Centripetal/angular acceleration

1. Mar 1, 2005

UrbanXrisis

I was doing a physics problem and realized that the formula for angular acceleration was the same as the formula for centripetal acceleration (in terms of angular speed)

They both are $$\omega^2r$$
where w is angular speed and r is the radius

Why is that so? When I tried to derive this I got...

$$a_{centripetal}=\frac{v^2}{r}$$
$$a_{centripetal}=\frac{v}{r}v$$
since v/r=w then...
$$a_{centripetal}=\omega v$$

how are they equal? $$\omega v=\omega^2r$$

2. Mar 1, 2005

stunner5000pt

You've got it wrong ,check your book

true centripetal acceleration is v^2 /r BUT ANGULAR acceleration is the something spinning up faster or spinning down slower. Angular acceleration is $$\alpha = \frac{\Delta \omega}{\Delta t} = \frac{v}{rt}$$

3. Mar 1, 2005

UrbanXrisis

Last edited by a moderator: May 1, 2017
4. Mar 1, 2005

stunner5000pt

$$\omega$$ is the ANGULAR VELOCITY (or frequency) and $$\omega = \frac{v}{r}$$
the CENTRIPETAL ACCELERATION is a = v^2 /r

since omega = v/r
a = omega r

5. Mar 1, 2005

UrbanXrisis

do you mean omega v? That's what I have on my first post... does this mean my book was wrong?

6. Mar 1, 2005

Andrew Mason

You have to work out the change in velocity as a function of its tangential speed, $v$ or $\omega r$.

Draw a diagram of the velocity vector of a unit mass at time 0. Then draw its velocity vector after a time dt. The mass turns through an angle $d\theta = ds/r = \frac{vdt}{r}$ in that time.

Also remember that $v = 2\pi r/T = \omega r$ and $d\theta = \omega dt$

Now, the new velocity vector at t=dt is the same length as at t=0 but pointed $d\theta$ to the original. The difference is the change in velocity or dv and is directed toward the centre of the circle along the radius. You can see from the diagram that:

$dv = vsin(d\theta)$ which approaches the limit of $dv = vd\theta$ as $d\theta \rightarrow 0$.

This means: $$dv = vd\theta = \omega r d\theta = \omega^2r dt$$ so

$$dv/dt = a_{centripetal} = \omega^2r$$

AM

Last edited by a moderator: May 1, 2017
7. Mar 1, 2005

UrbanXrisis

so the formula for angular acceleration is the same as the formula for centripetal acceleration?

8. Mar 1, 2005

UrbanXrisis

what is wrong with my method? I subbed in v/r for omega but got r*omega
I understand that if I subbed v=omega*r then the equation would come out correct

9. Mar 1, 2005

Andrew Mason

No. They are two distinct concepts; two quite different vector quantities with different directions.

For a mass moving in a curved path, centripetal acceleration is radial - toward the centre. Tangential acceleration - in the direction of travel gives rise to non-zero angular acceleration.

The centripetal acceleration ($a_c = -\omega^2r$) is always non-zero if there is circular motion.

AM

Last edited: Mar 1, 2005
10. Mar 1, 2005

UrbanXrisis

yes I understand the concepts are different, but both equations can be expressed as $$\omega^2 * r$$

is that correct?

11. Mar 2, 2005

Andrew Mason

No. Angular acceleration has nothing to do with $\omega$. It depends on torque not angular speed, just as acceleration is a function of force not velocity.

The definition of angular accelaration is $\alpha = a/r = f/mr = fr/mr^2 = \tau/mr^2$. So $\tau = m\alpha r^2$

AM

Last edited: Mar 2, 2005