Please help, derivation of a=v^2/r of the circular motion using calculus

[kevinet]

So ya, the question that I'm trying to solve is to derive a=v^2/r using calculus.

I am very intrigued when in class my teacher derive angular velocity \omega

he did it very elegantly by starting off with the equation

s=r\theta (s is the distance traveled)
then
ds/dt = d (r\theta)/dt (differentiating both sides by d/dt)
hence
v/r = d\theta/dt (which is how we derive angular velocity :D)

So I am intrigued to how they come up with a=v^2/r and try randomly any calculus methods that I have at my disposal (high school calculus)

Because I don't know what equation they started with, I decided to backtrack from the already established equation a= v^2/ r

if a=v^2/r

dv/dt = (v) (v/r) substituting v/r into d\theta/dt

dv/dt = (v) (d\theta/dt) cancelling out dt from both equation and rearranging

(1/v) dv = d\theta integrating both sides (separable differential equation)

ln v = \theta rearranging

v = e^\theta Which is a relationship that doesn't make any sense?


Because I'm doubtful of that equation I decided to differentiate back with respect to dt
checking answer:

v= e^\theta

dv/dt = d (e^\theta)/dt

a= (e^\theta) (d\theta/dt) chain rule

a= (v) (v/r) substituting e^\theta=v and d\theta/dt = v/r

hence we came up with the original equation a=v^2/r which mathematically works?


I am sure that I have made a mistake somewhere along the way, but I checked and rechecked, and I can't figure out where, the math works, so it should be okay?

Lots of thanks for those who could point out where it went wrong, even more thanks if you could show me how a=v^2/r is derived using calculus...

Cheers ^o^

Kevin

 

[bp_psy]

There is a slight problem a=(v^2)/r does not refer to the angular or tangential acceleration
$\alpha=\frac{d\omega }{dt}$ and a_{tan}=r\alpha (It's just this) .Your formula is for the centripetal acceleration which is a totally different thing and can't be derived as you did.

 

[HallsofIvy]

Be careful exactly how you state this- that formula is not true for general circular motion- it is true only for circular motion at a constant speed.

Set up a coordinate system in the plane of motion with origin at the center of rotation. Take t= 0 at the instant when object is at point (r, 0). Then circular motion, at the constant angular speed \omega, is given by x= r cos(\omega t), y= r sin(\omega t) or position vector
\vec{p}= r cos(\omega t)\vec{i}+ r sin(\omega t)\vec{j}.

Then the velocity vector is given by
\vec{v}= \vec{p}/dt= -r\omega sin(\omega t)\vec{i}+ r\omega cos(\omega t)\vec{j}
which has magnitude
\sqrt{r^2 \omega^2(sin^2(\omega t)+ cos^2(\omega t))}= r\omega
the "tangential speed" of the object.

Differentiating again,
\vec{a}= d\vec{v}/dt= -r\omega^2 cos(\omega t)\vec{i}- r\omega^2 sin(\omega t)\vec{j}
The magnitude of that vector, the scalar acceleration, is
\sqrt{r^2\omega^4(cos^2(\omega t)+ sin^2(\omega t))}= r\omega^2
Now, since the "tangential speed", v (the magnitude of the velocity vector), was v= r\omega, we have \omega= v/r so |\vec{a}|= r\omega^2= (r)(v/r)^2= v^2/r.

Notice also that the acceleration vector is -\omega^2 times the position vector- the acceleration is pointing opposite to the postion vector, back toward the center of rotation. (Again, that is only true for circular motion at a constant speed.

 

[Cleonis]

Interestingly, in the case of uniform circular motion the high symmetry of the problem allows a fast derivation.

Let the period of one revolution be called 'T'.

I will use the symbol \Delta for a change over the duration of going through a full circle.

We have the following relation:

v = \frac{2 \pi r}{T}

In the case of uniform circular motion we have the following:

\frac{dr}{dt} = \frac{\Delta r}{T}

That is, with uniform circular motion dr/dt is uniform and equal to the total change of position divided by the duration of a whole revolution.

\frac{\Delta r}{T} = \frac{2 \pi r}{T}

When the position vector goes through a 360 turn then the total change of the position vector is 2 \pi r . The 2 \pi comes from that 360 turn.

Next step:
In the case of uniform circular motion the velocity vector is at all times perpendicular to the position vector. It goes through the same 360 turn. So the expression for the total change of the velocity vector must have the same form as the one for the total change of the position vector.

\frac{\Delta v}{T} = \frac{2 \pi v}{T}

In the case of uniform circular motion the magnitude of the acceleration is uniform:

a = \frac{dv}{dt} = \frac{\Delta v}{T}

Removing 'T' from the expression with a subsitution:

\frac{\Delta v}{T} = \Delta v \frac{v}{2 \pi r} = \frac{v^2}{r}

Conclusion: in the case of uniform circular motion we have:

a = \frac{v^2}{r}

 

[matonski]

When the position vector goes through a 360 turn then the total change of the position vector is 2 \pi r . The 2 \pi comes from that 360 turn.

Next step:
In the case of uniform circular motion the velocity vector is at all times perpendicular to the position vector. It goes through the same 360 turn. So the expression for the total change of the velocity vector must have the same form as the one for the total change of the position vector.

Isn't the change in the velocity and position vectors after a full 360 degree turn actually zero?

 

[Cleonis]

Isn't the change in the velocity and position vectors after a full 360 degree turn actually zero?

You are referring to position relative to some initial point. Sure, after you've turned through 360 degrees you are back at the position where you started. Counting things that way is not what I intended in the derivation I wrote down.

In retrospect I should have used the expression 'accumulated change of position' instead of 'the total change of position'.

 

[Cleonis]

I'm presenting a reformulation of the derivation of a=v^2/r in the case of uniform circular motion. It's somewhat more in the direction of a mathematically rigorous derivation.

I will use the letter 'T' to stand for the total duration of going through a full circle.
I define a shorthand notation: \sum is to be read as summation over precisely a circle. In that shorthand notation: \sum \Delta t = T

When 'r' and 'v' are used as scalar quantities we have the relation :

v = \frac{2 \pi r}{T}

Now I write down a counterpart of that with 'r' and 'v' taken as vector quantities.

|\vec v| = \lim_{\Delta t \to 0} \frac{\sum | \Delta \vec r|}{T}

(I take it the notation |\vec x| for the length of the vector x is known.)
The position vector \vec r is a radial vector with a fixed origin. \vec r sweeps out a circle, \Delta \vec r is the change of \vec r during time interval \Delta t.

We have:

\lim_{\Delta t \to 0} \sum | \Delta \vec r| = 2 \pi | \vec r|

Hence:

|\vec v| = \frac{2 \pi |\vec r|}{T} There is a one-on-one relation between the instantaneous position vector and the instantaneous velocity vector. The two are all times perpendicular to each other.

Likewise, there is a one-on-one relation between the acceleration vector and the velocity vector; the two are perpendicular at all times. (And we know that the acceleration vector is at all times opposite in direction to the position vector)

Acceleration relates to velocity as velocity relates to position; the operation is taking the time derivative.

That is sufficient to imply the following:

|\vec a| = \lim_{\Delta t \to 0} \frac{\sum |\Delta \vec v|}{T}We are accustomed to thinking of the set of all position vectors relative to some point of origin as a space. If you will, a position space.
The necessary step is to transfer that thinking to velocity. Think of the set of all velocity vectors as a space. If you will, a velocity space

Any operation that is valid for position space is valid for velocity space.

In the limit of summation of infinitisimal parts we therefore have:

|\vec a| = \frac{2 \pi |\vec v|}{T}

Simplifying to an expression with scalar quantities:

a = \frac{2 \pi v}{T} Removing 'T' from the expression with a substitution:

a = 2 \pi v \frac{v}{2 \pi r} = \frac{v^2}{r}The key points:
- The set of all velocity vectors is the same kind of vector space as position space.
- The set of all acceleration vectors is the same kind of vector space as position space.
- In the case of uniform circular motion there is a one-on-one relation between the instantaneous position vector, the instantaneous velocity vector, and the instantaneous acceleration vector