Angular Displacement: $\omega$=$\frac{dQ}{dt}$

Angular acceleration :$\alpha = \frac{d \omega}{dt}$

$\alpha: \text { constant }$

$\frac{d \omega}{d t}=\alpha$= constant

Integrate: $\omega$ = $\alpha t +c$

at $t=t_0 , \omega=\omega_0 \Rightarrow c=w_0$

$\omega=\omega_0+\alpha t$

t = $\frac {\omega-\omega_0}{\alpha }$

$\frac {d\theta }{dt} = \omega_0 + \alpha t$

Integrate: $\theta = \omega_0 t + \frac{\alpha t^2}{2}$ + c (c: a-constant)

$\theta (t=0) = \theta_0 \Rightarrow c= \theta_0$

$\theta = \theta_0 + \omega_0 t + \frac {\alpha t^2}{2}$

v = r $\omega$

where: $(\vec{v} = \vec{\omega} \times \vec{r})$

v = $\frac{dr}{dt} = r \frac{d \theta}{dt}$

$\vec {PQ} = \vec{a} (\vec{A})$

$a_t = \frac{dv}{dt} = \frac{d}{dt} (r \omega)$

$= r \frac{d \omega }{dt} = r \alpha$

$(\vec \alpha \times \vec{r} = \vec{A}_r )$

$a_r = \frac {v^2}{r}$ (centripetal acceleration)

$= \frac{(r \omega)^2}{r} = r \omega^2 = r (\frac {d \theta }{dt})^2 = r \theta ^2$

$\vec{A}_r=-r \dot{\theta}^2 \hat{e}_r$

$\vec{a} = \vec{A} = A r \hat{e_r} + A_t \hat{e_\theta}$ :

a = $|\vec{a}|$ = $\sqrt {a_t^2 + a_r^2}$ = $\sqrt {(r \alpha)^2 + (r \omega^2 )^2}$

a = r$\sqrt{ {\alpha}^2+ {\omega}^4}$

$F_ t=m a_t$

The torque about the origin due to force $F_t$

$\tau={F_ t}r=(m a_t) r$

$a_t=r \alpha$

$\therefore tau=(m r^2) \alpha= I \alpha$

$\tau=I \alpha \Rightarrow \tau \propto \alpha$

I: Proportionality constant

$d {F_ t} = (dm) a_t$

Torque d $\tau$ due to the force $d{F_T}$ about the origin:

$d\tau = r d{F_ t} = (r dm) a_t$

• = (r dm) $(r \alpha)$

• $\tau = \alpha \int {r^2 dm}$

• $\Rightarrow \tau = I \vec{\alpha}$

$\vec{\tau} =\vec{r} \times \vec{F}$

${[\vec{\tau}] }$ = Dimension work or Energy

$\vec{\tau}$ rotates the object $d \theta$, then the work done for this infinitedecimal rotation:

$d \omega=\tau d \theta(F d x)$

$\frac{d \omega}{d t}=\tau \frac{d \theta}{d t}=\tau \omega$

Rate at which the work is done is callas instantaneous power.

P = $\tau \omega$

("P = F v")

Work energy theorem in rotational motion

we have

$\tau = I \alpha = I \frac{d \omega }{dt} =I \frac {d \omega }{dt} \frac {d \theta }{dt} = I \omega \frac {d \omega}{d \theta}$

So, $\tau d \theta = I \omega d \omega = d W$

Integrate,

$w=\int_{\theta_0}^\theta \tau d \theta=\int_{\theta_0}^\theta I \omega d$ $\omega=\frac{I}{2}(\omega^2-\omega_0^2)$

$\theta - \theta_0$ : the change in angular displacement.

$\omega-\omega_0$ : the corresponding change in angular speed.

W = $\frac{3}{2} ({\omega^2} - {\omega_0}^2)$

W = $\frac{m}{2} ({v^2} - {\omega}^2) \text{in linear motion }$

Rotational Motion

• 1. Angular velocity: $\omega = \frac {d \theta}{dt} = \dot{\theta}$
• 2. Angular acceleration: $\alpha = \frac {d \omega} {dt} = \ddot{\theta}$
• 3. Torque: $\tau = I \alpha$

Linear Motion

• 1. Linear velocity: v= $\frac {dx}{dt}$
• 2. Linear acceleration: a = $\frac {dv}{dt} = \ddot{x}$
• 3. Force: $F = M a$