解:Fn=m(,F12=m2a Fl=mra=ma fn=m M=0 ∑ 8=ma-m28 2D)△ma=0 由∑mr=C∑m知=mm TON Fo 解得a= mi-m g mtm tm INI, f En2解: F m a F m a I1 1 I 2 2 = , = F m r m a , i i t Ii =  = MO = 0, (m1 g −m1 a −m2 g −m2 a)r −mi ar = 0 由 m ar = (m )ar = mar i i 解得 g m m m m m a + + − = 1 2 1 2 r v F mi n Ii 2 =