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Unformatted text preview: Goldstein Chapter 1 Derivations Michael Good June 27, 2004 1 Derivations 1. Show that for a single particle with constant mass the equation of motion implies the follwing differential equation for the kinetic energy: dT dt = F · v while if the mass varies with time the corresponding equation is d ( mT ) dt = F · p . Answer: dT dt = d ( 1 2 mv 2 ) dt = m v · ˙ v = m a · v = F · v with time variable mass, d ( mT ) dt = d dt ( p 2 2 ) = p · ˙ p = F · p . 2. Prove that the magnitude R of the position vector for the center of mass from an arbitrary origin is given by the equation: M 2 R 2 = M X i m i r 2 i 1 2 X i,j m i m j r 2 ij . Answer: M R = X m i r i 1 M 2 R 2 = X i,j m i m j r i · r j Solving for r i · r j realize that r ij = r i r j . Square r i r j and you get r 2 ij = r 2 i 2 r i · r j + r 2 j Plug in for r i · r j r i · r j = 1 2 ( r 2 i + r 2 j r 2 ij ) M 2 R 2 = 1 2 X i,j m i m j r 2 i + 1 2 X i,j m i m j r 2 j 1 2 X i,j m i m j r 2 ij M 2 R 2 = 1 2 M X i m i r 2 i + 1 2 M X j m j r 2 j 1 2 X i,j m i m j r 2 ij M 2 R 2 = M X i m i r 2 i 1 2 X i,j m i m j r 2 ij 3. Suppose a system of two particles is known to obey the equations of mo tions, M d 2 R dt 2 = X i F ( e ) i ≡ F ( e ) d L dt = N ( e ) From the equations of the motion of the individual particles show that the in ternal forces between particles satisfy both the weak and the strong laws of ac tion and reaction. The argument may be generalized to a system with arbitrary number of particles, thus proving the converse of the arguments leading to the equations above. Answer: First, if the particles satisfy the strong law of action and reaction then they will automatically satisfy the weak law. The weak law demands that only the forces be equal and opposite. The strong law demands they be equal and oppo site and lie along the line joining the particles. The first equation of motion tells us that internal forces have no effect. The equations governing the individual particles are ˙ p 1 = F ( e ) 1 + F 21 ˙ p 2 = F ( e ) 2 + F 12 2 Assuming the equation of motion to be true, then ˙ p 1 + ˙ p 2 = F ( e ) 1 + F 21 + F ( e ) 2 + F 12 must give F 12 + F 21 = 0 Thus F 12 = F 21 and they are equal and opposite and satisfy the weak law of action and reaction. If the particles obey d L dt = N ( e ) then the time rate of change of the total angular momentum is only equal to the total external torque; that is, the internal torque contribution is null. For two particles, the internal torque contribution is r 1 × F 21 + r 2 × F 12 = r 1 × F 21 + r 2 × ( F 21 ) = ( r 1 r 2 ) × F 21 = r 12 × F 21 = 0 Now the only way for r 12 × F 21 to equal zero is for both r 12 and F 21 to lie on the line joining the two particles, so that the angle between them is zero, ie the magnitude of their cross product is zero....
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 Spring '12
 atkin
 Energy, Kinetic Energy, Mass, dt, ∂t, Lagrangian mechanics

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