PHYSICS 44 {CLASSICAL MECHANICS 2003 SUMMER FINAL EXAM { SOLUTIONS

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1 PHYSICS 44 {CLASSICAL MECHANICS 3 SUMMER FINAL EXAM { SOLUTIONS Problem 1 Consider the following Atwood machine comosed of the two masses m 1 and m and attached to a stand with the hel of a sring of strength k. The equilibrium oint of the sring is ataheighth below the stand and the generalized coordinates describing the state of the mechanical system are x and y as shown in the Figure below. (a) Construct the Lagrangian L(x; y; _x; _y) for this mechanical system. (b) Derive the Euler-Lagrange equations for x and y. Solution (a) The deths of masses m 1 and m below the stand are z 1 = h + y + x and z = h + ` + y x; resectively. From these exressions, the resective velocities are _z 1 = _y +_x and _z = _y _x; so that the kinetic energy of the mechanical system is K(_x; _y) = m 1 (_y +_x) + m (_y _x) 1

2 The sources of otential energy in this roblem are gravitational and elastic U(x; y) = m 1 gz 1 m gz + k y = m 1 g (y + x) m g (y x) + k y ; where we have eliminatedconstantterms. The Lagrangian for this mechanical system is, therefore, given by the exression L(x; y; _x; _y) = m 1 (_y +_x) + m (_y _x) k y + m 1 g (y + x) +m g (y x) (b) The Euler-Lagrange equation for x is _x = m 1 (_y +_x) m (_y _x)! d dt = (m 1 m )g (m 1 m )Äy +(m 1 + m )Äx = (m 1 m )g = (m 1 m )Äy +(m 1 + m )Äx By freezing the degree of freedom associatedwith y (i.e., _y = = Äy), we recover the Atwood machine roblem (m 1 + m )Äx =(m 1 m )g. or The Euler-Lagrange equation for y _y = m 1 (_y +_x) +m (_y _x)! d _y = (m 1 + m )Äy +(m 1 m )Äx = ky +(m 1 + m ) g (m 1 + m )Äy +(m 1 m )Äx = ky +(m 1 + m ) g By freezing the degree of freedom associated with x (i.e., _x = = Äx), we recover the sring-block roblem (m 1 + m )(Äy g)+ky =.

3 Problem Aarticle of mass m and angular momentum ` is observed to undergo eriodic motion with its distance r(µ) from the center of force given by the relation which describes a lemniscate of amlitude a. r (µ) = a cos( µ); (a) Show that theotential energyu(r) leading to this eriodic motion is given by the attractive otential U(r) = ` a 4 m r 6 (b) Solve the di erential equation for µ(t) givenby _ µ = `=mr for µ<¼=4subject to the initial condition µ =attimet =. Solution (a) From s(µ) =a 1 q sec(µ), we nd s = a 1 sin(µ)=[cos(µ)] 3= and s = 1 a à [cos(µ)] +3 sin (µ) 1= [cos(µ)] 5=! = 1 a Ã! 1 [cos(µ)] + 3 ; 1= [cos(µ)] 5= so that s + s = 3a 4 s 5 du ds! U(s) = 1 a4 s 6 3

4 Hence, the otential energy is U(r) = ` m U(1=r) = ` m a 4 r 6 Note that the e ective otential is V (r) = ` à 1 m r! a4 r 6 ; with a single local maximum at r =3 1 4 a, with V (r ) = ` 3 mr and V (a) = Hence, since r = a is a turning oint (see Figure above), this motion is characterized by a total energy E =andanin nite radial velocity as r!. If we now solve for s(µ) forthecasee =,we nd the integral solution µ µ 1 = = 1 Z 1 s Z ¼= d¾ a 4 ¾ 6 ¾ = 1 Z 1 s a 4 x 1 dã = 1 µ ¼ sec 1 (a s ) sec 1 (a s ) ; where we have usedthetrigonometric substitution x = a exression. Hence, we nd a s = secµ ¼ (µ 1 µ) = csc[(µ 1 µ)] We now choose s = a 1 at µ =,sothat µ 1 = ¼=4 and nally a s = sec(µ)! r = a cos(µ) sec à to obtain the nal (b) From the equation µ _ = `=mr,we ndcos(µ)dµ =(`=ma ) dt and, thus, we obtain the solution sin(µ) = ` ma t Here, since a quarter of the eriodic orbit takes a time of T=4 =ma =`, wereadily nd that the eriod is T =ma =`. 4

5 Problem 3 Consider the following scattering roblem where a article of mass m and total energy E > aroachesacenter of force with angular momentum ` and imact arameter b = `= me subject to the attractive otential where b = R sin <R. U(r) = 8 >< > U for r<r for r>r The scattering rocess is characterized by the distance of closest aroach ½<bat an angle  (see Figure below), from which the CM de ection angle is de nedas = ¼. (a) Find the distance of closest aroach ½. (b) Show that the imact arameter b( ) satis es the equation µ b( ) = nr sin ; where n = q 1+U =E > 1. (c) Show that themaximum de ection angle max is given as max = arccos ³ n 1 5

6 Solution (a) The turning oint r = ½ is a solution of the equation 1 = b ½ U E! ½ = b q 1+U =E = b n (b) The angle  corresonding to the distance of closest aroach ½ is given by the integral equation Z R (b=r Z ) dr n  = + q n (b=r) = + n x ; ½ where we usedthesubstitution x = b=r. Using the trigonometric substitution x = n sin Ã, the Â-integral becomes Z ¼=  = + dã = + ¼ arcsin(n 1 sin arcsin ³ n 1 sin Since b = R sin and  =(¼ + )=, we nd the imlicit exression for the imact arameter µ sin = n sin sin µ! b( ) = nr sin Since is a function of the imact arameter b, wecanalso ndthefollowing exlicit exression for theimact arameter b( ) = nr sin( =) q 1+n n cos( =) ; from which we derivetheexression db d = nr [n cos( =) 1] [n cos( =)] [1 + n n cos( =)] 3= (c) On the one hand,when =,we nd  = ¼= and min =. Ontheother hand, when = ¼=, we nd b = R and 1 = n sin µ ¼ max = n cos( max =)! max = arccos ³ n 1 Note that when = max,we nd that db=d vanishesand, therefore, the di erential cross section vanishes ¾( max )=,whileat =,we nd¾() = n R =4. 6

7 Problem 4 Consider a thin homogeneous rectangular late of mass M and area ab (with b>a) that lies on the (x; y)-lane. (a) Show that the inertia tensor (calculated in the reference frame with its origin at oint O in the Figure above)takestheform J = A C C B A + B and nd suitable exressions for A, B, andc in terms of M, a, andb. 1 C A ; (b) Calculate the inertia tensor I in the CM frame by using the Parallel-Axis Theorem and show thati is diagonal with comonents I xx = Mb 1 ; I yy = Ma 1 ; and I zz = M 1 ³ b + a Hint The osition of the center of mass is given by the vector ½ =(a bx + b by)=. (c) From the result in Part (b), determine the rincial moments of inertia (I 1 ;I ;I 3 )and the rincial axes of inertia (be 1 ; be ; be 3 )from(i xx ;I yy ;I zz )and(bx; by; bz). 7

8 Problem 4 (continued) (d) Show that the external torque N needed to rotate the rectangular late about a diagonal with constant angular velocity! (see Figure below) is N = M Ã a b! ab! be 1 a + b 3 Hint The unit vector b! is directed along the diagonal (as shown in the Figure) and the comonents! 1 and! are de ned as! 1 =! be 1 and! =! be. Solution (a) The mass density for the rectangular late is ½(x; y; z) =(M=ab) ±(z), where the delta function ±(z) indicates that mass is concentrated on the(z = )-lane. The diagonal comonents of the inertia tensor J are calculated as follows J xx = M ab J yy = M ab J zz = M ab = M 3 Z Z Z dz ±(z) dz ±(z) dz ±(z) ³ a + b dy ³ y + z = M ab dy ³ x + z M = ab dy ³ x + y = M ab dy y = M 3 b dy x = M 3 a dy ³ x + y while the non-diagonal comonents are J xy = M ab Z dz ±(z) dy x y = M ab dy x y = M 4 ab = J yx 8

9 with J xz = J yz ==J zx = J zy. (b) The dislacement vector ½ from the origin O to the center of mass (CM) is ½ = (a bx + b by)= sothat M ³ j½j 1 ½½ = M 4 h b bxbx + a byby +(a + b ) bzbz ab (bxby + bybx) i ; and thus, according totheparallel-axistheorem, the CM inertia tensor I = J M ³ j½j 1 ½½ has the comonents I xx = J xx M 4 b = M 1 b ; I yy = J yy M 4 a = M 1 a ; I zz = J zz M 4 (a + b ) = M 1 (a + b ); I xy = J xy + M 4 ab = (c) The rincial axes of inertia are be 1 = bx, be = by, andbe 3 = bz, sothati 1 = I xx, I = I yy, and I 3 = I zz. (d) From the unit vector b! b! = (a be 1 + b be ) a + b ; the frequency comonents are with! 3 =.! 1 =!a a + b and! =!b a + b ; Since! =! 1 be 1 +! be is constant, the Euler equations become! 1! (I I 1 ) = N 3 = M 1 Ã a b b + a! ab! 9