19.19解取=0°为零势点 TG=0,V0=2 mg=SIn60°+k2=mgl(1+√3),41=l (1)=30°时 T=J4°a ooc V=2 mg -sin30°+k2,l2=(2-√3) 1|3(53-g,转向为顺时针 Tn+V=T+V1→O=2V21 (2)6=0°时 T2=JA048+)Jp o8c+lJcoc3120AB(Vc=0=0c=0) 2=0 7o+1o=72+H2→o 133+g,转向为顺时针 BCk mgl l l = mg + = + 1 = 2 0 1 (1 3) , 2 1 2 1 sin 60 2 2 λ λ o 转向为顺时针 （ , 2 3(5 3 7) 2 1 , (2 3) 2 1 sin 30 2 2 , 3 4 2 1 2 1 2 1 0 0 1 1 2 2 1 2 2 2 2 2 2 1 g l T V T V k l l V mg r l T J J J ml AB C AB AB BC C AB C BC P BC AB P AB A − + = + ⇒ = = + = − = + + = = = ω λ λ ω ω ω ω ω ω ω ω Q o 19.19 解 取θ =0o 为零势点 T0=0 ,V (1) θ=30o 时 (2) θ =0o 时 ,转向为顺时针 2 3( 3 1) 2 1 0 ( 0 0) 3 1 2 1 2 1 2 1 0 0 2 2 2 2 2 2 2 2 2 g l T V T V V T J J J ml v AB C AB C C C BC P BC AB P AB A + + = + ⇒ = = = + + = = ⇒ = ω ω ω ω ω ω Q Q ωC ωAB PC ωBC PBC