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

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Unformatted text preview: 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 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 ~ 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 and similarly for ~ ` 2 . 8.11 We start from equation (8.15) and examine the x component: ~ r =- ~ U ( ~ r ) = x =- d dx 1 2 k ( x 2 + y 2 + z 2 ) = x =- kx And similarly for the y and z components. Thus we have harmonic oscillators in all three dimensions. Recall that the general solution of an harmonic oscillator is: x ( t ) = x cos t + v sin t If we choose our coordinates so that the initial position and initial velocity both lie in the x- y plane, then the z motion becomes z ( t ) = 0. Let: x ( t ) = A cos t + B sin t y ( t ) = cos t + sin t Lets evaluate By- x and x- Ay . By- x = B cos t + B sin t- A cos t- B sin t = ( B- A ) cos t x- Ay = A cos t + B sin t- A cos t- A sin t = ( B- A ) sin t Then: ( By- x ) 2 + ( x- Ay ) 2 = ( B- A ) 2 cos 2 t + ( B- A ) 2 sin 2 t = ( B- A ) 2 We expand the left hand side to obtain: ( B- A ) 2 = ( By- x ) 2 + ( x- Ay ) 2 = ( 2 + 2 ) x 2- 2 ( A + B ) xy + ( B 2 + A 2 ) y 2 Matching against ax 2 + 2 bxy + cy 2 = k we obtain: 3 a = 2 + 2 b =- A- B c = A 2 + B 2 k = ( B- A ) 2 Note that a , c , and k are non-negative since they are sums of squares. Furthermore: ac- b 2 = ( 2 + 2 )( A 2 + B 2 )- (- A- B ) 2 = 2 A 2 + 2 A 2 + 2 B 2 + 2 B 2- 2 A 2- 2 AB- 2 B 2 =...
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This note was uploaded on 04/01/2008 for the course PHYSICS 105 taught by Professor Edgarknobloch during the Fall '07 term at University of California, Berkeley.

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

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