# centripetal/angular acceleration

by UrbanXrisis
Tags: acceleration
 P: 1,214 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? $$\omega v=\omega^2r$$
 P: 1,445 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 $$\alpha = \frac{\Delta \omega}{\Delta t} = \frac{v}{rt}$$
P: 1,445

## centripetal/angular acceleration

$$\omega$$ 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
 P: 1,214 do you mean omega v? That's what I have on my first post... does this mean my book was wrong?
HW Helper
P: 6,567
 Quote by UrbanXrisis Here is the question...http://home.earthlink.net/~suburban-...physics001.jpg the answer is A. why?
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
 P: 1,214 so the formula for angular acceleration is the same as the formula for centripetal acceleration?
P: 1,214
 Quote by 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? $$\omega v=\omega^2r$$
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
HW Helper
P: 6,567
 Quote by UrbanXrisis so the formula for angular acceleration is the same as the formula for centripetal acceleration?
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
 P: 1,214 yes I understand the concepts are different, but both equations can be expressed as $$\omega^2 * r$$ is that correct?
 Quote by UrbanXrisis yes I understand the concepts are different, but both equations can be expressed as $$\omega^2 * r$$ is that correct?
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 $\tau = m\alpha r^2$