r = acosωti + bsinωtj
v = dr/dt = -aωsinωti + bωcosωtj
角動(dòng)量
L = r×p = r×mv
= m(acosωti + bsinωtj)×(-aωsinωti + bωcosωtj)
= m(abωcos^2ωt + abωsin^2ωt)k
= mabωk 常量
質(zhì)點(diǎn)所受對(duì)原點(diǎn)的力矩M為
M = dL/dt = 0w不是常數(shù)嗎？為什么還要對(duì)其運(yùn)算沒有啊...不好意思了，我看錯(cuò)題目了沒事，能幫到你就好！(^_^)........