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19.15βsin23)(2,gQPQPa++=19.1119.1019.14211121212121222178280418778216)(08116721mmkhgmakhvgvmavmavmmmkhghmvkhghmvmvmAAAAAAAAAA+−==+−+∗+−=∗=+−+得）式两边求导对（19.18lmglklmg=+=+=1210,)31(212160sin22λλo转向为顺时针（,2)735(321)32(,2130sin22,3421212111002221222221glVTVTlklmgVrlmlJJJTABABCBCABABCCPBCBCPABABA−=⇒+=+−=+====++=ωλλωωωωωωωωQo19.19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)13(3210)00(3121212122002222222glVTVTVvmlJJJTABCCABCCPBCBCPABABA+=⇒+=+==⇒==++=ωωωωωωQQωCωABPCωBCPBC