• The change in angle $\Delta\theta$ is the angular displacement

$$ \begin{aligned} \text{Angular v ; rigid body: } \vec\omega &=\frac{\text{d}\vec\theta}{\text{d}t} \\ \text{for 1 direction: } \omega_{z}&=\frac{\text{d}\theta_z}{\text{d}t} \\ \text{Angular acceleration: }\vec\alpha&=\frac{\text{d}\vec\omega}{\text{d}t}=\frac{\text{d}^2\vec\theta}{\text{d}t^2} \end{aligned} $$

💼 Case:

• constant angular acceleration:

$$ \begin{aligned}\omega &=\omega_0+\alpha t\\\theta &=\theta_0+\omega_0 t+\frac{1}{2}\alpha t^2\\\omega^2&=\omega_0^2+2\alpha (\theta-\theta_0)\end{aligned} $$

The linear velocity of a particle is its tangent to the circular path so:

$$ v_t = \frac{\text{d} s_i}{\text{d} t}=r_i \frac{\text{d} \theta}{\text{d} t}=r_i\omega $$

The tangantial acceleration is:

$$ \begin{aligned}a_t&=\frac{\text{d} v_t}{\text{d} t}=r_i \frac{\text{d} \omega}{\text{d} t}\\a_t&=r_i\alpha\end{aligned} $$

The radial acceleration is the centripetal acceleration which is facing inward:

$$ a_r = \frac{v_t^2}{r_i}=\frac{(r_i\omega)^2}{r_i}=\omega^2r_i $$

Vector Product (cross product):

📝 Definition:

$$ \text{if }\left\{\begin{aligned} a&=a_x\hat{i}+a_y\hat{j}+a_z\hat{k} \\ a&=b_x\hat{i}+b_y\hat{j}+b_z\hat{k} \end{aligned} \right . \\ \begin{aligned} \vec a \times \vec b&=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ a_x & a_y & a_z \\ b_x & b_y & b_z \end{vmatrix} \\&=(a_y b_z -a_z b_y)\hat{i} \\&\;+(a_z b_x -a_x b_z)\hat{j} \\&\;+(a_x b_y -a_y b_x)\hat{k} \\\vert\vec a \times \vec b \vert&=\vert \vec a \vert\vert \vec b \vert \sin \theta_{ab} \end{aligned} $$

Properties:

$$ \begin{aligned} \hat{i} \times \hat{i} &=0&\hat{j} \times \hat{j} &=0&\hat{k} \times \hat{k}&=0\\ \hat{i} \times \hat{j} &=\hat{k}&\hat{j} \times \hat{k}&=\hat{i} &\hat{k}\times \hat{i} &=\hat{j} \\ \hat{j} \times \hat{i} &=-\hat{k}&\hat{k} \times \hat{j} &=-\hat{i} &\hat{i} \times \hat{k}&=-\hat{j} \,\\ \end{aligned} $$

$$ \vec a \times \vec a =0\quad \vec a \times \vec b =-\vec b \times \vec a $$

Rotational Kinetic Energy:

⚠️ only works for a fixed axis

• Each particle has a Kinetic energy $K_i=\frac{1}{2}m_iv^2_i$ so for a rotation $K_i=\frac{1}{2}m_ir^2_{\perp\,i}\omega^2$ so:

$$ K=\frac{1}{2}\left(\sum_i m_i r_{\perp i}^2\right)\omega^2 $$