19.11 19.10 19.15 βsin 23 )(2 , g QP QP a + + = 19.14 121 21 2 1 2 1 2 2 2 1 0 4 1 8 7 78 216 )(0 8 1 16 7 2 1 khvgvmavmavm mm khghm v khghmvmvm AAAAAA A AA =+?+? + ? = ?=+?+ ）式两边求导对（ 19.18 1 28 khgm a A ? =得 21 78 mm + lmglk l mg =+=+= 1 2 10 ,)31( 2 1 2 1 60sin 2 2 λλ o 转向为顺时针 （ , 2 )735(3 2 1 )32(, 2 1 30sin 2 2 , 3 4 2 1 2 1 2 1 1100 2 2 21 22222 1 g l VTVT lk l mgV r l mlJJJT AB ABCBCABABC C PBC BC PAB AB A ? =?+=+ ?=+= ===++= ω λλ ωωωωωωωω Q o 19.19 解 取θ =0 o 为零势点 T 0 =0 ,V (1) θ=30 o 时 (2) θ =0 o 时 转向为顺时针, 2 )13(3 2 1 0 )00( 3 1 2 1 2 1 2 1 2200 2 22222 2 g l VTVT V vmlJJJT AB CCABC C PBC BC PAB AB A + =?+=+ = =?==++= ω ωωωωω Q Q ω C ω AB P C ω BC P BC