Effect of centripetal acceleration on angular acceleration

When working with tangential and normal directions/coordinate system in the circular motion, you can thing of:

• tangential acceleration: the change in the magnitude of velocity
• normal acceleration: the change in the velocity direction

In a case of uniform circular motion, the magnitude of the tangential acceleration is zero, because the magnitude of the velocity does not change.

The derivation for uniform circular motion can be found:

$$vec{a} = frac{dvec{v}}{dt} = vec{a}_t + vec{a}_n$$ $$vec{a} = vec{a}_t + vec{a}_n$$ $$vec{a} = vec{alpha}times vec{r} + vec{omega}times left(vec{omega}times vec{r}right)$$

can be found on any textbook.

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$

You don't use the centripetal acceleration because there is an increase in the magnitude of velocity. As I mentioned above the tangential component is responsible for the changes in the magnitude of the velocity.

Doesn’t centripetal acceleration also affect the angular acceleration. We have ω in the formula of centripetal acceleration which is stated as amount of θ covered per unit time.

The angular acceleration is independent of the rotational velocity. The presence of ω in the formula does not mean that the angular acceleration $dot{omega}$ is affected. In the same way that a car can accelerate with 1[m/s^2] independently of whether its velocity is 5[m/s] or 50[m/s] or 500000[m/s].