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SOLUTIONS MANUALIntroduction to Classical MechanicsWith Problems and SolutionsDavid MorinCambridge University Press

TO THE INSTRUCTOR: I have tried to pay as much attention to detail inthese exercise solutions as I did in the problem solutions in the text. Butdespite working through each solution numerous times during the variousstages of completion, there are bound to be errors. So please let me know ifanything looks amiss.Also, to keep this pdf file from escaping to the web, PLEASE don’t distributeit to anyone, with the exception of your teaching assistants. And please makesure they also agree to this. Once this file gets free, there’s no going back.In addition to any comments you have on these solutions, I welcome anycomments on the book in general. I hope you’re enjoying using it!David [email protected](Version 2, April 2008)c David Morin 2008

Chapter 1Strategies for solvingproblems1.8. Pendulum on the moonThep only way to get units of time from , g, and m is through the combination /g. Therefore,p /gMTM p TE /gErgEgM TM 6 TE 7.3 s.(1)1.9. Escape velocity(a) Using M ρV , we haverv p2G · (4/3)πR3 ρ (8/3)πGR2 ρ.R (b) We see that v R ρ. Therefore, RJ ρJvJ1 11 · 5.5. vERE ρE4(2)(3)1.10. Downhill projectileThe angle β is some function of the form, β f (θ, m, v0 , g). In terms of units, wecan write 1 f (1, kg, m/s, m/s2 ). We can’t have any m dependence, because thereis nothing to cancel the kg. And we also can’t have any v0 or g dependence, becausethey would have to appear in the ratio v0 /g to cancel the meters, but then secondswould remain. Therefore, β can depend on at most θ. (And it clearly does dependon θ, because β 90 for θ 0 or 90 , but β 6 90 for θ 6 0 or 90 .)1.11. Waves on a stringThe speed v is some function of the form, v f (M, L, T ). In terms of units, we canwrite m/s f (kg, m, kg m/s2 ). We needp to get rid of the kg’s, so we must use theratio T /M . We then quickly see that LT /M has the correct units of m/s. Notepthat this can also be written as T /ρ, where ρ is the mass density per unit length.1.12. Vibrating water dropThe frequency ν is some function of the form, ν f (R, ρ, S). In terms of units, wecan write 1/s f (m, kg/m3 , kg/s2 ). Wepneed to get rid of the kg’s, so we must usethe ratio S/ρ. We then quickly see that S/ρR3 has the correct units of 1/s. Notepthat this can also be written as S/M , where M is the mass of the water drop.1

2CHAPTER 1. STRATEGIES FOR SOLVING PROBLEMS1.13. Atwood’s machine(a) This gives a1 0. (Half of m2 balances each of m1 and m3 .)(b) Ignore the m2 m3 terms, which gives a1 g. (Simply in freefall.)(c) Ignore the terms involving m1 , which gives a1 3g. (m2 and m3 are in freefall.And for every meter they go down, a total of three meters of string appearsabove them, so m1 goes up three meters.)(d) Ignore the m1 m3 terms, which gives a1 g. (m2 goes down at g, and m1 andm3 go up at g.)(e) This gives a1 g/3. (Not obvious.)1.14. Cone frustumThe correct answer must reduce to the volume of a cylinder, πa2 h, when a b. Onlythe 2nd, 3rd, and 5th options satisfy this. The correct answer must also reduce tothe volume of a cone, πb2 h/3, when a 0. Only the 1st, 3rd, and 4th options satisfythis. The correct answer must therefore be the 3rd one, πh(a2 ab b2 )/3.1.15. Landing at the cornerThe correct answer must go to infinity for θ 90 . Only the 2nd and 3rd optionssatisfy this. The correct answer must also go to infinity for θ 45 . Only the 1stand 2nd options satisfy this. The correct answer must therefore be the 2nd one.1.16. Projectile with dragUsing the Taylor series for e αt , we have³y(t) ³ 1ggtv0 sin θ 1 (1 αt α2 t2 /2 · · ·) ααα³ ³ ggt2v0 sin θ t αt /2 αα1 2 1 2(v0 sin θ)t gt αt v0 sin θ.22(4)If α ¿ g/(v0 sin θ), then the third term is much smaller than the second, and weobtain the desired result. So α ¿ g/(v0 sin θ) is what we mean by “small α.”However, we also assumed αt ¿ 1 in the expansion for e αt above, so we shouldcheck that this doesn’t necessitate a stricter upper bound on α. And indeed, the totaltime of flight is less than 2v0 sin θ/g (because this t makes the above y(t) negative),so the condition α ¿ g/(v0 sin θ) implies αt ¿ (g/v0 sin θ)(2v0 sin θ/g) 2. Soαt ¿ 1 is guaranteed by α ¿ g/(v0 sin θ).1.17. PendulumHere is a Maple program that does the job:q: 3.14159/2:q1: 0:e: .0001:i: 0:while q 0 doi: i 1:q2: -(9.8)*sin(q)/1:q: q e*q1:q1: q1 e*q2:end do:i*e;###########initial θ valueinitial θ speeda small time intervali will count the number of time stepsrun the program while θ 0increase the counter by 1the given equationhow q changes, by definition of q1how q1 changes, by definition of q2the Maple command to stop the do loopprint the value of the timeThis yields a time of t 0.5923 s. If we instead use a time interval of .00001 s, weobtain t 0.59227 s. And a time interval of .000001 s gives t 0.592263 s.

31.18. Distance with dampingIn the ẍ Aẋ case, we have the following Maple program:x: 0:x1: 2:T: 1:e: .001:for i to T/e dox2: -(1)*x1:x: x e*x1:x1: x1 e*x2:end do:x;##########initial x valueinitial x speedthe total timea small time intervalrun the program for a time Tthe given equationhow x changes, by definition of x1how x1 changes, by definition of x2the Maple command to stop the do loopprint the value of the positionTo run the program for different times, we simply need to change the value of T inthe 3rd line. Letting T equal 1 gives a final position of 1.264. Letting T equal 10 and100 gives final positions of 1.99991 and 1.9999996, respectively. These approach 2.In the ẍ Aẋ2 case, the only change in the entire program is in the 6th line,where we now have the square of x1:x2: -(1)*x1 2:# the given equationLetting T equal 1, 10, 100, 1000, and 10000, gives final positions of 1.099, 3.044,5.302, 7.600, and 9.903, respectively. Looking at the successive differences betweenthese values, we see that they approach roughly 2.3. This constant difference forinputs of powers of 10 implies a log dependence on the time.

4CHAPTER 1. STRATEGIES FOR SOLVING PROBLEMS

Chapter 2Statics2.20. Block under an overhangLet’s break up the forces into components parallel and perpendicular to the overhang. Let positive Ff point up along the overhang. Balancing the forces paralleland perpendicular to the overhang gives, respectively,Ff M g sin β M g cos β,N M g sin β M g cos β.and(5)N must be positive, so we immediately see that β must be at least 45 if there isany chance that the setup is static.The coefficient µ tells us that Ff µN . Using Eq. (5), this inequality becomesM g(sin β cos β) µM g(sin β cos β) µ 1 tan β.µ 1(6)We see that we must have µ 1 in order for there to exist any values of β thatsatisfy this inequality. If µ , then β can be as small as 45 , but it can’t be anysmaller.2.21. Pulling a blockThe Fy forces tell us that N F sin θ mg 0 N mg F sin θ. Andassuming that the block slips, the Fx forces tell us that F cos θ µN . Therefore,F cos θ µ(mg F sin θ) F µmg.cos θ µ sin θ(7)Taking the derivative topminimize this then gives tan θ µ. Plugging this θ backinto F gives F µmg/ 1 µ2 . If µ 0, we have θ 0 and F 0. If µ , wehave θ 90 and F mg.2.22. Holding a coneLet F be the friction force at each finger. Then the Fy forces on the cone tell usthat 2F cos θ 2N sin θ mg 0. But F µN . Therefore,2µN cos θ 2N sin θ mg 0 N mg.2(µ cos θ sin θ)(8)This is the desired minimum normal force. When µ tan θ, we have N . Soµ tan θ is the minimum allowable value of µ.2.23. Keeping a book upThe result of Problem 2.4 is F mg/(sin θ µ cos θ), assuming that sin θ µ cos θis positive (that is, tan θ µ). If it is negative, there is no solution for F . To findthe maximum force, consider two cases:5

6CHAPTER 2. STATICS(a) Your force is directed upward (θ 0): Then Ff points downward in the maximal F case. So Fy gives F sin θ Ff mg 0 Ff F sin θ mg. ButFf µN µ(F cos θ), so we havemgF sin θ mg µF cos θ F ,(9)sin θ µ cos θassuming that sin θ µ cos θ is positive (that is, tan θ µ). If it is negative,then F (sin θ µ cos θ) mg is true for any F , so there is no upper bound inthis case.(b) Your force is directed downward (θ 0): Then Ff points upward in the maximal F case. So Fy gives (note that sin θ is negative here) F sin θ Ff mg 0 Ff F sin θ mg. But Ff µN µ(F cos θ), so we havemg,(10) F sin θ mg µF cos θ F sin θ µ cos θassuming that sin θ µ cos θ is positive (that is, tan θ µ). If it is negative,then F (sin θ µ cos θ) mg is never true, so there is no solution for F . This isthe same result as in Problem 2.4, so it doesn’t actually yield an upper boundon F .Putting all this together (along with the results from Problem 2.4): As a functionof θ, and for a generic value of µ less than 1, the values of F that keep the book upare signified by the shaded region in Fig. 1.Fmgsinθ µcosθF mgsinθ µcosθmg1 µ2F mgF mg/µθ π/2tanθ -µtanθ µtanθ 1/µπ/2Figure 12.24. Bridges(a) Looking at the Fx forces on the car, we see that the two inner diagonal beamsmust have equal tensions.Then Fy with these two beams tells us that the tensions are each mg/ 3. Then Fy on one of the the upper (massless) hingesgives the compression in the outer diagonal beams as mg/ 3. Then Fx withone of the upper hinges gives the tension in the top beam as mg/ 3.(b) Using Fy , we can start in the middle and work our way out along the diagonalbeams to show that they all have equal forces of mg/ 3, alternating tensionand compression. We can then work our way back in along the top beams(using Fx at the hinges) to show that the outer ones have mg/ 3 compression,and the middle one has 2mg/ 3 compression. Likewise, the outer bottombeams have mg/2 3 tension, and the inner bottom beams have 3mg/2 3tension.

7(c) This is similar to part (b) for the diagonal beams; they all have equal forces ofmg/ 3, alternating tension and compression. For the top beams, starting atthe outsidein (using Fx at the hinges), they have compressions and working of mg/ 3, 2mg/ 3, 3mg/ 3, and so on. For the bottom beams, startingat the outside andin (using Fx at the hinges), they have tensions of working mg/2 3, 3mg/2 3, 5mg/2 3, and so on.2.25. Rope between inclinesLet x be the length in contact with one of the platforms, and let be half thelength of the rope. The normal force on the x part is N ρxg cos θ, and so thefriction force satisfies Ff ρxg cos θ. Balancing the vertical forces on one half of therope gives N cos θ Ff sin θ ρ g. Using the above values of N and Ff , this givesρxg(cos2 θ sin θ cos θ) ρ g. The minimum x occurs when the function of θ hereis maximum. Setting the derivative equal to zero yields tan 2θ 1, so θ 22.5 .Plugging this back in and simplifying (using cos2 θ (1 cos 2θ)/2 and sin θ cos θ (sin 2θ)/2) gives the desired maximum fraction as ( x)/ 3 2 2 0.172. Notethat the setup isn’t possible if θ 45 , because the above inequality gives x ,which is by definition not allowed.2.26. Hanging chain(a) Let F and T be the tensions at the wall and the lowest point, respectively.Looking at the y forces on half of the chain gives F cos θ (M/2)g, andlooking at the x forces gives F sin θ T . These yield T (M/2)g tan θ.(b) The slope of the chain is y 0 sinh αx, which is approximately αx for smallx. Consider a small piece that goes from x to x. The weight is essentiallyρ(2x)g. The upward component of the tensions at the two ends is essentially2T y 0 2T (αx). Balancing these gives T ρg/α.Now let’s find α. The length from the bottom, as a function of a general valueof x, equalsZx0pZ1 y 02 dx xcosh αx (1/α) sinh αx.(11)0Therefore, M/2 (ρ/α) sinh αx0 , where x0 is the location of the wall. Butthe slope at the wall is sinh α0 x 1/ tan θ. So M/2 ρ/(α tan θ) ρ/α (M/2) tan θ. Plugging this into the above T gives T (M/2)g tan θ, in agreement with part (a).2.27. Gravitational torqueA small mass element is ρ(dx)g. So the torque around the end isρgL2 /2 (ρL)gL/2 M g(L/2), as desired.RL0ρ(dx)g · x 2.28. Linear functionLetting b a gives 2f (a) f (2a). Using this in the case where b 2a gives f (a) f (2a) f (3a) 3f (a) f (3a). Repeating this process yields the general result,n1 f (a1 ) f (n1 a1 ), for any number a1 and any integer n1 . Likewise, n2 f (a2 ) f (n2 a2 ) for any number a2 and any integer n2 .Given n1 , n2 , and a1 , choose a2 so that n1 a1 n2 a2 . Then n1 f (a1 ) n2 f (a2 ),and so (n1 /n2 )f (a1 ) f (a2 ) (n1 /n2 )f (a1 ) f (n1 /n2 · a1 ) for any number a1and any integers n1 and n2 . Equivalently, rf (x) f (rx) for any number x and anyrational number r, as desired.2.29. Direction of the force(1) Look at the torque around one end. If the stick is massless, then there is nogravitational force, so the only possible force providing a torque is the hinge at theother end. If this force doesn’t point along the stick, it will result in a nonzerotorque. This then implies a nonzero angular acceleration of the stick (infinite, in

8CHAPTER 2. STATICSfact, because the stick is massless), which contradicts the fact that the system isstatic.(2) If the stick is massive, there is now a torque from gravity (unless the stick ishanging vertically). This can cancel a nonzero torque from the hinge at the otherend.2.30. Ball on a wallLet T be the tension in the string. Then the friction force Ff from the wall mustalso be T , to provide zero net torque around the center. So if N is the normalforce from the wall, then balancing the x forces quickly gives N T sin θ. ButFf µN T µ(T sin θ) µ 1/ sin θ. Interestingly, this equals 1 forθ 90 .2.31. Cylinder and hanging massIf T is the tension in the string, then T mg. If F is the friction force from theplane, then balancing torques around the center of the cylinder gives F T , so Falso equals mg. If N is the normal force from the plane, then balancing horizontalforces on the cylinder gives N sin θ F cos θ N mg/ tan θ. Finally, balancingvertical forces on the cylinder gives³N cos θ F sin θ M g T 0 mgcos θ (mg) sin θ mg M gtan θ ³sin θM.(12)m 1 sin θIf θ 0, then m 0. And if θ 90 , then m . These make sense.Alternatively, once we know that T mg, we can just use torque around the contactpoint on the plane, which doesn’t require knowing F or N . The lever arm for theM g force is R sin θ, and the lever arm for the T force is R(1 sin θ). Balancing thetorques around the contact point therefore gives (M g)R sin θ (mg)R(1 sin θ), inagreement with the above result.2.32. Ladder on a cornerIf Nc is the normal force from the corner, then balancing torques around the top endof the ladder gives Nc (3L/4) M g(L/2) cos θ Nc (2/3)M g cos θ. And if Nwis the normal

SOLUTIONS MANUAL Introduction to Classical Mechanics With Problems and Solutions David Morin Cambridge University Press. ... then F(sinµ+„cosµ) ‚ mg is never true, so there is no solution for F. This is the same result as in Problem 2.4, so it doesn’t actually yield an upper bound