我1個(gè)直徑1M的圓柱,每分鐘轉(zhuǎn)100圈,每秒轉(zhuǎn)1.7圈.所以,這個(gè)圓柱的角速度就是 w = 100/60 ≈ 1.7 r/s.
如果,從0到勻速時(shí)為2秒,那么,該段時(shí)間的角加速度就是：b = (1.7-0)/2 = 0.85 r/s2.
這個(gè)圓柱的轉(zhuǎn)動(dòng)慣量,可以通過(guò)積分的方法計(jì)算出來(lái),J = 0.5 m r^2 = 0.5x40kg x 0.5^2 = 5 kg r/s2.
這段時(shí)間的平均轉(zhuǎn)矩 M = J x b = 5x 0.85 = 4.25 Nm.（假設(shè)圓柱勻質(zhì)、旋轉(zhuǎn)加速均勻,轉(zhuǎn)動(dòng)軸在幾何軸上）??????и???????????2?????/???????????????2??2??*1.7R/S--------------=5.38 2??????ü???2?У??????????????????????豸40KG????????????? 100RPM/MIN????????????2??????????????5*5.38=27NM?????????????????????????0.1??5*106.76=533.8NM???????????????????????????100RPM/MIN????????????????????????