(r0

r

)12 2

(r0r

)6]

(a.) We nd the minimum of this potential bydierentiating it with respect to r and settingthe results equal to zero:

dU

dr= 12$

r

[(r0r

)12(r0r

)6]= 0

This is easy to solve:(r0r

)12=(r0r

)6 r = r0

The depth of the potential well is just U(r0) =$. Thus the potential well has a depth $.(b.) We nd the frequency of small oscillationsby making a Taylor expansion of the potentialabout r = r0. Read section 4.10 in K&K formore information on this. We can write thepotential as follows:

U(r) = U(r0) +(dU

dr

)r=r0

(r r0)

+12

(d2U

dr2

)r=r0

(r r0)2 +

We know that dU/dr = 0 at r = r0, so we dropthe middle term.

U(r) $+ 12

(d2U

dr2

)r=r0

(r r0)2

This is exactly the form of the potential of a masson a spring. We only have to identify the springconstant. Remembering that Uspring = kx2/2,we make the identication

k =(d2U

dr2

)r=r0

7For the Lennard-Jones potential, we alreadyknow the rst derivative, so we need to dieren-tiate once more.

d2U

dr2=

12$r2

[13(r0r

)12(7r0r

)6]

Plugging in r = r0, we nd the eective springconstant for this potential

k =72$r20

We now consider two identical masses m on theends of this spring. Their (coupled) equationsof motion are:

mr1 = k(r r0) mr2 = k(r r0)

where r = r2 r1 is the distance between themasses. Subtracting these two equations, we get

mr = 2k(r r0)

The frequency of oscillation is then 2 = 2k/m.(Note that we could have obtained the sameresult by considering the two-mass system tobe a single mass of reduced mass mreduced =m1m2/(m1 +m2)). Plugging in the above valuefor the eective spring constant k,

= 12

$

r20m

University of California, BerkeleyPhysics H7A Fall 1998 (Strovink)

PROBLEM SET 6

1. K&K problem 6.1 Show that if the totallinear momentum....

2. K&K problem 6.3 A ring of mass M andradius R lies....

3. Expansion of the previous problem:

(a.) Let the azimuth of the bug on the ring be that is, is zero when the bug startswalking, and 360 when the bug makes onerevolution. Assume (for part (a.) only) thatthe ring is xed. Calculate the angular mo-mentum l of the bug about the pivot in theprevious problem, as a function of . Checkthat your result is consistent with what youused in the previous problem when was180.

(b.) Now assume that the bug is xed at someazimuth on the ring, but that the ringitself is not xed, having angular velocity about the pivot (opposite to the angularvelocity of the bug when the bug was mov-ing). Calculate the angular momentum l ofthe bug as a function of and .

(c.) Now assume that neither the bug nor thering are xed. By requiring that the totalangular momentum l + l of the bug aboutthe pivot be balanced by the angular mo-mentum of the ring, obtain an expressionfor the angular velocity of the ring, as afunction of .

(d.) Get an integral expression for the angle through which the ring rotates, as a functionof time, assuming that d/dt = = con-stant. You need not evaluate the integral.Note that the system is bootstrapping itsway around the pivot!

4. K&K problem 6.5 A 3,000-lb car is parkedon a....

5. A man begins to climb up a 12-ft ladder (seegure). The man weighs 180 lb, and the ladder20 lb. The wall against which the ladder restsis very smooth, which means that the tangential

(vertical) component of force at the contact be-tween ladder and wall is negligible. The foot ofthe ladder is placed 6 ft from the wall. The lad-der, with the mans weight on it, will slip if thetangential (horizontal) force at the contact be-tween the ladder and ground exceeds 80 lb. Howfar up the ladder can the man safely climb?

6. K&K problem 6.8 Find the moment ofinertia of a uniform sphere....

7. K&K problem 6.14 A uniform stick of massM and length L is....

8. K&K problem 6.18 Find the period of apendulum....

1University of California, BerkeleyPhysics H7A Fall 1998 (Strovink)

SOLUTION TO PROBLEM SET 6Composed and formatted by E.A. Baltz and M. Strovink; proofed by D. Bacon

1. K&K problem 6.1

(a.) We know that the total linear momentumof the system is zero. (This would occur, forexample, if we were in the center of mass frame.)

P =i

pi = 0

Examine the total angular momentum of thesystem. The vector from the origin to point i isdenoted ri. The angular momentum in generaldepends on where the origin is:

L =i

ri pi

We now want to nd the angular momentumabout a new origin whose position vector is Rin the current coordinate system. In this newsystem, the position vector of point i becomesri R. Each point has its position changed bythe same amount. The new value of the angularmomentum is

Lnew =i

(ri R) pi

Expanding,

Lnew =i

ri pi i

R pi

Because R is the same for all points, we can pullit outside of the sum:

Lnew =i

ri pi Ri

pi

=i

ri pi RP

We know that P = 0, so we are done.

Lnew = L

(b.) The proof for this part is identical if angu-lar momentum is replaced by torque and linearmomentum is replaced by force.

2. K&K problem 6.3

This problem and the next concern the samesystem that of a bug walking along a hoopthat is free to pivot around a point on its edge.The hoop lies at on a frictionless surface. Thering has mass M and radius R, and the bug hasmass m and walks on the ring with speed v.

The key idea in this problem is conservationof angular momentum. About the pivot there isno net torque on the system, so the total angularmomentum about that point is conserved. Thering starts at rest with the bug on the pivot, andthe bug starts walking at speed v. Immediatelyafter the bug starts walking, the total angularmomentum measured about the pivot point con-tinues to be zero. The ring is not yet moving, soit has no angular momentum; the bug has begunto move, but it is at r = 0, so it has no angularmomentum yet.

We want to nd the angular velocity ofthe ring when the bug is opposite to the pivot.The bug is moving at speed v on the ring, butthe ring is also moving. The bug is at a distance2R from the pivot, so the velocity of that por-tion of the ring which is under the feet of the

2bug is 2R. The total velocity of the bug is thusv + 2R. Next we need to know the moment ofinertia of the hoop. A hoop has moment of in-ertia I = MR2 about its center of mass. We usethe parallel axis theorem to nd the moment ofinertia about a point on the edge.

I = ICM +Md2

The distance d from the center of mass to thedesired axis in this case is just R, so the momentof inertia of the hoop about a point on the edgeis I = 2MR2. We can now nd an expression forthe total angular momentum of the system. Forthe hoop we use L = I and for the bug we useL = mvr sin . The angle between the positionvector and the velocity vector of the bug in thiscase is simply /2, so sin is just 1. We nowwrite the angular momenta of the two pieces

Lbug = 2mR(v + 2R) Lhoop = 2MR2

Since angular momentum about the pivot is con-served throughout the motion, We know thatLbug + Lhoop = 0. This gives the following ex-pression:

2mvR+ 4mR2 + 2MR2 = 0

We solve this equation for in terms of v and get

= mvMR+ 2mR

Note that the minus sign means that the hooprotates in a direction opposite to that in whichthe bug moves. This makes sense because thetotal angular momentum about the pivot pointmust vanish.

3. We now study the bug and hoop system inmore detail. See the diagram in the previousproblem.

(a.) In this part we assume that the ring is xed.We want to calculate the angular momentum ofthe bug about the pivot point. The rst step isto nd the distance r between the bug and thepivot. We can do this using the law of cosines.Consider the triangle made by the line betweenthe bug and the pivot, and the two radial lines

extending from the center of the hoop to the bugand pivot, respectively. This is an isosceles tri-angle, with two equal sides of length R havingan angle between them. If we dene the az-imuth of the bug on the hoop to be zero at thepivot, the angle is simply the azimuth of thebug on the hoop. The length r of the third sideis found using the law of cosines:

r2 = R2+R22R2 cos r2 = 2R2(1cos)Using the trigonometric identity 1 cos =2 sin2 2 , we can get a simple result for r:

r = 2R sin

2

Now we know v and r. The only thing left todetermine is the angle between the position andvelocity vectors. The rst step is to nd the an-gle between the position vector r of the bug andthe line from the center of the circle to the bug.The isosceles triangle (like any triangle) has atotal angle , and its central angle is . The re-maining two angles are equal, so they must be( )/2 each. Thus the three angles add up to radians. Because the velocity vector v is tan-gent to the circle, the angle between r and v is/2 minus this angle. Thus the angle between rand v is /2. We can now get the angular mo-mentum of the bug. Assuming that the ring isxed, l = mvr sin from the bug alone, so

l = 2mvR sin2

2

(b.) In this part we assume that the ring is ro-tating with angular velocity , but that the bugis xed on the ring. The velocity of the bug isjust r, where r is the same as was calculated inthe previous part. The velocity in this case is al-ways perpendicular to its position vector. Thiscan be seen by remembering that the bug isntmoving on the ring, so it must be in uniformcircular motion about the pivot, with a velocitythat is tangent to its present position. Thereforethe angular momentum l of the bug is simplymvr, yielding

l = mr2 = 4mR2 sin2

2

3(c.) We now allow both the bug and the ringto move. The total angular momentum of thebug is l+ l from parts (a.) and (b.) respectively.To this we must add the angular momentum ofthe ring to get the total angular momentum ofthe system. From problem 2. we know that thetotal angular momentum must be zero. The an-gular momentum of the ring is I, so we get thefollowing equation.

4mR2 sin2

2+ 2mvR sin2

2+ 2MR2 = 0

We solve this for in terms of . The result is

= mv sin2 2

MR+ 2mR sin2 2

This agrees with the result of problem 2. whenthe bug is at = , opposite to the pivot.

(d.) Finally, we want to nd an expression forthe angle through which the ring rotates. Weknow that is related by a simple dierentialequation to the angular velocity of the hoop

=d

dt

but we want to express in terms of so thatwe can use the fact that Rd/dt, the speed v ofthe bug with respect to the rim of the hoop, isconstant. We apply the chain rule to get

=d

dt=

d

d

d

dt

Substituting d/dt = v/R, where v is constant,we can write an integral for :

= Rv

o

mv sin2(/2)MR+ 2mR sin2(/2)

d

We can simplify this a little, but doing the inte-gral is hard, which is why you werent asked toevaluate it. Setting the initial bug azimuth 0to zero and using the fact that d/dt = v/R is aconstant so that = vt/R,

(t) = vt/R0

sin2(/2)(M/m) + 2 sin2(/2)

d

4. K&K problem 6.5

A car of mass m is parked on a slope of angle facing uphill. The center of mass is a distance dabove the ground, and it is centered between thewheels, which are a distance l apart. We wantto nd the normal force exerted by the road onthe front and rear tires.

It is easiest to do this problem choosing theorigin as the point on the road directly below thecenter of mass. About this point there are threetorques. The normal force on the front (Nf ) andrear (Nr) set of wheels provides a torque, andalso gravity provides a torque mgd sin becausethe car isnt horizontal. However, the forces offriction on the tires dont provide any torque be-cause they are in line with the direction to theorigin. The torque from the front wheels and thetorque due to gravity tend to want to ip thecar over backwards, while the torque on the rearwheels opposes this tendency. We want the sumof the torques to vanish, because the (static) caris not undergoing any acceleration, angular orlinear:

0 = Nfl

2+mgd sin Nr l2

Nr Nf = 2mgdl

sin

We can get one more condition from the factthat the car is not undergoing linear accelera-tion perpendicular to the road. This means thatthe normal forces exactly cancel gravity:

Nr +Nf = mg cos

We can take the sum and dierence of thesetwo equations to get expressions for Nf and Nr.These are

Nr = mg(12cos +

d

lsin

)Nf = mg

(12cos d

lsin

)

Plugging in = 30, mg = 3000 lb, d = 2 ft, andl = 8 ft, we get Nr =1674 lb and Nf =924 lb.

45. We will solve this problem symbolically andwait until the end to plug in numbers. Thisis always good practice because it makes it alot easier to check the units of the result and toexplore whether the result is reasonable when theinputs have limiting values. We takeM to be themass of the man (Mg = 180 lb) and m to be themass of the ladder (mg = 20 lb). The length Hof the ladder is 12 ft, and its point of contact withthe wall is d = 6 ft from the wall. The angle thatit makes with the wall is = arcsin (d/H) = 30.Finally, the force of friction on the ladder fromthe ground is Ff Fmaxf , where Fmaxf = 80 lb.

There are ve forces to consider in thisproblem. They are the two normal forces onthe ladder, Ng from the ground and Nw fromthe wall; the force Ff of friction at the base ofthe ladder; and the two forces of gravity, Mgon the man and mg on the ladder. This isa torque balance problem, so choosing a goodorigin makes it a lot easier. With this choice ofthe point of contact with ground, two of the veforces contribute no torque about that point.Not bad! As a sanity check we evaluate , thecoecient of friction between the ladder and theground. The normal force Nf from the oor isequal and opposite to (M +m)g, the sum of theweights of the ladder and the man. We are giventhe maximum frictional force Fmaxf , and we knowthat Fmaxf = N , so = F

maxf /((M +m)g) =

80/200 = 0.4, a reasonable value.

We now calculate the torques. To nd themaximum height h to which the man can climbwithout the ladder slipping, we assume that theladder is about to slip. This means that the nor-mal force Nw from the wall is equal and oppositeto Fmaxf , exactly countering the maximum forceof friction: since these two forces are the onlyforces in the horizontal direction they must sumto zero. The torque from the wall is then w =Fmaxf H cos , where the minus sign indicatesthat this torque pushes clockwise. The torquefrom the weight of the ladder is exerted at themidpoint of the ladder, its center of mass. Thevalue of this torque is m = mgH2 sin . Similarly,the torque exerted by the weight of the man, whois a distance h up the ladder, is M = Mgh sin .

Requiring these three torques to sum to zero,

0 = M + m + w

= Mgh sin +mgH

2sin Fmaxf H cos

Solving for h,

h =Fmaxf H cos mgH2 sin

Mg sin h

H=

FmaxfMg

cot m2M

=(M +m)

Mcot m

2M

=(M +m) cot (m/2)

M

=0.4(200)

3 10

180= 0.7142

h = 8.571 ft

6. K&K problem 6.8

Because of the spherical symmetry, we work inspherical polar coordinates. To nd the momentof inertia we need to evaluate the integral

I =

r2dv

where in these coordinates r = r sin is theperpendicular distance to the axis and dv =r2dr d(cos ) d is the element of volume. Theintegral to evaluate is thus

I = M

R0r4 dr

11 sin

2 d(cos ) 20

d R0r2 dr

11 d(cos )

20

d

where the denominator is the volume V of thesphere, needed to evaluate its density = M/V .Substituting u r/R,

I

MR2=

10u4 du

11 sin

2 d(cos ) 20

d 10u2 du

11 d(cos )

20

d

In both the numerator and the denominator, allthree integrals have limits that do not dependon the other variables, so each integral can be

5evaluated independently. The integrals cancel,and the u integrals have the ratio

1/51/3

=35

The integrand in the cos integral in the numer-ator can be rewritten

sin2 d(cos ) = (1 cos2 )d(cos )= d(cos ) d

(13cos3

)Therefore the ratio of the integrals is

2 232

=23

Putting it all together,

I = MR2 35 2

3=

25MR2

7. K&K problem 6.14

When the stick is released, there are twoforces acting on it, gravity at the midpoint, andthe normal force at the point B. We use thepoint B as the origin, so the only torque aboutthis point is pro