曲線運動 物理求解

Question

一圓盤(r = 20 cm)由靜止到轉速180 rpm 需時為2 秒，求(a) 2 秒內圓盤共轉

了多少圈？ (b) 又當轉速為120 rpm 時，圓盤邊緣之點的加速度為何？

Answer 1

最佳解答

一圓盤r=20cm,由靜止到轉速w=180rpm,需時為2秒，求(a) 2 秒內圓盤共轉了多少圈？ w = 180*π/60 = 3π (rad/s)α = w/t = 3π/2 (rad/s^2)θ = 0.5α*t^2= 0.5(w/t)*t^2= w/2t= 3π/2*2= 3π/4 (rad)

(b) 又當轉速為w=120 rpm 時，圓盤邊緣之點的加速度為何？ w = 120*π/60 = 2π (rad/s)α = 3π/2 = constantθ = w^2/α = 4π^2*2/3π = 8π/3= 2π + 2π/3= 120°=> n = (cos210°,sin210°) = (-√3/2, -1/2)=> a = r*α*n= 0.2*(3π/2)*(-√3/2,-1/2)= 0.3π*(-√3/2,-1/2) = 0.3π*[(-√3/2)*i - (1/2)*j] (m/s)