同济大学理论力学孙杰习题解答7(练习册P39P46).ppt

* * 习题解答 练习册 动力学 2 P37 习题：1 A O m m1 m2 a M 动量矩定理： w v v LO = JOw +( m1v)r+(m2v)r 2 1 =( m + m1 + m2)rv SMO (Fi )= M – m1gr + m2gr 其中： v = rw JO = mr2 2 1 dt dLO = SMO (Fi ) 2 1 ( m + m1 + m2)ra = M – m1gr + m2gr ∴ M = 2.89 kN·m 对系统： 取： g = 9.8 m/s2 dt dLO = SMO (Fi ) 研究对象 m1g m2g P37 习题：2 F Ⅰ O G Ⅱ O P G w FOx FOy FOx FOy SMO (Fi )= Fr 刚体定轴转动微分方程： JOa = SMO (Fi ) ∴ JO Fr a = JO Pr = 质心运动定理： S Fx = 0 FOx = 0 S Fy = 0 FOy - G - F = 0 ∴ FOy = G + F = G + P a 动量矩定理： LO = JOw + vr g P = (JO + r2)w g P SMO (Fi )= Pr dt dLO = SMO (Fi ) (JO + r2)a = Pr g P Pr a = JO + r2 g P JOg + Pr2 Pr = g 质心运动定理： S Fx = 0 FOx = 0 FOy – G – F = - a g P FOy = G + JOg + Pr2 PJOg v = rw w a v S Fy = - a g P JO P38 习题：4 A aC mg F a C 刚体平面运动微分方程： S Fx = 0 S Fy = maC SMC (Fi ) = JCa aCx = 0 mg – F = maC aC = ra Fr = JCa 其中： JC = mr2 2 1 ∴ 3r 2g a = 3 2 aC = g 3 1 F = mg 平面运动 P39 习题：1 q C A a aC mg q FT F FN x y 刚体平面运动微分方程： S Fx = 0 S Fy = maC SMC (Fi ) = JCa FN – mgcosq = 0 FN = mgcosq F = fFN = fmgcosq mgsinq – FT – F = maC FT = mg(sinq – fcosq ) – mra (FT – F )r = JCa aC = ra 其中： JC = mr2 2 1 ∴ 3r 2g a = (sinq – 2fcosq ) 3 2 = (sinq – 2fcosq )g aC = ra FT = mra + fmgcosq 2 1 平面运动 图! 摩擦力：f ? f ：动摩擦系数 P40 习题：3 A O h l SMA (Fi ) = JAaA 正方形板： A 相对于质心的动量矩定理 mg aA = 0 SMA (Fi ) = 0 ∴ 整体： 相对于转轴O的动量矩定理 FOx FOy FOx FOy mg mg dt dLO = SMO(Fi ) SMO(Fi )= mgl + mgl 2 1 LO = JOwO + mvAl = mlvA 3 4 = mgl 2 3 mlaA 3 4 3 1 JO = ml 2 wO = l vA = mgl 2 3 ∴ aA = g 8 9 质心运动定理： S Fx = 0 FOx = 0 aA aC aC = g 16 9 FOy – 2mg = – 3maC S Fy = – maA – maC FOy = mg 16 5 wO S Fy = – 2maE aE P40 习题：4 q A C B aCx aCy FA aA mg a 刚体平面运动微分方程： S Fx = maCx S Fy = maCy SMC (Fi ) = JCa 运动学： aCx = aA – a 2 √3 2 l aCy = aA 2 1 FA = m( ) aA – a 2 √3 2 l 2 1 × =( ml 2)a 12 1 2 FA 2 l mg – FA = m aA 2 1 2 √3 FA = mla 3 1 aA =