1D Kinematics¶

Constant Acceleration¶

$v_f=v_i+at$

$x_f=x_i+v_it+\frac{1}{2}at^2$

$v_f^2-v_i^2=2a(x_f-x_i)$

$x_f=x_i+v_{avg}t$

$a_c=\frac{v^2}{r}=r\omega^2$ , $a_c$ points to the centre of the circle

Period: $T=\frac{2\pi r}{v}$

Angular velocity: $\omega=\frac{2\pi}{T}=\frac{v}{r}$ ( $\omega$ in rad/s)

Equation	Explain
$\tau = rF\sin\theta$	$r$ : distance between the point where force is applied and the rotating center $F$ : raw amount of force $\theta$ : angle between $F$ and $r$
$\alpha=\frac{\tau}{I}$	$\alpha$ : angular acceleration $I$ : moment of inertia / rotational inertia
$\alpha=\frac{a}{r}$
$I=mr^2$	moment of inertia of a point mass
$I=\frac{1}{3}mr^2$	moment of inertia of a rod about its end
$I=\frac{1}{2}mr^2$	moment of inertia of a circle

Equation	Explain
$U_s=\frac12kx^2$	Elastic Potential Energy
$U_g=mgh$	Gravitational Potential Energy

Equation	Explain
$p=mv$	$p$ : linear momentum ( $\frac{kg\times m}{s}$ )
$I = \Delta p=F\Delta t$ (for constant force)	$\Delta p / I$ : impulse
$I\approx\int_{t_i}^{t_f}F_{\text{contact}}$	contact inpulse approximation (ignore all other forces)
$E_k=\frac{p^2}{2m}$
$\Sigma F = \frac{dp}{dt}$	Net external force = rate of change of momentum
$m_1v_1+m_2v_2=(m_1+m_2)v_f$	perfectly inelastic collision (conservation of momentum)
$v_{1i}-v_{2i}=v_{2f}-v_{1f}$	perfectly elastic collision
$m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}$	perfectly elastic collision (conservation of momentum)
$m_1v_{1i}^2+m_2v_{2i}^2=m_1v_{1f}^2+m_2v_{2f}^2$	perfectly elastic collision (conservation of kinetic energy)
$v_{1f}=v_{1i}\times\frac{m_1-m_2}{m_1+m_2}$ $v_{2f}=2v_{1i}\times\frac{m_1}{m_1+m_2}$	perfectly elastic collision when $v_{2i}=0$

Equation	Explain
$Mv_{CM}=\sum m_iv_i=p_{total}$	Total mass x speed of CM = sum of momentum of all particles = momentum of the object
$\sum(\text{external forces})=M_{total}a_{CM}$

Equation	Explain
$\omega = \frac{v}{R}$	$\omega$ : rotating speed
$\alpha=\frac{a_{CM}}{R}$	$\alpha$ : angular acceleration
$E = \frac12mv^2+\frac12I(\frac{v}{R})^2$	kinetic energy for rolling motion

Equaiton	Explain
$L=I\omega$	$L$ : Angular momentum ( $\frac{kg\times m^2}{s}$ )
$	L
$I_f\omega_f=I_i\omega_i$	conservation of angular momentum

Last update: December 12, 2021
Created: September 14, 2021