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### Analytical Mech Homework Solutions 55

Course: PHYSICS 4360C, Fall 2011
School: UNF
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Word Count: 156

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= k , with = 0 r = ber , so ( r ) = b 2 er v = ve , so v = ver From eqn 5.2.14, a = a + 2 v + ( r ) and putting in terms from above v 2 2 v b 2 ar = b For no slipping F s mg , so a s g v 2 + 2 v + b 2 s g b 2 vm + 2 bvm + b 2 2 b s g = 0 vm = b 2b 2 b 2 2 + b s g Since v was defined positive, the +square root is used. vm = b + b s g (b) v ve = v = + ver 2 v + 2...

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= k , with = 0 r = ber , so ( r ) = b 2 er v = ve , so v = ver From eqn 5.2.14, a = a + 2 v + ( r ) and putting in terms from above v 2 2 v b 2 ar = b For no slipping F s mg , so a s g v 2 + 2 v + b 2 s g b 2 vm + 2 bvm + b 2 2 b s g = 0 vm = b 2b 2 b 2 2 + b s g Since v was defined positive, the +square root is used. vm = b + b s g (b) v ve = v = + ver 2 v + 2 v b 2 ar = b 2 v 2 v + b 2 s g b vm = b + b s g 2 5.9 V V i As in Example 5.2.2, = and A = j For the point at the front of the wheel: 2 V r = j and v = V k b =0 V Vb j i r = k b = ( r ) = v = ( ) 2 V Vb V b k i = j ( ) V k V k = 0 2 2 2 V V V b a = r + ( r ) + A = i + + 2 j b 5
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UNF - PHYSICS - 4360C
5.10(See Example 5.3.3)m 2 x = mxx ( t ) = Ae t + Be tx ( t ) = Ae t Be tBoundary Conditions:lx2 ( 0) = = A + B2x ( 0 ) = 0 = ( A B )A=l4B=lx ( t ) = cosh t2lllx (T ) = + = cosh T222(a)(b) cosh T = 2lx ( t ) = sinh t2when the bea
UNF - PHYSICS - 4360C
v = zv yi + zvx yvx k jj( v )horiz = zvyi + zvx ( v )horiz112222= ( z2v y + z2vx ) 2 = z ( vx + v y ) 2 = zvFcor = 2m v(F )cor5.13horiz= 2m ( v )horiz = 2m z v , independent of the direction of v .From Example 5.4.1 11 8h 3 2 = xh
UNF - PHYSICS - 4360C
da = r + r + r + 2 r + 2 r dt rot()() a = r + ( r ) + 2 ( r ) + ( r ) + ( r ) + r + r Now( r )is to and r . Let this define a direction n : r = r nSince n , ( r ) is in the plane defined by and r and ( r ) = n r = r .Since ( r ) ( r ) = 2
UNF - PHYSICS - 4360C
1z ( t ) = gt 2 + v t sin + v t 2 cos cos = 022v sin 2v sin t=org 2 v cos cos gWe have ignored the second term in the denominatorsince v would have to beimpossibly large for the value of that term to approach the magnitude gFor example, for = 4
UNF - PHYSICS - 4360C
(jHence: a = a 2 v = 3 2 x i 2 k ix + ySo5.19)ja = ix + y = 3 2 xi + 2 yi 2 xj2x 2 y 3 x = 0y + 2 x = 0(mr = qE + q v BEquation 5.2.14)r = r + r + 2 v + ( r )v = v + r Equation 5.2.13q =B so = 02mqmr q B v B ( r ) = qE + q ( v + r
UNF - PHYSICS - 4360C
Collecting terms and neglecting terms in 2 :gg x + x cos t + y + y sin t = 0llgg x + x sin t y + y cos t = 0ll24hourssin 24T== 73.7 hourssin 195.21T=5.22Choose a coordinate system with the origin at the center of the wheel, the x and
UNF - PHYSICS - 4360C
r = 0Vb VtVt VtVtVt Vt k sin cos + V sin i cos j k sin cosbbbbbb22V Vt Vt V Vt Vt+ k cos 2 j cos ( r ) = k sin 2+ i sinbb bbb22VV ( r ) = k nbSince the origin of the coordinate system is traveling in a circle of r
UNF - PHYSICS - 4360C
CHAPTER 6GRAVITATIONAL AND CENTRAL FORCES6.14m = V = rs331 3m 3rs = 4 F=Gmm( 2rs )222Gm 2 4 3 G 4 3 43== m4 3m 4 3 2FFGm 2 4 3 3==mW mg4g 3 121F 6.672 1011 N m 2 kg 2 4 11.35 g cm 3 1 kg 106 cm3 3=3 (1 kg ) 3W4 9.8
UNF - PHYSICS - 4360C
The gravitational force on the particle is due only to the mass of the earth that isinside the particles instantaneous displacement from the center of the earth, r.The net effect of the mass of the earth outside r is zero (See Problem 6.2).4M = r334
UNF - PHYSICS - 4360C
6.56.6GMm mv 2GMv2 ==so2rrrfor a circular orbit r, v is constant.2 rT=v4 2 r 2 4 2 3T2 =r r3=2vGM2 rvFrom Example 6.5.3, the speed of a satellite in circular orbit is (a) T =1 gR 2 2v= e r3T=2 r 21g 2 Re1 T 2 gRe2 3r =
UNF - PHYSICS - 4360C
This is the same expression as derived in Problem 6.3 for a particle dropped into ahole drilled through the earth. T 1.4 hours.6.8The Earths orbit about the Sun is counter-clockwise as seen from, say, the northstar. Its coordinates on approach at the
UNF - PHYSICS - 4360C
4M d = r33GMm 4F ( r ) = 2 mGr3r6.101 1 k=errduk= e kdru=d 2u k 2 k= e = k 2ud 2 rFrom equation 6.5.10 d 2u1+ u = k 2u + u = 2 2 f ( u 1 )2duml u122f ( u ) = ml ( k + 1) u 3f (r ) = ml 2 ( k 2 + 1)r3The force varies as th
UNF - PHYSICS - 4360C
kIf 1 + 2 &lt; 0 , ml kIf 1 + 2 = 0 , ml du= C1du = c1 + c21r=c1 + c2kIf 1 + 2 &gt; 0 , ml u = A cos(c + d 2u cu = 0 ,d 2d 2u=0d 2)d 2u+ cu = 0 ,d 2kr = A cos 1 + 2 + ml6.12c &gt; 0 , for which u = aeb is a solution.c&gt;0111=r
UNF - PHYSICS - 4360C
For r = a , and the force to be central, try = bt nf ( ) = m 2ab 2 n 2t 2 n 2 + ab 2 n ( n 1) t 2 n 2 For a central field f ( ) = 02n + ( n 1) = 0n=131 = bt 3(a)6.14Calculating the potential energy 4 a2 dv = f ( r ) = k 3 + 5 drrr 2 a2
UNF - PHYSICS - 4360C
1r = a cos ( r = a @ = 0 )33(b) at =r 0 the origin of the force. To find how long it takes 2avvl= 2==11ra 2 cos 2 a cos 2 33a1dt = cos 2 dv3ThusT=320a3a 23 a212cos d = cos d = 4vv3v01 9k 2Since v = 2 2a 113 2
UNF - PHYSICS - 4360C
dV V 1 r (b) dr1 = 2r1 dV = 2 60 1% = 120% !(a)() dr1 2 r1 r1 r V r1 The approximation of a differential has broken down a correct result can be obtained bycalculating finite differences, but the implication is clear a 1% error in boost causes
UNF - PHYSICS - 4360C
6.17From Example 6.10.1 12vr2 2 = 1 + q 2 ( qd sin ) where q =and d =dveaeare dimensionless ratios of the comets speed and distance from the Sun in terms of theEarths orbital speed and radius, respectively (q and d are the same as the factor
UNF - PHYSICS - 4360C
But the total orbital energy iskkE=SoE = 2 a2a2aSince planetary orbits are nearly circularkkV ~andT~a2aaE aand=Thus, E TaTaav=2We obtainav6.20 (a) V =1Vdt0krFrom equation 6.5.4, l = r 2dlr 2 d= 2 or dt =dt rl2 kr
UNF - PHYSICS - 4360C
Integrate LHS by parts111 mr r mr 2 dt = F rdt0 00The first term is zero since the quantity has the same value at 0 and .denote time average of the quantity within brackets.Thus 2 T = F r wherebut r F = r V = rdVk== Vdrrhence 2 T = Vbu
UNF - PHYSICS - 4360C
k 2 1m RE a and solving for aREa=noting that 2 RE2 mvkkGM E== gRE(3)mRERERERa== E = 4.49 103 km2v 1.4262gRE v2 =The eccentricity can be found from the angular momentum per unit mass, l, equation6.5.19, and the data on ellipses de
UNF - PHYSICS - 4360C
6.23From Problem 6.9, f ( r ) = GMm 4 mGr3r22GMm 4 mG3r344 a 32GMma 3 mG2 +f (a)33M==4f ( a ) GMma 2 mGa 4 a 3 a 1 +33M f (r ) =f (a) From equation 6.14.3, = 3 + af (a) 4 a 2 + 3M = 3 +4 a 31+3M3121 1+ c 2 = , 1 +
UNF - PHYSICS - 4360C
6.25f (r ) =2k 4+r3 r5From equation 6.13.7, the condition for stability is f ( a ) +k a 2k 4 4 + 3 + 5 &lt;02aa3 aak 2 + 4 &lt;03a 3aaf (a) &lt; 03a2&lt;k1 2a&gt; k6.26 (a)e brr2e br b 2f ( r ) = ke br 2 3 = k 2rrrf ( r ) = k2b + r
UNF - PHYSICS - 4360C
a (1 2 )1 + cos and the data on ellipses in Figure 6.5.1 p = a (1 ) so(1 + )r=p1 + cos For a parabolic orbit, = 1The comet intersects earths orbit at r = a .2pa=1 + cos 2pcos = 1 +a6.27(See Figure 6.10.1) From equation 6.5.18a r =6.28(See
UNF - PHYSICS - 4360C
3T=2 p2a32d+ (1 + cos )From a table of integrals,2yrdx (1 + cos x )2=1x1xtan + tan 3226232 p 2 1 3 T=tan + tan yr232 a 1x 1 cos x 2tan = 2 1 + cos x 12 p 21 2 a a p 2tan = =2 2p p a1332 p 2 a p 2 1 a p 2 T
UNF - PHYSICS - 4360C
dV k 3k k 2=+= 4 ( r + 3 )dr r 2 r 4r2k 12k2k 2f ( r ) = 3 5 = 5 ( r + 6 )rrr2f (a)2 r + 6 = 2f (a)a r + 3 f (r ) = f (a) From equation 6.14.3, = 3 + af (a) ForFor6.30121 r 2 + 6 2r 2 + 3 = 3 2 2 = 2r 3 r + 3 2 = RR , R =
UNF - PHYSICS - 4360C
E1 + m c 2 =1k 1 + m c 2 E + a m c2 + E =3+ af ( a ) m c2 + E + k af (a)f (a) = 3 + af (a) 121k2a = 1 +2 mc +Ea (1 2 )(Here is the polar angle of conic6.31 From equation 6.5.18a r =1 + cos section trajectories as illustrated
UNF - PHYSICS - 4360C
1212sin = 2 ( RV 2 sin 2 ) + 2 ( RV 2 sin 2 ) 1Again from Example 6.5.3 122 2 = 1 + V 2 ( RV sin ) R 2 = 1 + ( RV 2 sin ) 2 RV 2 sin 2 212221sin = ( RV 2 sin ) ( RV 2 sin 2 ) 1sin = RV 2 sin cos cos RV 2 sin 2 1=sin RV 2 sin cos 2
UNF - PHYSICS - 4360C
Letting r1 = RE + h1 and r2 = RE + h 2 where h1 and h2 are the heights of the 2 satellitesabove the ground. Inserting these into the above gives 33h1 + h 2 2h1 + h 2 22+Tt =RE 2 +=RE RE RE gRE g From Example 6.6.2, RE = 6371 km, h1 = 200 mi
UNF - PHYSICS - 4360C
Before evaluating this integral, we need to find umax ( = rmin 1 ) , in other words, thedistance of closest approach to the scattering center.11k1E = T ( rmin ) + V ( rmin ) = mv 2 += mv0 2222 rmin2But, the angular momentum per unit mass l is
UNF - PHYSICS - 4360C
Chapter 7Dynamics of Systems of Particles7.1z1 mi rimiFrom eqn. 7.1.1, rcm =(3)11jj( r1 + r2 + r3 ) = i + + + k + k331rcm = i + 2 + 2kj3d11vcm = rcm = ( v1 + v2 + v3 ) = 2i + + i + + kjjdt331vcm = 3i + 2 + kj3From eqn 7.1.
UNF - PHYSICS - 4360C
vg = 7.4v1+ v mv = m + Mvblk21mvblk = v=M2221 2 1 v 1 v Ti T f = mv m + M 22 2 2 2 Momentum is conserved:1 2 1 1 2 = mv 1 M24 4mTi T f 3 =44Ti7.5v0 / 2At the top of the trajectory:vv = iv cos 60 = i2Momentum is conserved
UNF - PHYSICS - 4360C
Total distance = h + 2 2 h + 2 4 h + = h 1 + 2 2 n n=0a, r &lt;1.Now ar n =1 rn=021+ 21 + 2 2 n = 1 +=1 2 1 2n=0 1+ 2 total distance = h 2 1 For the first fall,122hgt = h , so t =2g2h2h=ggAccounting for equal rise and fall times:
UNF - PHYSICS - 4360C
7.81112From eqn. 7.2.15, T = mi vi2 = m1v12 + m2 v222i2Meanwhile:21 2 1 2 1 m1v1 + m2 v2 1 m1m22mvcm + v = m ( v1 v2 )+m222 2 m1 + m21222 m12 v12 + m2 v2 + 2m1m2 v1 v2 + m1m2 ( v12 + v2 2v1 v2 ) =2m112= m1v12 + m2 v2221212
UNF - PHYSICS - 4360C
2m1 2v14T1 T1m==1T1mm1v1227.11From eqn. 7.2.14, L = rcm mvcm + ri mi vir m viiii= r1 m1v1 + r2 m2 v2i m m + m2 m1From eqn. 7.3.2, R = r1 1 + 1 = r1 1r1= m2 m2 mSince from eqn. 7.3.1, r1 = 2 r2m1mR = 2 r2r m viiii R m1
UNF - PHYSICS - 4360C
(1 )b) V ( x , y ) = 12( x ) + y ( x + 1 ) + y V (1 ) ( x ) ( x + 1 )=+x33x2222[]2212x2 + y 22(7.4.13)[]Now x = 0.5 at L4 and L5also, each bracket term in the denominator equals 1 at L4 , L5V= (1 ) ( 0.5 ) + ( 0.5 + 1 ) ( 0.5
UNF - PHYSICS - 4360C
2v 2v 2 + 60v 2 v2 62=1010vp &gt; 0 , so the positive square root is used.(vp =)vp = 0.9288 vvpx = vpy =vp2= 0.657 v1112( v vp2 ) 2 = v2 (1 .9288 2 ) 22v = 0.1853vvpvp.92882=tan ==vp2v vp2 .9288v21 = tan 1.9134 = 62.41v x = v
UNF - PHYSICS - 4360C
= tan 1 1.2638 = 51.65v x = v cos = 0.110 vv y = v sin = 0.140 v7.16sin 1 and are the scattering angles in + costhe Lab and C.M. frames respectively.mFrom eqn. 7.6.16, for Q = 0, = 1m2sin tan 45 =1 =1+ cos 41+ cos = sin and squaring 4
UNF - PHYSICS - 4360C
P17.18Conservation of momentum:P = P cos + P2 cos ( 11P1)0 = P sin P2 sin ( )1From Appendix B for sin ( + ) and cos ( + ) :P2P = P cos + P2 ( cos cos + sin sin )110 = P sin P2 ( sin cos cos sin )1P 2 = P2 cos 2 + P22 ( cos 2 cos 2 + 2 cos
UNF - PHYSICS - 4360C
vcm1m1==Equation 7.6.11v1 m2 + m1 1 + Thus1212v1212r2r=+=+22v1 (1 + )2 (1 + )2 (1 + )1 + ) (1 + ) (1 + )(Simplifying2 1 1 rr2 +=01+ 1+ Let x 2 = r and solving the resulting quadratic for x11 2 (1 2 ) 2+x=1+ 1+ Squar
UNF - PHYSICS - 4360C
1=m1m2( 2T m1 ) 212( 2 m1 ) (1 + m1Tm2 ) Q (1 + m1 m2 )1212Finally=7.21m1m211 Q (1 + m1 m2 ) 21 TThe time of flight, = constantso =rbut from problem 7.19 abovev1v1 + 2 + 2 11+As an example, let v1 = 1 and we haver1 =
UNF - PHYSICS - 4360C
7.234m = r33m = 4 r 2 r r 2 zwhere v = zr = kzk a constant of proportionalityr = r + kzFrom eqn. 7.7.6, mg = mv + vm434 r g = r 3 z + 4 r 2 ( kz ) z3323kzg = z+r3z 2z=grz+k3z 2For r = 0 , z = g zA series solution is used for thi
UNF - PHYSICS - 4360C
7.24From eqn. 7.7.6, mg = mv + vm , where m and v refer to the portion of the chainhanging over the edge of the table.m = z and v = zwhere is the mass per unit length of chainm = z and v = zdz dz dzdz 1 d ( z 2 )= =z =dt dz dtdz 2 dz2 1 d (z )
UNF - PHYSICS - 4360C
At t = 0 , z = 0 , and z = ba 2 gb0= 2 +3b2a = gb332 b3 2z 2 = g 2 + gz3z32b3 2 gAt z = a , z 2 = g a 2 = 2 ( a 3 b3 )a 3a31 2g2z = 2 ( a 3 b3 ) 3a7.25Initially, the upward buoyancy force balances the weight of the balloon and sand
UNF - PHYSICS - 4360C
1 2 gtg1gt ln (1 kt ) 1 + dtkk21 kt 1gtgt g= C gt 2 ln (1 kt ) + + 2 ln (1 kt )kkk21ggt= C gt 2 + 2 (1 kt ) ln (1 kt ) +kk2but y = 0 at t = 0 so C = 0gt 1gy = gt 2 + 2 (1 kt ) ln (1 kt )k2kand at t = t=C(a)(b)(c)Mgt 2 2
UNF - PHYSICS - 4360C
dvVk=+dtm ktThusdt= V ln ( m k t ) + C1m ktNow, v = 0 @ t = 0 so C1 = V ln mv = Vk where C1 is a constant. ( m kt ) m0t Hence v = V ln ,kmLet y be the displacement at the time t so ( m kt ) y = V ln dt + C 2mIntegrating the above
UNF - PHYSICS - 4360C
v V = mm dvm dv dmdvv === mmm dtm dm dtdmdmdv= m v Vv1 m dmdv=m m 0 v V1 m v V ln = ln m V 1 m v+1 =Vm 1 m 1 v = V m 7.28From eqn. 7.7.5, mg = mv Vmsince V is opposite in direction from v , mg = mv + Vm mgdt = m
UNF - PHYSICS - 4360C
1 g m = exp 1mp k kve m mfFor V = kve ,ve 11From chap. 2, Section 2.3 For m = 0.01 m s 1and k =kms1:49.8= exp 4 11mp(11, 000 ) ( .01) 4mf= 77mpmf7.29 We can use Equation 7.7.9 to calculate the final velocity attained by the io
UNF - PHYSICS - 4360C
1CHAPTER 8MECHANICS OF RIGID BODIES:PLANAR MOTIONmand centered at3 b bb b , , ( 0, 0 ) , and , 2 2 2 21 b m b m xcm = + 0 + = 0m 2 3 2 3 8.1 (a) For each portion of the wire having a massybycm =x1 b m b m b 2 3 + 0 + 2 3 = 3m
UNF - PHYSICS - 4360C
2bzcm=0 z bzdzb0 bzdzb z dz= zdz20b02zcm = b3(e)The center of mass is on the z-axis. is the half-angle of the apex of the cone. r is the radius ofthe base at z = 0 and r is radius of a circle at some arbitrary z ina plane parallel
UNF - PHYSICS - 4360C
3zcm =a1422m b b 8.4 (a) I z = mi R = + 0 + 3 2 2 i2iIz =mb 26(b)ds = rd dr , R = r sin I z = R 2 dsIz = r =br =0Iz =b44r rdr 2=4 = sin42sin 2 d4 d4 sin 2 2 sin d = 2 4 4b 1 Iz =4 4 21m = b 24mb 2Iz =( 2
UNF - PHYSICS - 4360C
4R = (x + y2122) = ( bz )1211 b 2dR = dz2 z 12bI z = R dv = bz 2 ( bz )2011b2( b z ) dz2 z b1I z = b 2 ( bz z 2 ) dz = b506bm = dv = 2 ( bz )12011b2( b z ) dz2 z b1m = b ( b z ) dz = b3021I z = mb 23(e) is the
UNF - PHYSICS - 4360C
58.5Consider the sphere with the cavity to be made of a (i) solid sphere of radius aand and mass M s , with its center of mass at z = 0 , and (ii) a solid sphere the size of theacavity, with mass M c and center of mass at z = . The actual sphere with
UNF - PHYSICS - 4360C
68.7For a rectangular parallelepiped:dv = hdxdyyR = ( x2 + yh is the length of the box in the z-direction122)I z = R 2 dv = x=a (xy =bx = ay = b2+ y 2 ) hdxdy2b3 4222I z = h 2bx + dx = hab ( a + b )a33m = ( 2a ) ( 2b ) h = 4 ab
UNF - PHYSICS - 4360C
7m = h ( ab )Iz =m2( a + b2 )4For an ellipsoid:R2 = x2 + y 2dv = hdxdy ,and on the surface,x2 y 2 z 2++ =1a 2 b2 c21 x2 y2 2z = c 1 2 2 bay12bx x2 y2 2h = 2 z = 2c 1 2 2 bax2 y 2In the xy plane 2 + 2 = 1ab1 y2 2x = a 1 2
UNF - PHYSICS - 4360C
88.8(See Figure 8.4.1)Note that l + l is the distance from 0 to 0 , defined as d2From eqn. 8.4.13, kcm = ll 2kcm + l 2 = ll + l 2 = l ( l + l )2kcm + l 2 = ld2From eqn. 8.4.9b, k 2 = kcm + l 2k 2 = ldFrom eqn. 8.4.6, T = 2T = 2k2gldgPer
UNF - PHYSICS - 4360C
912 22 2 2 1 Ma (1 ) 1 + 5 + 3 (1 ) T = 2 Mga 1 2 2(a)(b)8.10T=T212(1 ) 1 + 2 + (1 )53 1 1 2 2M = 1kga = 1.27 m = 0.0394m = 10 g = 0.01T1 1 = 0.999212T1 to 1st order in 12b = 5cm(actually 0.9994 using complete e
UNF - PHYSICS - 4360C
1040 ln lmm=2M lm 1gLetting lm = 1.3m ; l = 1.0m ; we obtain 1.15 1038.11(a) I cm = ma 2 (all mass in rim)I rim = ma 2 + ma 2 = 2ma 2T = 2(b)I z = I x + I y = 2 I cm = ma 2 I cm =I rim2a= 2mgag( = I cm )ma 22ma 23+ ma 2 = ma 2
UNF - PHYSICS - 4360C
11l3Rxcm = =2m 3R mg R = m = 3RmmgR=46 R 6 mg =xend = l = lml m 4 3xend = g28.14For a solid sphere:42M s = a 3 and I s = M s a 235222( kcm )s = 5 aFor subscript c representing a solid sphere the size of the cavity, from eqn.
UNF - PHYSICS - 4360C
12Energy is conserved:a8.15m2m18.162111 xm1 x 2 + m2 x 2 + I + m2 gx m1 gx = E222 2Im1 xx + m2 xx + 2 xx + m2 gx m1 gx = 0am1 m2 ) g(x=Im1 + m2 + 2axWhile the cylinders are in contact:mv 2bf r = cm = mg cos Rr2Rmvcmr = a +
UNF - PHYSICS - 4360C
131212mvcm + I + mgy = mgy22llx = cos ,x = sin ,22lly = sin ,y = cos ,22x=l2 cos + sin 2ly = 2 sin + cos 2()(2222l l l2vcm = x 2 + y 2 = sin + cos =42 22mlI=, and = 121 l 2 2 1 ml 2 2llm+ + mg sin = mg sin
UNF - PHYSICS - 4360C
14m l 2 2 1 ml 2 2l+ = mg (1 cos )242 1222l= g (1 cos )31 3g2 = (1 cos ) l11 3g3g 2 3g = (1 cos ) sin = sin 2 l2l lml 3g 3gRx =( sin ) l (1 cos ) + cos 2l sin 23mgRx =sin ( 3 cos 2 )4ml 3g 3gRy = mg cos l (1 cos