7=J101+JCc02+m2v 式中 R2 R1 R R 代入后得 T=(2m1+3m,)2 应用动能定理7-70=∑W 得 (2m1+3m2)v2-0=(M R 2g·Sn6)s 所以,得 NCS/(M-m2 8R, sin O)s R1(2m1+3m2) 式(1)两边求导,(2m+3m)eac=(Mp-m28SinO)nc 解得 2(M-m,gR sin 0 R1(2m1+3m2)式中 2 2 2 2 2 1 1 2 1 2 1 2 1 C C T = J ω + J ω + m v 2 2 2 2 2 1 1 1 2 1 J = m R , J = m R 2 1 2 1 1 , , R s R v R v C C ，ω = ω = ϕ = 代入后得 2 1 2 ( 2 3 ) 4 1 C T = m + m v 应用动能定理 T − T0 = ∑ W m g s R m m v C M sin ) 1 ( 2 3 ) 0 ( 4 1 2 1 2 得 1 + 2 − = − ⋅ θ (1) ( 2 3 ) ( sin ) 2 1 1 2 2 1 R m m M m gR s v C + − = θ 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 + − = θ 所以，得 式(1)两边求导 解得