解:J=mR2=60×0.252 2 =3.75kgm2 1000 f t=0Oo=2n=2× 60 =1047r/s t=5=0 O-0-104.7 C =-20.9r/S 5 F(l1+l2)-Nl1=0 fR=J=NR口N R 1da=314N l,t12 uR 结束目录=3.75kg.m2 t =0 1000 60 ω0 = 2πn= 2π× =104.7 r/s t =5 ω =0 f N N F f l 1 l 2 J mR 2 = = 60×(0.25) 解： 2 ω 104.7 20.9 r/s2 5 0 a t = = = ω0 F (l 1 + l 2) N l 1= 0 f R= J a = m NR l 1 F = l 1 + l 2 mR J a = 314N m N = R J a 结束 目录