In most physics books ( even the University ones) i looked at, they use geometry to derive that the acceleration is [math]\frac{v^2}{r}[/math],where v is the constant speed and r is the radius (magnittute of the position vector r) of the motion.

 

Can we not use pure vector analysis to derive the above fact??

It would still be geometry/trig. What is your concern with it?

Then we can use a dimensional analysis. Time is about r/v and velocity is v so the acceleration is v/t = v^2/r.

It would still be geometry/trig. What is your concern with it?

 

proof without using geometry at all ,but only vector analysis.

 

In the proof that i read in most books they use the properties of similar triangles .

 

In proving that r.r=r^2 you can do that without using the fact that cosθ =1

Okay so this is what I was able to do:

 

[math] {\mathbf{r}} = r \hat{r} = r ( cos\theta\hat{\mathbf{x}} + sin\theta\hat{\mathbf{y}}) [/math]

[math] {\mathbf{v}} = v (-sin\theta\hat{\mathbf{x}} + cos\theta\hat{\mathbf{y}}) [/math]

 

[math] \mathbf{a} = \frac{d\mathbf{v}}{dt} = v (-\dot{\theta} cos\theta\hat{\mathbf{x}} - \dot{\theta}sin\theta\hat{\mathbf{y}}) [/math]

 

1) [math] \mathbf{a} = - v\dot{\theta}\hat{\mathbf{r}} [/math]

 

2) [math] \dot{\theta} = \frac{d( \omega t)}{dt} = \omega [/math]

 

[math] \dot{\mathbf{r}} = \mathbf{v} = r \dot{\theta}(-sin\theta\hat{\mathbf{x}} + cos\theta\hat{\mathbf{y}}) [/math]

 

3) [math] |\mathbf{v}| = v = \sqrt{ (r \dot{\theta})^2((-sin\theta\hat{\mathbf{x}} + cos\theta\hat{\mathbf{y}}))^2} = r \dot{\theta} [/math]

 

Then using 1), 2) and 3 you get [math] a = \frac{v^2}{r} [/math]

 

Is this what you were wanting?

Okay so this is what I was able to do:

 

[math] {\mathbf{r}} = r \hat{r} = r ( cos\theta\hat{\mathbf{x}} + sin\theta\hat{\mathbf{y}}) [/math]

[math] {\mathbf{v}} = v (-sin\theta\hat{\mathbf{x}} + cos\theta\hat{\mathbf{y}}) [/math]

 

[math] \mathbf{a} = \frac{d\mathbf{v}}{dt} = v (-\dot{\theta} cos\theta\hat{\mathbf{x}} - \dot{\theta}sin\theta\hat{\mathbf{y}}) [/math]

 

1) [math] \mathbf{a} = - v\dot{\theta}\hat{\mathbf{r}} [/math]

 

2) [math] \dot{\theta} = \frac{d( \omega t)}{dt} = \omega [/math]

 

[math] \dot{\mathbf{r}} = \mathbf{v} = r \dot{\theta}(-sin\theta\hat{\mathbf{x}} + cos\theta\hat{\mathbf{y}}) [/math]

 

3) [math] |\mathbf{v}| = v = \sqrt{ (r \dot{\theta})^2((-sin\theta\hat{\mathbf{x}} + cos\theta\hat{\mathbf{y}}))^2} = r \dot{\theta} [/math]

 

Then using 1), 2) and 3 you get [math] a = \frac{v^2}{r} [/math]

 

Is this what you were wanting?

 

 

Thanks ,but you have to explain each step because i am a bit confused:

 

How do you get :

 

[math] {\mathbf{v}} = v (-sin\theta\hat{\mathbf{x}} + cos\theta\hat{\mathbf{y}}) [/math]

Thanks ,but you have to explain each step because i am a bit confused:

 

How do you get :

 

[math] {\mathbf{v}} = v (-sin\theta\hat{\mathbf{x}} + cos\theta\hat{\mathbf{y}}) [/math]

 

yea we will go step by step, just keep asking questions.

 

So this is just vector v, which consists of its magnitude and direction, since v is moving in a circular path the sin and cos show which direction it is pointing:

 

 

For example if it is at the angle [math] \theta = 90 [/math] (at the top) the vector is [math] \mathbf{v} = v (-\hat{\mathbf{x}}) [/math] pointing only left in the x direction.

yea we will go step by step, just keep asking questions.

 

So this is just vector v, which consists of its magnitude and direction, since v is moving in a circular path the sin and cos show which direction it is pointing:

 

 

For example if it is at the angle [math] \theta = 90 [/math] (at the top) the vector is [math] \mathbf{v} = v (-\hat{\mathbf{x}}) [/math] pointing only left in the x direction.

 

This is my main objection , how do we prove that v is perpendicular to r

This is my main objection , how do we prove that v is perpendicular to r

The problem at hand is circular motion, so the length of the position vector with respect to the center of rotation is constant. The time derivative of any constant length vector is either zero or normal to the vector.

This is my main objection , how do we prove that v is perpendicular to r

 

I proved it here:

 

[math]\dot{\mathbf{r}} = \mathbf{v} = r \dot{\theta}(-sin\theta\hat{\mathbf{x}} + cos\theta\hat{\mathbf{y}}) [/math]

 

and then you can take a dot product of v and r to see they are perpendicular.

I proved it here:

 

[math]\dot{\mathbf{r}} = \mathbf{v} = r \dot{\theta}(-sin\theta\hat{\mathbf{x}} + cos\theta\hat{\mathbf{y}}) [/math]

 

and then you can take a dot product of v and r to see they are perpendicular.

 

Correct:

 

What is the use now of bringing in ω ?? we simply have now:

 

v=[math]r\dot{\theta}[/math] and

 

[math]

\mathbf{a} = \frac{d\mathbf{v}}{dt} = v (-\dot{\theta} cos\theta\hat{\mathbf{x}} - \dot{\theta}sin\theta\hat{\mathbf{y}})

[/math]

 

And |a|=a =[math]r\dot{\theta}^2 = \frac{v^2}{r}[/math]

[math]\omega[/math] is just another name for [math]\dot{\theta}[/math].

Yes indeed i did not notice

Also because it makes it a function of time, since [math] \theta = \omega t [/math] this is a more convenient notation when working with a physical system.

The question now is what is going to be, a if v is not constant

Then you won't have uniform circular motion. If the motion is still circular, the velocity vector will still be normal to the radial vector because, as I mentioned in post #9, the time derivative of any constant-length vector is necessarily either zero or normal to that constant-length vector.

Edited April 6, 201015 yr by D H
The vector might be stationary, in which case the derivative is zero

i mean is still :

 

a = [math]\frac{v^2}{r}[/math]??

No. That result pertains to uniform circular motion only. In the case of non-uniform circular motion, the acceleration vector must necessarily have some non-zero component normal to the radial vector (or parallel to the velocity vector, same thing).

But remember it just really depends how v is changing, because here v is NOT constant. The magnitude is constant, but the direction is always changing, otherwise we would not have an acceleration. Now you could get a v where the the magnitude was changing, but the direction was the same (circular motion) and [math] a = \frac{v^2}{r}[/math] would still hold, but the v (magnitude) could be a function of something else. Either way the conditions will not be exactly the same.

Not true. In uniform circular motion the magnitude of the velocity vector is, by definition, constant. In non-uniform circular motion the magnitude of the velocity vector is not constant.

 

In terms of polar coordinates, the position vector is [math]\vec r = r \hat r[/math]. Differentiating with respect to time, [math]\vec v = r\dot{\theta}\hat{\theta}[/math]. Note that there is no [math]\dot r \hat r[/math] term because r (the magnitude of the position vector) is constant. Differentiating velocity to obtain the acceleration vector, [math]\vec a = r \ddot{\theta}\hat{\theta} - r \dot{\theta}^2 \hat r[/math].

 

In uniform circular motion, [math]\dot{\theta}[/math] is constant, so the acceleration vector is purely radial, and its magnitude is equal to [math]v^2/r[/math]. In non-uniform circular motion, [math]\dot{\theta}[/math] varies with time. While the radial component of the acceleration vector is still related to the velocity vector via [math]v^2/r[/math], the acceleration vector by necessity has a non-zero transverse component when the motion is not uniform.

Not true. In uniform circular motion the magnitude of the velocity vector is, by definition, constant. In non-uniform circular motion the magnitude of the velocity vector is not constant.

 

I meant to say the vector v is not constant, while of course the magnitude is. And that is what I said but nonetheless we are both correct in what we have stated.

DH answer is absolutely right. Acceleration is a vector and may have not only a radial but also a tangent component even with constant r. Its absolute value may be different from v^2/r.

Edited April 6, 201015 yr by Bob_for_short

I meant to say the vector v is not constant, while of course the magnitude is.

The magnitude of the velocity vector is constant in uniform circular motion, but not in non-uniform circular motion. Think of a vehicle performing a loop-de-loop on a vertically-oriented circular track. The vehicle might slow down as it climbs toward the top of the track and speed up once it has passed the top. The motion is still circular (the vehicle is following a circular track), but the motion is not uniform.

I know what you are saying, and that is not what I meant. I'm not meaning to argue, because of course you are correct. I merely meant the magnitude could be changing and still be following a circular path with radius r, for example: if the angular frequency was changing, but the distance from the center. It's really a mute point and no longer helps the poster.

Look for the following derivation where the circular motion is not uniform.

 

WE have r.v=0

 

Differentiate with respect to time and:

 

[math]\frac{d({\mathbf{r.v}})}{dt}= \frac{d{\mathbf{r}}}{dt}.{\mathbf{v}} + {\mathbf{r.a}}[/math]=

 

= v.v +r.a=0 => r.a= [math]-v^2[/math] =>

 

=> [math]r\hat{\mathbf{r}}[/math].a=[math]-v^2[/math] => [math]\hat{\mathbf{r}}[/math].a = [math]\frac{-v^2}{r}[/math]

 

multiply by :[math]\hat{\mathbf{r}}[/math] both sides of the equation and:

 

 

a= -[math]\hat{\mathbf{r}}\frac{v^2}{r}[/math]