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本文章旨在分享基础大物知识
Preface# 转动惯量I可以类比为质量，皆如下表所示：
物理概念 平动 转动 惯性量度 质量m m m 转动惯量I I I 速度量度 速度v v v 角速度ω \omega ω 核心动力学方程 F = m a F = ma F = ma τ = I α \tau = I\alpha τ = I α 动能表达 E k = 1 2 m v 2 E_k = \frac{1}{2}mv^2 E k = 2 1 m v 2 E k = 1 2 I ω 2 E_k = \frac{1}{2}I\omega^2 E k = 2 1 I ω 2 做功形式 W = ∫ F d x W = \int F \mathrm{d}x W = ∫ F d x W = ∫ τ d θ W = \int \tau \mathrm{d}\theta W = ∫ τ d θ 状态量度 动量：p = m v p = mv p = m v 角动量：L = I ω L = I\omega L = I ω 守恒定律条件合外力 F = 0 ⟹ F=0 \implies F = 0 ⟹ 合外力矩 τ = 0 ⟹ \tau=0 \implies τ = 0 ⟹
刚体转动惯量# 转动惯量# I = ∫ r 2 d m I = \int r^2 \mathrm{d}m I = ∫ r 2 d m 定义原因# 假设一个刚体绕固定轴转动，角加速度为 α \alpha α d m \mathrm{d}m d m r r r a = r α a = r\alpha a = r α F = m a F=ma F = ma
d F = d m ⋅ a = d m ⋅ ( r α ) \mathrm{d}F = \mathrm{d}m \cdot a = \mathrm{d}m \cdot (r\alpha) d F = d m ⋅ a = d m ⋅ ( r α ) 力矩的定义是力臂乘以力（τ = r F \tau = rF τ = r F
d τ = r ⋅ d F = r ⋅ ( d m ⋅ r α ) = ( d m ⋅ r 2 ) α \mathrm{d}\tau = r \cdot \mathrm{d}F = r \cdot (\mathrm{d}m \cdot r\alpha) = (\mathrm{d}m \cdot r^2) \alpha d τ = r ⋅ d F = r ⋅ ( d m ⋅ r α ) = ( d m ⋅ r 2 ) α 刚体受到的总力矩 τ \tau τ α \alpha α
τ = ∫ d τ = ∫ ( d m ⋅ r 2 ) α = ( ∫ r 2 d m ) α \tau = \int \mathrm{d}\tau = \int (\mathrm{d}m \cdot r^2) \alpha = \left( \int r^2\,\mathrm{d}m \right) \alpha τ = ∫ d τ = ∫ ( d m ⋅ r 2 ) α = ( ∫ r 2 d m ) α τ ⃗ = r ⃗ × F ⃗ \vec{\tau} = \vec{r} \times \vec{F} τ = r × F 注意是矢量，并且单位是N·m不能写做J，J是标量;
转动定律# τ = I α \tau = I \alpha τ = I α 动能定理# W = ∫ θ 1 θ 2 τ d θ = 1 2 I ω 2 2 − 1 2 I ω 1 2 W = \int_{\theta_1}^{\theta_2} \tau \mathrm{d}\theta = \frac{1}{2}I\omega_2^2 - \frac{1}{2}I\omega_1^2 W = ∫ θ 1 θ 2 τ d θ = 2 1 I ω 2 2 − 2 1 I ω 1 2 角动量定理# 微分形式:
τ = d L d t \tau = \frac{\mathrm{d}L}{\mathrm{d}t} τ = d t d L 积分形式/冲量矩定理:
∫ t 1 t 2 τ d t = L 2 − L 1 = I ω 2 − I ω 1 \int_{t_1}^{t_2} \tau \mathrm{d}t = L_2 - L_1 = I\omega_2 - I\omega_1 ∫ t 1 t 2 τ d t = L 2 − L 1 = I ω 2 − I ω 1 角动量守恒定律# 一个旋转系统不受外力矩作用,那么角动量守恒,即:
L = I ω L = I \omega L = I ω 是一个常数
转动矢量# 不好解释,找个被投影成此轨迹的圆,也就是找那个被投影的点的转动角,即:
Δ t = Δ θ ω \Delta t = \frac{\Delta \theta}{\omega} Δ t = ω Δ θ 以下为AI:(多个方向同时做简谐运动也能用)ω x = ω y = ω \omega_x = \omega_y = \omega ω x = ω y = ω Δ ϕ = ϕ y − ϕ x \Delta\phi = \phi_y - \phi_x Δ ϕ = ϕ y − ϕ x Δ ϕ = 0 \Delta\phi = 0 Δ ϕ = 0 π \pi π Δ ϕ = π 2 \Delta\phi = \frac{\pi}{2} Δ ϕ = 2 π 90 ∘ 90^\circ 9 0 ∘ A 3 = A 4 A_3 = A_4 A 3 = A 4 ω x ≠ ω y \omega_x \neq \omega_y ω x = ω y x x x y y y
ω x ω y = Y 方向的切点数 X 方向的切点数 \frac{\omega_x}{\omega_y} = \frac{Y\text{方向的切点数}}{X\text{方向的切点数}} ω y ω x = X 方向的切点数 Y 方向的切点数 直观理解:x x x 2 π 2\pi 2 π x x x X X X y y y y y y Y Y Y t t t x x x N x = ω x ⋅ t 2 π N_x = \frac{\omega_x \cdot t}{2\pi} N x = 2 π ω x ⋅ t y y y N y = ω y ⋅ t 2 π N_y = \frac{\omega_y \cdot t}{2\pi} N y = 2 π ω y ⋅ t
平行轴定理# I = I C + m d 2 I = I_C + md^2 I = I C + m d 2 Proof:
建立坐标系：C C C ( 0 , 0 ) (0,0) ( 0 , 0 ) d m \mathrm{d}m d m ( x , y ) (x, y) ( x , y )
I C = ∫ ( x 2 + y 2 ) d m I_C = \int (x^2 + y^2)\,\mathrm{d}m I C = ∫ ( x 2 + y 2 ) d m 这条轴要与实际的轴平行( d x , d y ) (d_x, d_y) ( d x , d y ) d m \mathrm{d}m d m
r 2 = ( x − d x ) 2 + ( y − d y ) 2 r^2 = (x - d_x)^2 + (y - d_y)^2 r 2 = ( x − d x ) 2 + ( y − d y ) 2 I = ∫ ( x 2 − 2 x d x + d x 2 + y 2 − 2 y d y + d y 2 ) d m I = \int \left( x^2 - 2x d_x + d_x^2 + y^2 - 2y d_y + d_y^2 \right) \mathrm{d}m I = ∫ ( x 2 − 2 x d x + d x 2 + y 2 − 2 y d y + d y 2 ) d m I = ∫ ( x 2 + y 2 ) d m + ∫ ( d x 2 + d y 2 ) d m − 2 d x ∫ x d m − 2 d y ∫ y d m I = \int (x^2 + y^2)\,\mathrm{d}m + \int (d_x^2 + d_y^2)\,\mathrm{d}m - 2d_x \int x\,\mathrm{d}m - 2d_y \int y\,\mathrm{d}m I = ∫ ( x 2 + y 2 ) d m + ∫ ( d x 2 + d y 2 ) d m − 2 d x ∫ x d m − 2 d y ∫ y d m 第一项就是 I C I_C I C d x 2 + d y 2 = d 2 d_x^2 + d_y^2 = d^2 d x 2 + d y 2 = d 2 d 2 ∫ d m d^2 \int \mathrm{d}m d 2 ∫ d m m d 2 md^2 m d 2
