$a_t$ = $\frac{dv}{dt}$ = $\frac{d(r*\omega)}{dt}$ = $r $* $\alpha$.
$a_c$ = r*$\omega^2$ = $\frac{v^2}{r}$
These are almost all the variables and values associated with centripetal and tangential acceleration in case of circular motion.
The Q 1 : is to find angular acceleration of particle when it’s speed changes from $2$m/s to$ 4 $m/s in 4 seconds where r of the circle =0.5m.
$a_t$ = $\frac{dv}{dt}$ where $dv=4-2$ and $dt=2$.Therefore , $ a_t$=$0.5m/$ $s^2$.
Then used $r $* $\alpha$ = $a_t$= 1 rad /$ s^2$
My Q’s are that why did we not need to use the formula of centripetal acceleration here( in Q:1)to find angular acceleration. But instead used formula of tangential acceleration.
If I take a situation in which there are 3 different values of centripetal acceleration. For example , 4 , 5 ,6 m/$s^2$ and take the same values of the above Q 1 only for dv,dt and r. Then , still Angular acceleration would be having same result since there is no mention of centripetal acceleration.
Why do I think it is not possible :
- Doesn’t centripetal acceleration also affect the angular acceleration. We have $\omega$ in the formula of centripetal acceleration which is stated as amount of $\theta$ covered per unit time. Then , it means there has to be centripetal acceleration relation with $\alpha$ or angular acceleration also.
My confusion in brief is that , we know net acc = $a_t$ + $a_c$ . Right. So , I am thinking that both formulas when combined together should give angular acceleration
EDIT: Regarding a= $\frac{dv}{dt}$.