105Sol5 - Physics 105 Solutions to Problem Set 5 By Stephen...

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Physics 105 Solutions to Problem Set 5 By Stephen Glassman, Fall 2007 8.1 Recall problem (3.18) from problem set 2. There we showed that for a mass m at a and a mass M at b : r CM - a = M m + M b - a Note that b - a is (up to an arbitrary sign choice) the relative position vector, so that our previous result reduces to the result here when we make the replacements r CM R , b - a → - r , a r 1 , M m 2 , and m + M M . The total kinetic energy is then: KE = 1 2 m 1 ˙ r 2 1 + 1 2 m 2 ˙ r 2 2 = 1 2 m 1 ˙ R + m 2 M ˙ r 2 + 1 2 m 2 ˙ R - m 1 M ˙ r 2 = 1 2 m 1 ˙ R 2 + m 1 m 2 M ˙ R · ˙ r + 1 2 m 1 m 2 2 M 2 ˙ r 2 + 1 2 m 2 ˙ R 2 - m 1 m 2 M ˙ R · ˙ r + 1 2 m 2 m 2 1 M 2 ˙ r 2 = 1 2 ( m 1 + m 2 ) ˙ R 2 + 1 2 m 1 m 2 2 + m 2 1 m 2 M 2 ˙ r 2 = 1 2 ( m 1 + m 2 ) ˙ R 2 + 1 2 m 1 + m 2 M m 1 m 2 M ˙ r 2 = 1 2 M ˙ R 2 + 1 2 μ ˙ r 2 as required. 8.6 We reuse our result from (8.1). In the CM frame R = 0 and ˙ R = 0. Then: r 1 = R + m 2 M r = m 2 M r 1
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˙ r 1 = ˙ R + m 2 M ˙ r = m 2 M ˙ r and 1 = r 1 × m 1 ˙ r 1 = m 2 M r × m 1 m 2 M ˙ r = m 1 m 2 2 M 2 r × ˙ r Similarly: r 2 = - m 1 M r ˙ r 2 = - m 1 M ˙ r = 2 = + m 2 1 m 2 M 2 r × ˙ r where we carefully note that the two minus signs cancel. The total angular momentum is: L = 1 + 2 = m 1 m 2 2 + m 2 1 m 2 M 2 r × ˙ r = m 1 + m 2 M m 1 m 2 M r × ˙ r = m 1 m 2 M r × ˙ r Finally: 1 = m 1 m 2 2 M 2 r × ˙ r = m 2 M m 1 m 2 M r × ˙ r = m 2 M L 2
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and similarly for 2 . 8.11 We start from equation (8.15) and examine the ˆ x component: μ ¨ r = -∇ U ( r ) = μ ¨ x = - d dx 1 2 k ( x 2 + y 2 + z 2 ) = μ ¨ x = - kx And similarly for the ˆ y and ˆ z components. Thus we have harmonic oscillators in all three dimensions. Recall that the general solution of an harmonic oscillator is: x ( t ) = x 0 cos ωt + v 0 ω sin ωt If we choose our coordinates so that the initial position and initial velocity both lie in the x - y plane, then the ˆ z motion becomes z ( t ) = 0. Let: x ( t ) = A cos ωt + B sin ωt y ( t ) = α cos ωt + β sin ωt Let’s evaluate By - βx and αx - Ay . By - βx = αB cos ωt + βB sin ωt - βA cos ωt - βB sin ωt = ( αB - βA ) cos ωt αx - Ay = αA cos ωt + αB sin ωt - αA cos ωt - βA sin ωt = ( αB - βA ) sin ωt Then: ( By - βx ) 2 + ( αx - Ay ) 2 = ( αB - βA ) 2 cos 2 ωt + ( αB - βA ) 2 sin 2 ωt = ( αB - βA ) 2 We expand the left hand side to obtain: ( αB - βA ) 2 = ( By - βx ) 2 + ( αx - Ay ) 2 = ( β 2 + α 2 ) x 2 - 2 ( αA + βB ) xy + ( B 2 + A 2 ) y 2 Matching against ax 2 + 2 bxy + cy 2 = k we obtain: 3
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a = α 2 + β 2 b = - αA - βB c = A 2 + B 2 k = ( αB - βA ) 2 Note that a , c , and k are non-negative since they are sums of squares. Furthermore: ac - b 2 = ( α 2 + β 2 ) ( A 2 + B 2 ) - ( - αA - βB ) 2 = α 2 A 2 + β 2 A 2 + α 2 B 2 + β 2 B 2 - α 2 A 2 - 2 αAβB - β 2 B 2 = β 2 A 2 + α 2 B 2 - 2 αAβB = ( αB - βA ) 2 = k Since k is non-negative, ac b 2 , with ac = b 2 ⇐⇒ k = 0. Furthermore, a = 0 implies that α = β = 0 which implies that k = 0. Similarly, c = 0 = k = 0. Therefore, our only degenerate case is k = 0. ax 2 + 2 bxy + cy 2 = 0 = ( a x + c y ) 2 + ( b - ac ) xy If k = 0, then the coefficient of xy above is zero. We conclude that a x + c y = 0 for the orbit, so that the motion ( x, y ) is perpendicular to the vector ( a , c ).
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