设：滑轮的角加速度为：α,则有：m1加速度为：a1=αr1,a2=αr2
则由动量矩定理可得：
(J1+J2)α+m1αr1^2+m2αr2^2=m1gr1-m2gr2
解得：α=（m1r1-m2r2）/[(J1+J2)+m1r1^2+m2r2^2]
故：a1=αr1=r1（m1r1-m2r2）/[(J1+J2)+m1r1^2+m2r2^2]
a2=r2（m1r1-m2r2）/[(J1+J2)+m1r1^2+m2r2^2]
拉力：T1=m1g+m1a1=m1g+m1r1（m1r1-m2r2）/[(J1+J2)+m1r1^2+m2r2^2]
T2=m2g+m2a2=m2g+m2r2（m1r1-m2r2）/[(J1+J2)+m1r1^2+m2r2^2]
注意加速度a1、a2为矢量,方向相反.