Exam (elaborations) TEST BANK FOR Introduction to Classical Mechanics By David Morin (Solution manual)

Press Chapter 1 Strategies for solving problems 1.8. Pendulum on the moon The only p way to get units of time from `, g, and m is through the combination `=g. Therefore, TM TE = p `=gM p `=gE = r gE gM =) TM ¼ p 6 TE ¼ 7:3 s: (1) 1.9. Escape velocity (a) Using M = ½V , we have v = r 2G ¢ (4=3)¼R3½ R = p (8=3)¼GR2½: (2) (b) We see that v / R p ½. Therefore, vJ vE = RJ p ½J RE p ½E = 11 ¢ 1 p 4 = 5:5: (3) 1.10. Downhill projectile The angle ¯ is some function of the form, ¯ = f(µ; 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 µ. (And it clearly does depend on µ, because ¯ = 90± for µ = 0 or 90±, but ¯ 6= 90± for µ 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 write m=s = f(kg;m; kgm=s2). We need to get rid of the kg's, so we must use the ratio T=M. We then quickly see that p LT=M has the correct units of m=s. Note that this can also be written as p 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 can write 1=s = f(m; kg=m3; kg=s2). We need to get rid of the kg's, so we must use the ratio S=½. We then quickly see that p S=½R3 has the correct units of 1=s. Note that this can also be written as p 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 µ ! 90±. Only the 2nd and 3rd options satisfy this. The correct answer must also go to in¯nity for µ ! 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 y(t) = 1 ® ³ v0 sin µ + g ® ´ ³ 1 ¡ (1 ¡ ®t + ®2t2=2 ¡ ¢ ¢ ¢) ´ ¡ gt ® ¼ ³ v0 sin µ + g ® ´ ³ t ¡ ®t2=2 ´ ¡ gt ® = (v0 sin µ)t ¡ 1 2 gt2 ¡ 1 2 ®t2v0 sin µ: (4) If ® ¿ g=(v0 sin µ), then the third term is much smaller than the second, and we obtain the desired result. So ® ¿ g=(v0 sin µ) 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 °ight is less than 2v0 sin µ=g (because this t makes the above y(t) negative), so the condition ® ¿ g=(v0 sin µ) implies ®t ¿ (g=v0 sin µ)(2v0 sin µ=g) = 2. So ®t ¿ 1 is guaranteed by ® ¿ g=(v0 sin µ). 1.17. Pendulum Here is a Maple program that does the job: q:=3.14159/2: # initial µ value q1:=0: # initial µ speed e:=.0001: # a small time interval i:=0: # i will count the number of time steps while q0 do # run the program while µ 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 . s gives t = 0: s. 3 1.18. Distance with damping In the xÄ = ¡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., respectively. These approach 2. In the xÄ = ¡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. 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 Mg(sin ¯ + cos ¯) · ¹Mg(sin ¯ ¡ cos ¯) =) ¹ + 1 ¹ ¡ 1 · tan ¯: (6) 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 µ ¡ mg = 0 =) N = mg ¡ F sin µ. And assuming that the block slips, the Fx forces tell us that F cos µ ¹N. Therefore, F cos µ ¹(mg ¡ F sin µ) =) F ¹mg cos µ + ¹ sin µ : (7) Taking the derivative to minimize this then gives tan µ = ¹. Plugging this µ back into F gives F ¹mg= p 1 + ¹2. If ¹ = 0, we have µ = 0 and F 0. If ¹ ! 1, we have µ 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 µ ¡ 2N sin µ ¡ mg = 0. But F · ¹N. Therefore, 2¹N cos µ ¡ 2N sin µ ¡ mg 0 =) N ¸ mg 2(¹ cos µ ¡ sin µ) : (8) This is the desired minimum normal force. When ¹ = tan µ, we have N = 1. So ¹ = tan µ is the minimum allowable value of ¹. 2.23. Keeping a book up The result of Problem 2.4 is F ¸ mg=(sin µ + ¹ cos µ), assuming that sin µ + ¹ cos µ is positive (that is, tan µ ¡¹). If it is negative, there is no solution for F. To ¯nd the maximum force, consider two cases: