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※ 引述《bpbpbp (= =)》之銘言: : A particle of mass M moves in the xy plane. Its coordinates as a function of : time are given by x(t)=At^3; y(t)=Bt^2-Ct, where A,B,C are constants. : Find its angular momentum about the origin. : 答案是 (2C-Bt)MAt^3 : 我想不太通,請版上高手解答。感恩! → → → → → L= x ╳ p =Mx ╳v → x= (At^3,Bt^2-Ct,0) → v =(3At^2,2Bt-C,0) → → → → → L= x ╳ p =Mx ╳v =M(0,0,[At^3][2Bt-C]-[Bt^2-Ct][3At^2]) =M(0,0,2ABt^4-ACt^3-3ABt^4+3ACt^3) =(0,0,MA(2C-Bt)t^3) --
reyhua:[問題]SoBad到底有幾線? 05/12 18:19
MIOisMyBride: 前列腺 05/12 18:21
champions: __護腺 05/12 18:26
sbboky: 扁條線 發炎中 更.... 05/12 18:32
Keo:拍拍  05/12 18:34
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ntust661:推~~ 07/27 00:21
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bpbpbp:讚 07/27 00:22
dreamingaway:來秀簽名檔的 07/27 07:35