Period of three oscillating charged coins

[timetraveller123]

Homework Statement

Homework Equations

$u = 1/2 a x^2\\ ke = 1/2 b {\dot x }^2\\ then\\ \omega = \sqrt{\frac{a}{b}}$

The Attempt at a Solution

$U = \frac{2 k q ^2}{l} + \frac{k q^2}{2 l cos \frac {\theta}{2}}$
from that it is easily seen that equilibrium position is when the system is straight
so i was thinking the middle mass would move upwards and the left and right mass would move outwards
is that correct is my approach even correct in the first place? thanks

 

[BvU]

Left and right move downwards in your picture -- the driving force is somewhat outwards, but only the tangential component works toward equilibrium.
(actually, things happen in a frictionless horizonta plane)
For the kinetic energy you want to consider the motion of the center of mass -- or its absence ...
Compare with a simple pendulum.

Very nice exercise !

 

[timetraveller123]

no i am confused about the motion of masses
the left mass would move leftwards and the right mass would move rightwards and the bottom mass would move upwards is this right?

 

[BvU]

no i am confused about the motion of masses

Is the answer to what question ?

The masses are constrained to move on circles with the middle mass as center point. The electrostatic force is a restoring force that tries to align all three masses -as you found already.

so i was thinking the middle mass would move upwards and the left and right mass would move outwards

would that keep the center of mass in the same position ?

 

[timetraveller123]

If the centre of mass is supposed to remain where it is then the middle mass can't be fixed the middle mass moves upwards and the upper masses move down to meet the lower mass at line of centre of mass before overshooting and continuing all over
Is this correct I can't just visualize the lower mass being fixed

 

[BvU]

Yes: the center of mass stays in place if the middle mass moves 'up' (in the drawing) twice as much as each of the outer masses moves 'down'. This is called a 'normal mode' for the system.

 

[timetraveller123]

Yes that's what I meant so the upper mass tangential velocity can be approximated mostly as vertical right

 

[BvU]

Correct: the so-called small angle approximation (just like with a simple pendulum)

 

[timetraveller123]

so this is what i have come up with so far
if the upper mass moves down with velocity v then the middle mass would move up with velocity 2v
$y = l sin \frac{\theta}{2}\\ y \approx l \frac{\theta}{2}\\ \dot y = l \frac{\dot \theta}{2}\\ Ke = 2 \frac{1}{2} m (l \frac{\dot \theta}{2})^2 + \frac{1}{2}m (l \dot \theta)^2\\ Ke = m (l \frac{\dot \theta}{2})^2 + \frac{1}{2}m (l \dot \theta)^2\\ Ke = \frac{3m l^2 {\dot \theta}^2}{4}$
now ke is in the form of 1/2 a v²
is it correct
how to make the potential energy in the form U = U_o + 1/2 kX²
$U = \frac{2kq^2}{l} + \frac{kq^2}{2 l cos \frac{\theta}{2}}$
how to make the second into the form i tried using the taylor expansion for cos but to no avail

 

[BvU]

Well, what did you try, given that $\cos\epsilon = 1-{\epsilon^2\over 2} + {\mathcal O}(\epsilon^4)$ and what did you find ?

 

[TSny]

so this is what i have come up with so far
if the upper mass moves down with velocity v then the middle mass would move up with velocity 2v

OK

$y = l sin \frac{\theta}{2}\\ y \approx l \frac{\theta}{2}\\ \dot y = l \frac{\dot \theta}{2}\\ Ke = 2 \frac{1}{2} m (l \frac{\dot \theta}{2})^2 + \frac{1}{2}m (l \dot \theta)^2\\$

$y = l sin \frac{\theta}{2}$ is the $y$-displacement between the middle coin and the end coins. But $\dot y$ is not the velocity of any of the coins.

 

[timetraveller123]

OK

$y = l sin \frac{\theta}{2}$ is the $y$-displacement between the middle coin and the end coins. But $\dot y$ is not the velocity of any of the coins.

yup that's right been so oblivious to it sorry

so decided to do it in centre of mass frame but how to do it there is so many changing variables around the centre of mass how to do it the distance from centre of mass to the upper mass is also constantly changing

how to set up variables

 

[BvU]

center of mass is in a fixed position. $\theta$ or $\theta/2$ is your oscillating variable. Why do you use $\theta/2$ -- probably because it was in the problem statement ? And why is that ?

Am I still waiting for a reply to #10 ?

 

[timetraveller123]

center of mass is in a fixed position. θθ\theta or θ/2θ/2\theta/2 is your oscillating variable. Why do you use θ/2θ/2\theta/2# -- probably because it was in the problem statement ? And why is that ?

no the question only states displacing one coin by a small angle so i though that would be similar to displacing each of the coins by theta/2

Am I still waiting for a reply to #10 ?

i am sorry i didnt mean to keep you waiting but as pointed by TSny the previous approach was flawed hence my equations were wrong

still need help to set up the new approach picking com as the origin theta as the angle between the line joining the com and the upper mass and the x axis.
how to get velocities of masses from this

 

[TSny]

Yes, put the origin at the center of mass in your figure of post #12. But keep $\theta/2$ as the angle of either string from horizontal. Actually, as BvU suggested, get rid of the factor of 1/2 by letting, say, $\phi = \theta / 2$.

Can you express the vertical coordinates of the coins ($y_1$, $y_2$, $y_3$) in terms of $\phi$?

 

[timetraveller123]

that is exactly what i am not able to do i am missing some piece of geometrical information

what is the additional information i need
the distance of upper masses from com is constantly varying

meanwhile i was thinking since the previous result derived was flawed but i was wondering since the expression for $\dot y$ for the vertical mases is relative to the lower mass and we know the velocity of lower mass is half that of the upper masses can we work out the actual velocity of upper masses using the concept of relative masses

 

[BvU]

If the outer masses move by $\ \Delta y = l\sin \phi \approx l\phi\$, how much does the central mass have to move to keep the center of mass in place ?

 

[timetraveller123]

If the outer masses move by $\ \Delta y = l\sin \phi \approx l\phi\$, how much does the central mass have to move to keep the center of mass in place ?

but the distance from centre of mass to outer mass is not l

 

[BvU]

I agree.

 

[timetraveller123]

thats where i am stuck will my method of relative velocity approach in post 16 work

 

[BvU]

But for small angles, and that's what we are working with, it is.

 

[timetraveller123]

so does it work or not i am not understanding

 

[BvU]

 

[timetraveller123]

i think i have made a little progress help me check if it would work
once again using my old method

this time the velocity of the upper mass as viewed from the lower mass is $- l \dot \theta$ but let the lower mass be moving up with velocity u respective to stationary frame then the velocity of upper mass is $-l \dot \theta +u$ but from the stationary frame outside using conservation of momentum and small angle approximation the upper mass is moving with -u/2 hence equating the two velocities we have $-l \dot \theta + u = - \frac{u}{2} \\ \frac{3u}{2} = l\dot \theta$ then the energy of system can be given by
$E = \frac{1}{2} m ((\frac{2}{3} l \dot \theta)^2 + 2 (\frac{1}{3} l \dot \theta)^2) + 2\frac{k q^2}{l} + \frac{kq^2}{2lcos\theta}\\$
simplifiying differentiating and cancelling yields
$\ddot \theta = \frac{3}{4} \frac{kq}{m l^3 (cos \theta)^2} sin \theta \\ \ddot \theta \approx \frac{3}{4} \frac{kq}{m l^3 } \theta \\$
would this work?

 

[haruspex]

i think i have made a little progress help me check if it would work
once again using my old method
View attachment 212992
this time the velocity of the upper mass as viewed from the lower mass is $- l \dot \theta$ but let the lower mass be moving up with velocity u respective to stationary frame then the velocity of upper mass is $-l \dot \theta +u$ but from the stationary frame outside using conservation of momentum and small angle approximation the upper mass is moving with -u/2 hence equating the two velocities we have $-l \dot \theta + u = - \frac{u}{2} \\ \frac{3u}{2} = l\dot \theta$ then the energy of system can be given by
$E = \frac{1}{2} m ((\frac{2}{3} l \dot \theta)^2 + 2 (\frac{1}{3} l \dot \theta)^2) + 2\frac{k q^2}{l} + \frac{kq^2}{2lcos\theta}\\$
simplifiying differentiating and cancelling yields
$\ddot \theta = \frac{3}{4} \frac{kq}{m l^3 (cos \theta)^2} sin \theta \\ \ddot \theta \approx \frac{3}{4} \frac{kq}{m l^3 } \theta \\$
would this work?

Looks right, except that you lost the exponent on q, and there's a sign error.

 

[TSny]

I agree with haruspex's remarks.

 

[timetraveller123]

ay yes sorry about that so it is
$\ddot \theta = -\frac{3}{4}\frac{k q^2}{m l^3} \theta\\$
thanks for the help

 

[haruspex]

ay yes sorry about that so it is
$\ddot \theta = -\frac{3}{4}\frac{k q^2}{m l^3} \theta\\$
thanks for the help

You are welcome.