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카이스트 고전역학 족보 고전역학2 midterm solution

2015 Classical Mechanics II Midterm Exam
2015. 10. 23 09:00 ~ 15:00 (6 hours)

1. (a) Show that the total kinetic energy of a system of particles is equal to the sum of the kinetic energy of a particle of mass moving with the velocity of the center of mass and the kinetic energy of motion of the individual particles relative to the center of mass. Total kinetic energy of the system 1. The work done on the particle by the net resultant force in transforming the particles from condition 1 ...

2015 Classical Mechanics II Midterm Exam
2015. 10. 23 09:00 ~ 15:00 (6 hours)

1. (a) Show that the total kinetic energy of a system of particles is equal to the sum of the kinetic energy of a particle of mass moving with the velocity of the center of mass and the kinetic energy of motion of the individual particles relative to the center of mass. Total kinetic energy of the system 1. The work done on the particle by the net resultant force in transforming the particles from condition 1 to condition 2 is given as
2

(4 points)

= ∫
1

2.

Summing the expression over ,
2 2 1 2 12 = ∑ ∫ = ∑ ∫ ( ) = 2 1 2 1 1

3. 4.

Using = R + ′ 2 2′ = = ( ′ ′) + 2 ( ′ ) + ( ) = + 2 ( ′ ) + 2 Then 1 1 1 2 2 ≡ ∑ = ∑ ′ + ∑ 2 + ∑ ′ 2 2 2

1 1 2 = ∑ ′ + 2 2 2 5.

The total kinetic energy of the system is equal to the sum of the kinetic energy of a particle of mass moving with the velocity of the center of mass and the k…(생략(省略))

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카이스트 고전역학 족보 고전역학2 midterm solution

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