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homework6sol.20100303.4b8eff99de8384.47843235

# homework6sol.20100303.4b8eff99de8384.47843235 - 5 Here T =...

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TAM 412, Homework 6, Selected answers Harry Dankowicz Mechanical Science and Engineering University of Illinois at Urbana-Champaign [email protected] 3. By Newton’s second law it follows that m 1 ¨ r 1 = r 1 U ( r 2 r 1 ) m 2 ¨ r 2 = r 2 U ( r 2 r 1 ) where r 1 U ( r 2 r 1 ) refers to the gradient of U with respect to the cartesian components of the position vector r 1 from the origin of some coordinate system to the fi rst particle. But r 2 U ( r 2 r 1 ) = r 1 U ( r 2 r 1 ) = r U ( r ) , where r = r 2 r 1 . Thus, with ( m 1 + m 2 ) r cm = m 1 r 1 + m 2 r 2 it follows that ¨ r cm = 0 , i.e., the acceleration of the center of mass is independent of U . Also, d 2 dt 2 m 2 r = r μ m 2 m 1 + 1 U ( r ) , 4. Here T = m 2 ¡ ˙ x 2 1 + ˙ y 2 1 ¢ + M 2 ¡ ˙ x 2 2 + ˙ y 2 2 ¢ and U = mgy 1 + Mgy 2 where x 1 = a sin θ , y 1 = a cos θ x 2 = x 1 + b sin ϕ , y 2 = y 1 b cos ϕ
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Unformatted text preview: 5. Here T = m 2 Â¡ Ë™ x 2 + Ë™ y 2 + Ë™ z 2 Â¢ and U = mgz where x = z cos Î¸ ,y = z sin Î¸ The equation of motion corresponding to Î¸ yields z 2 Ë™ Î¸ = h where h is some constant (which depends on initial conditions and the mass). 6. Here T = m 2 Â¡ Ë™ x 2 P + Ë™ y 2 P Â¢ + m 2 Ë™ h 2 1 + m 2 Ë™ h 2 2 and U = mgy P + mgh 1 + mgh 2 where x P = 4 b 2 + PA 2 âˆ’ PB 2 4 b ,y P = âˆ’ p PA 2 âˆ’ x 2 h 1 = âˆ’ ( a âˆ’ PA ) ,h 2 = âˆ’ ( a âˆ’ PB ) Note, the T in the question is the f rst term in the expression above. 1...
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