T=-,0+=J0+-m 式中 J,=m,RJ R2 C R1 9÷J 2-C R 代入后得 7=(2m1+3m2)v2 应用动能定理7=∑W 得 (2m1+3m2)v2-0=(M R 2g·Sn6)s 所以,得 (M-m2gR, sin e)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 1 1 1 2 1 J = m R , J C = 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