Question

飞轮半径为0.4m,自静止启动,其角加速度为0.2转每秒,求t=2s时边缘上

Answer 1

ω=ω0+a'tω0=0,t=2s,a'=0.2 × 2pi弧度/s^2=1.257弧度/s^2ω=a't=1.257弧度/s^2×2s=2.514弧度/s切向速度v=ωr=0.4mx1.257弧度/s=1m/s法向加速度a.=ω^2r=(2.514弧度/s)^2 × 0.4m=2.528m/s^2切向加速度a''=dv/dt=...