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Centripetal/angular acceleration

by UrbanXrisis
Tags: acceleration
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UrbanXrisis
#1
Mar1-05, 03:35 PM
P: 1,206
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 [tex]\omega^2r[/tex]
where w is angular speed and r is the radius

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

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

how are they equal? [tex]\omega v=\omega^2r[/tex]
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stunner5000pt
#2
Mar1-05, 04:58 PM
P: 1,440
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 [tex] \alpha = \frac{\Delta \omega}{\Delta t} = \frac{v}{rt} [/tex]
UrbanXrisis
#3
Mar1-05, 05:17 PM
P: 1,206
Here is the question...http://home.earthlink.net/~suburban-...physics001.jpg

the answer is A. why?

stunner5000pt
#4
Mar1-05, 05:21 PM
P: 1,440
Centripetal/angular acceleration

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

since omega = v/r
a = omega r
UrbanXrisis
#5
Mar1-05, 05:44 PM
P: 1,206
do you mean omega v? That's what I have on my first post... does this mean my book was wrong?
Andrew Mason
#6
Mar1-05, 06:36 PM
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Quote 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, [itex]v [/itex] or [itex] \omega r[/itex].

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 [itex]d\theta = ds/r = \frac{vdt}{r}[/itex] in that time.

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

Now, the new velocity vector at t=dt is the same length as at t=0 but pointed [itex]d\theta[/itex] 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:

[itex] dv = vsin(d\theta)[/itex] which approaches the limit of [itex] dv = vd\theta[/itex] as [itex]d\theta \rightarrow 0[/itex].

This means: [tex]dv = vd\theta = \omega r d\theta = \omega^2r dt[/tex] so

[tex]dv/dt = a_{centripetal} = \omega^2r[/tex]

AM
UrbanXrisis
#7
Mar1-05, 06:48 PM
P: 1,206
so the formula for angular acceleration is the same as the formula for centripetal acceleration?
UrbanXrisis
#8
Mar1-05, 10:23 PM
P: 1,206
Quote 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 [tex]\omega^2r[/tex]
where w is angular speed and r is the radius

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

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

how are they equal? [tex]\omega v=\omega^2r[/tex]
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
#9
Mar1-05, 10:41 PM
Sci Advisor
HW Helper
P: 6,654
Quote 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 ([itex]a_c = -\omega^2r[/itex]) is always non-zero if there is circular motion.

AM
UrbanXrisis
#10
Mar1-05, 10:49 PM
P: 1,206
yes I understand the concepts are different, but both equations can be expressed as [tex]\omega^2 * r[/tex]

is that correct?
Andrew Mason
#11
Mar2-05, 06:41 AM
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P: 6,654
Quote Quote by UrbanXrisis
yes I understand the concepts are different, but both equations can be expressed as [tex]\omega^2 * r[/tex]

is that correct?
No. Angular acceleration has nothing to do with [itex]\omega[/itex]. It depends on torque not angular speed, just as acceleration is a function of force not velocity.

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

AM


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