Centripetal/angular acceleration

[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? $$\omega v=\omega^2r$$

 

[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 $$\alpha = \frac{\Delta \omega}{\Delta t} = \frac{v}{rt}$$

 

[UrbanXrisis]

Here is the question...http://home.earthlink.net/~suburban-xrisis/physics001.jpg

the answer is A. why?

 

[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

 

[UrbanXrisis]

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

 

[Andrew Mason]

Here is the question...http://home.earthlink.net/~suburban-xrisis/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

 

[UrbanXrisis]

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

 

[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

 

[Andrew Mason]

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

 

[UrbanXrisis]

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

is that correct?

 

[Andrew Mason]

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 $\tau = m\alpha r^2$

AM