T=J@f+oJca2+om,vc 式中 =m1R123,0B2=c,如sd R 代入后得 T=(2m1+3m2) 应用动能定理=∑W 得 (2m,+3m, vdv =(M-m,g sin 0)ds R 上式两边同除以dt 2(2m,+3m2)vc.ac=(MI-m28 sinO)vc 解得 2(M-m2gR, sin 8 R1(2m1+3m2)2 2 2 2 2 1 1 2 1 2 1 2 1 C C T = J ω + J ω + m v 2 1 m1R1 J = 1 1 R v C ，ω = 2 1 2 ( 2 3 ) 4 1 C T = m + m v 2 2 2 2 1 , J C = m R 2 2 , R v C ω = 1 , R ds dϕ = 式中 代入后得 应用动能定理 dT = ∑ δW m g ds R m m v Cdv C M sin ) 1 ( 2 3 ) ( 2 1 2 1 得 1 + 2 = − ⋅ θ 上式两边同除以dt C C C m g v R m m v a M sin ) 1 ( 2 3 ) ( 2 1 2 1 1 + 2 ⋅ = − ⋅ θ ( 2 3 ) 2 ( sin ) 1 1 2 2 1 R m m M m gR a C + − = 解得 θ