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