Question

圆环转动惯量M*（R^2+r^2)/2 推导

Answer 1

转动惯量定义式:mr^2

圆环,半径不恒定(由内径r变化到外径R),不妨设为x,

在距离圆心x的位置的质量m = M*2πx/π(R^2-r^2)= 2Mx/(R^2-r^2)

PS:π(R^2-r^2)是圆环面积.

所以转动惯量为:

∫mx^2 dx (x从r到R)

=∫2Mx/(R^2-r^2)x^2 dx

=2M/(R^2-r^2)∫x^3 dx (1)

=2M/(R^2-r^2) * (R^4-r^4)/4 (2)

=M*（R^2+r^2)/2

PS：(1) ∫x^3 dx = x^4/4

(2) R^4-r^4 = (R^2+r^2)(R^2-r^2)