Effect of centripetal acceleration on angular acceleration

Question

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

1. 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= $\frac{dv}{dt}$.

Answer 1

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:

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

can be found on any textbook.

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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.

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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].

Answer 2

$a_t = \frac{dv}{dt} = \frac{d(r∗ω)}{dt} = r* α$. Here, $v$ is a vector.

$a_c = r*ω^2 = \frac{v^2}{r}$. Here $v$ is a scaler.

In your numerical example, both changes in velocity and time are given, so you go ahead to calculate $a_t$ using the information provided. However, there is no information for the $v$ and $\omega$, so, you need to figure them out by:

1. Find the centripetal force $F_c$.

2. Solve $a_c = F_c/mass$.

3. Then solving $v$ and $\omega$ by the equation $a_c = r*ω^2 = \frac{v^2}{r}$.

Now you can find the acceleration of the particle by combining $a_t$ & $a_c$

Please review the linked article, which may help you to better understand the current problem, and the other similar question of yours. http://getyournotes.blogspot.com/2011/07/calculating-circular-motion-questions.html