SOLUTIONS MANUAL Introduction to Classical Mechanics With Problems and Solutions David Morin Cambridge University Press TO THE INSTRUCTOR: I have tried to pay as much attention to detail in these exercise solutions as I did in the problem solutions in the text. But despite working through each solution numerous times during the various stages of completion, there are bound to be errors. So please let me know if anything looks amiss. Also, to keep this pdf ¯le from escaping to the web, PLEASE don't distribute it to anyone, with the exception of your teaching assistants. And please make sure they also agree to this. Once this ¯le gets free, there's no going back. In addition to any comments you have on these solutions, I welcome any comments on the book in general. I hope you're enjoying using it! David Morin [email protected] (Version 2, April 2008) °c David Morin 2008 Chapter 1 Strategies for solving problems 1.8. Pendulum on the moon pThe only way to get units of time from `, g, and m is through the combination `=g. Therefore, p r TM `=gM gE p = p = =) TM ¼ 6 TE ¼ 7:3 s: (1) TE `=gE gM 1.9. Escape velocity (a) Using M = ½V , we have r 2G ¢ (4=3)¼R3½ p v = = (8=3)¼GR2½: (2) R p (b) We see that v / R ½. Therefore, p v RJ ½J 1 J = p = 11 ¢ p = 5:5: (3) vE RE ½E 4 1.10. Downhill projectile The angle ¯ is some function of the form, ¯ = f(θ; m; v0; g). In terms of units, we can write 1 = f(1; kg; m=s; m=s2). We can't have any m dependence, because there is nothing to cancel the kg. And we also can't have any v0 or g dependence, because they would have to appear in the ratio v0=g to cancel the meters, but then seconds would remain. Therefore, ¯ can depend on at most θ. (And it clearly does depend on θ, because ¯ = 90± for θ = 0 or 90±, but ¯ 6= 90± for θ 6= 0 or 90±.) 1.11. Waves on a string The speed v is some function of the form, v = f(M; L; T ). In terms of units, we can 2 write m=s = f(kg; m; kg m=s ). We needp to get rid of the kg's, so we must use the ratio T=M. We then quickly seep that LT=M has the correct units of m=s. Note that this can also be written as T=½, where ½ is the mass density per unit length. 1.12. Vibrating water drop The frequency º is some function of the form, º = f(R; ½; S). In terms of units, we 3 2 can write 1=s = f(m; kg=m ; kg=s ). Wep need to get rid of the kg's, so we must use 3 the ratio S=½. We then quickly seep that S=½R has the correct units of 1=s. Note that this can also be written as S=M, where M is the mass of the water drop. 1 2 CHAPTER 1. STRATEGIES FOR SOLVING PROBLEMS 1.13. Atwood's machine (a) This gives a1 = 0. (Half of m2 balances each of m1 and m3.) (b) Ignore the m2m3 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 appears above them, so m1 goes up three meters.) (d) Ignore the m1m3 terms, which gives a1 = g.(m2 goes down at g, and m1 and m3 go up at g.) (e) This gives a1 = ¡g=3. (Not obvious.) 1.14. Cone frustum The correct answer must reduce to the volume of a cylinder, ¼a2h, when a = b. Only the 2nd, 3rd, and 5th options satisfy this. The correct answer must also reduce to the volume of a cone, ¼b2h=3, when a = 0. Only the 1st, 3rd, and 4th options satisfy this. The correct answer must therefore be the 3rd one, ¼h(a2 + ab + b2)=3. 1.15. Landing at the corner The correct answer must go to in¯nity for θ ! 90±. Only the 2nd and 3rd options satisfy this. The correct answer must also go to in¯nity for θ ! 45±. Only the 1st and 2nd options satisfy this. The correct answer must therefore be the 2nd one. 1.16. Projectile with drag Using the Taylor series for e¡®t, we have ³ ´ ³ ´ 1 g 2 2 gt y(t) = v0 sin θ + 1 ¡ (1 ¡ ®t + ® t =2 ¡ ¢ ¢ ¢) ¡ ³® ´® ³ ´ ® g gt ¼ v sin θ + t ¡ ®t2=2 ¡ 0 ® ® 1 1 = (v sin θ)t ¡ gt2 ¡ ®t2v sin θ: (4) 0 2 2 0 If ® ¿ g=(v0 sin θ), then the third term is much smaller than the second, and we obtain the desired result. So ® ¿ g=(v0 sin θ) is what we mean by \small ®." However, we also assumed ®t ¿ 1 in the expansion for e¡®t above, so we should check that this doesn't necessitate a stricter upper bound on ®. And indeed, the total time of flight is less than 2v0 sin θ=g (because this t makes the above y(t) negative), so the condition ® ¿ g=(v0 sin θ) implies ®t ¿ (g=v0 sin θ)(2v0 sin θ=g) = 2. So ®t ¿ 1 is guaranteed by ® ¿ g=(v0 sin θ). 1.17. Pendulum Here is a Maple program that does the job: q:=3.14159/2: # initial θ value q1:=0: # initial θ speed e:=.0001: # a small time interval i:=0: # i will count the number of time steps while q>0 do # run the program while θ > 0 i:=i+1: # increase the counter by 1 q2:=-(9.8)*sin(q)/1: # the given equation q:=q+e*q1: # how q changes, by definition of q1 q1:=q1+e*q2: # how q1 changes, by definition of q2 end do: # the Maple command to stop the do loop i*e; # print the value of the time This yields a time of t = 0:5923 s. If we instead use a time interval of .00001 s, we obtain t = 0:59227 s. And a time interval of .000001 s gives t = 0:592263 s. 3 1.18. Distance with damping In thex Ä = ¡Ax_ case, we have the following Maple program: x:=0: # initial x value x1:=2: # initial x speed T:=1: # the total time e:=.001: # a small time interval for i to T/e do # run the program for a time T x2:=-(1)*x1: # the given equation x:=x+e*x1: # how x changes, by definition of x1 x1:=x1+e*x2: # how x1 changes, by definition of x2 end do: # the Maple command to stop the do loop x; # print the value of the position To run the program for di®erent times, we simply need to change the value of T in the 3rd line. Letting T equal 1 gives a ¯nal position of 1.264. Letting T equal 10 and 100 gives ¯nal positions of 1.99991 and 1.9999996, respectively. These approach 2. In thex Ä = ¡Ax_ 2 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 equation Letting T equal 1, 10, 100, 1000, and 10000, gives ¯nal positions of 1.099, 3.044, 5.302, 7.600, and 9.903, respectively. Looking at the successive di®erences between these values, we see that they approach roughly 2.3. This constant di®erence for inputs of powers of 10 implies a log dependence on the time. 4 CHAPTER 1. STRATEGIES FOR SOLVING PROBLEMS Chapter 2 Statics 2.20. Block under an overhang Let's break up the forces into components parallel and perpendicular to the over- hang. Let positive Ff point up along the overhang. Balancing the forces parallel and perpendicular to the overhang gives, respectively, Ff = Mg sin ¯ + Mg cos ¯; and N = Mg sin ¯ ¡ Mg cos ¯: (5) N must be positive, so we immediately see that ¯ must be at least 45± if there is any chance that the setup is static. The coe±cient ¹ tells us that jFf j · ¹N. Using Eq. (5), this inequality becomes ¹ + 1 Mg(sin ¯ + cos ¯) · ¹Mg(sin ¯ ¡ cos ¯) =) · tan ¯: (6) ¹ ¡ 1 We see that we must have ¹ > 1 in order for there to exist any values of ¯ that satisfy this inequality. If ¹ ! 1, then ¯ can be as small as 45±, but it can't be any smaller. 2.21. Pulling a block The Fy forces tell us that N + F sin θ ¡ mg = 0 =) N = mg ¡ F sin θ. And assuming that the block slips, the Fx forces tell us that F cos θ > ¹N. Therefore, ¹mg F cos θ > ¹(mg ¡ F sin θ) =) F > : (7) cos θ + ¹ sin θ Taking the derivative top minimize this then gives tan θ = ¹. Plugging this θ back into F gives F > ¹mg= 1 + ¹2. If ¹ = 0, we have θ = 0 and F > 0. If ¹ ! 1, we have θ ¼ 90± and F > mg. 2.22. Holding a cone Let F be the friction force at each ¯nger. Then the Fy forces on the cone tell us that 2F cos θ ¡ 2N sin θ ¡ mg = 0. But F · ¹N. Therefore, mg 2¹N cos θ ¡ 2N sin θ ¡ mg > 0 =) N ¸ : (8) 2(¹ cos θ ¡ sin θ) This is the desired minimum normal force.