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# f5ch09 - CHAPTER 9 SYSTEMS OF PARTICLES 239 CHAPTER 9...

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CHAPTER 9 SYSTEMS OF PARTICLES 239 C HAPTER 9 Answer to Checkpoint Questions . ( a ) origin; ( b ) fourth quadrant; ( c ) on y axis below origin; ( d ) origin; ( e ) third quadrant; ( f ) origin . ( a ) to ( c ) at the center of mass, still at the origin (their forces are internal to the system and cannot move the center of mass) . ( a ) , , and then and tie (zero force); ( b ) . ( a ) ; ( b ) no; ( c ) negative . ( a km/h; ( b km/h; ( c km/h . ( a ) yes; ( b ) no Answer to Questions . point . ( a ) to ( d ) at the origin . ( a ) at the center of the sled; ( b ) L= , to the origin; ( c ) not at all (no net external force); ( d ) L= , to the left; ( e ) L ; ( f ) L= ; ( g ) L= . ( a ) less; ( b ) greater . ( a ) ac, cd, and bc; ( b ) bc; ( c ) bd and ad . ( a ) g ; ( b ) g . ( a ) N, rightward; ( b ) N, rightward; ( c ) greater than N, rightward . c, d, and then a and b tie . b, c, a . ( a ) all tie; ( b ) all tie; ( c ) all tie . ( a ) yes; ( b ) kg m/s in x direction; ( c ) can't tell

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240 CHAPTER 9 SYSTEMS OF PARTICLES . , , Solutions to Exercises & Problems E ( a ) Put the origin at the center of the Earth. Then the distance r cm between the origin and the center of mass of the Earth-Moon system is given by r cm = M m r m M m + M e ; where M m is the mass of the Moon, M e is the mass of the Earth, and r m is their separation. These values are given in Appendix C. The numerical result is r cm = ( : kg)( : m) : kg + : kg = : m : ( b ) The radius of the Earth is R e : m so r cm =R e : . E Denote the mass of the carbon atom as m c and that of the oxygen atom as m o . Let the separation between the carbon atom and the center of mass be x . Then m c x = m o ( : m x ). Thus the center of mass is located along the line joining the two atoms, a distance x from the carbon atom, where x = m o m c + m o ( : m) = : : + : ( : m) = : m : E ( a ) Let x (= ), y (= ) be the coordinates of one particle, x (= : m), y : m) be the coordinates of the second particle, and x : m), y : m) be the coordinates of the third. Designate the corresponding masses by m : kg), m : kg), and m : kg). Then the x coordinate of the center of mass is x cm = m x + m x + m x m + m + m = + ( : kg)( : m) + ( : kg)( : m) : : : kg : m
x L/ 2 L/ 2 M M O 3 M x O x 1 x 2 m 2 m 1 CHAPTER 9 SYSTEMS OF PARTICLES 241 and the y coordinate is y cm = m y + m y + m y m + m + m = + ( : kg)( : m) + ( : kg)( : m) : kg + : : kg = : m : ( b ) As the mass of the topmost particle is increased the center of mass shifts toward that particle. In the limit as the topmost particle is much more massive than the others, the center of mass is at the position of that particle. E Denote the location of the center of mass as O . From symmetry we see that O has to be equidistant from the two rods on the arms. To nd x (see gure), write Mx M ( L= x ), which gives x : L . E First, from symmetry we know that the y coordinate of the cemter of mass is y cm = . To get the x coordinate, imagine that the piece cut from the plate is rst re lled with its original mass m to recover the complete plate and then lled again with an equal amount of \negative" mass (= m ). The end result is just the same as having the piece cut from the square. If the mass of the whole plate is M , then the mass of the square piece cut from it should be m = : m : m M = M : The x coordinate of the center of mass of the whole larger square (with mass M ) is x = , while that of the smaller one (with mass m ) is x : m. Thus the x coordinate of the center of mass of the remaining piece is x cm = x M + x ( m ) M + m = ( : m) m m m = : m : P

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## This note was uploaded on 01/22/2012 for the course PHYS 2101 taught by Professor Grouptest during the Fall '07 term at LSU.

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f5ch09 - CHAPTER 9 SYSTEMS OF PARTICLES 239 CHAPTER 9...

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