#ed_op#DIV#ed_cl#角速度 = 角加速度*時間#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#轉過的弧度 = (1/2)*(角加速度)*(時間^2)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# = (1/2)*{[(角加速度)*(時間)]^2 /(角加速度)} #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# = (1/2)*[(角速度)^2/(角加速度)]#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#就是這樣#ed_op#/DIV#ed_cl#6213 寫到:而轉過的弧度 = (1/2)*[(角速度)^2/(角加速度)]#ed_op#BR#ed_cl##ed_op#BR#ed_cl#什麼意思呢?為什麼?
#ed_op#DIV#ed_cl#1. 題目要改吧。#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#2. 由題目得知,向心加速度大小/切線加速度大小 = 4/3,而#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# 向心加速度大小 = 半徑*(角速度)^2#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# 切線加速度 = 半徑*(角加速度)#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# 故向心加速度大小/切線加速度大小 = (角速度)^2/(角加速度) = 4/3#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# 而轉過的弧度 = (1/2)*[(角速度)^2/(角加速度)] = (1/2)*[4/3] = 2/3 #ed_op#/DIV#ed_cl#6213 寫到:#ed_op#DIV#ed_cl#當一個物體對一固定軸以#ed_op#FONT color=#ff0000#ed_cl#等角加速度#ed_op#/FONT#ed_cl#由靜止開始轉動,該點的合加速度與切線加速度成五十三度角的瞬間,此飛輪恰好轉過若干弧度?#ed_op#BR#ed_cl##ed_op#BR#ed_cl#Ans:2/3#ed_op#BR#ed_cl#可以給我詳解嗎?#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#或說明一下怎麼做都好#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl# #ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl#謝~#ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl#