Dipole Oscillation: Solving for Period

[indigojoker]

Homework Statement

#3 on this PDF

Homework Equations

\tau = Frsin(\theta) = I \alpha
I=mr^2

The Attempt at a Solution

Here's what I've done:
\tau = Frsin(\theta) = I \alpha
F=QE

2QEsin(\theta) \frac{A}{2} = 2M\left(\frac{A}{2}\right)^2 \alpha
simplify:
QEsin(\theta)A = \frac{1}{2}MA^2 \frac{d^2\theta}{dt^2}
\frac{d^2\theta}{dt^2} - \left(\frac{2QE}{MA}\right) \theta = 0

I'm not sure how to get the period.

I know that \omega = 2 \pi \frac{1}{T}

but I'm stuck here. Any help would be much appreciated.

 

[learningphysics]

You're almost there... \alpha here equals -\frac{d^2\theta}{dt^2}

once you fix that your equation is of the form:

\frac{d^2\theta}{dt^2} + {\omega}_0^2\theta = 0

so you can get {\omega}_0

 

[indigojoker]

why is alpha negative?
and is the form \frac{d^2\theta}{dt^2} + {\omega}_0^2\theta = 0 standard?

where did this equation come from?

also, how would i show the system to have small oscillations?

 

[learningphysics]

why is alpha negative?

The way you calculate torque... it is acting clockwise... therefore \alpha should be the angular acceleration in the clockwise direction... however as the object turns clockwise, your theta angle decreases... examine what happens to theta as the object rotates in the direction of the torque...

and is the form \frac{d^2\theta}{dt^2} + {\omega}_0^2\theta = 0 standard?

Yes, it is the standard equation for a harmonic oscillator.

where did this equation come from?

also, how would i show the system to have small oscillations?

The question asks you to assume small oscillations. ie approximate sin(theta) as theta... as you did.

 

[indigojoker]

I'm sorry, i did not mean to ask about small oscillations, I was wondering more about how to show the system exhibits simple harmonic motion, before ever going to get the period of oscillation.

and the torque will sometimes be negative, and sometimes be positive. How do you know alpha is negative?

 

[learningphysics]

I'm sorry, i did not mean to ask about small oscillations, I was wondering more about how to show the system exhibits simple harmonic motion, before ever going to get the period of oscillation.

and the torque will sometimes be negative, and sometimes be positive. How do you know alpha is negative?

The fact that the equation is in this form for small theta:
\frac{d^2\theta}{dt^2} + {\omega}_0^2\theta = 0

shows that there's simple harmonic motion...

As for the torque... let me ask this... what angle is \theta... and how does it change with positive torque... how does it change with negative torque...

when the torque is positive \alpha should be positive... when the torque is negative \alpha should be negative.

 

[indigojoker]

right, so by saying -\frac{d^2\theta}{dt^2}

this means the torque is always negative, but it isn't is it?

 

[learningphysics]

right, so by saying -\frac{d^2\theta}{dt^2}

this means the torque is always negative, but it isn't is it?

No, not when \frac{d^2\theta}{dt^2} is negative...

 

[indigojoker]

i'm just really dense right now. I still don't know why you would need a negative sign.
could you explain it differently? I'm not seeing the whole picture

 

[learningphysics]

i'm just really dense right now. I still don't know why you would need a negative sign.
could you explain it differently? I'm not seeing the whole picture

let's put this problem aside and look at a different one... suppose I have a wrench on the ground connected to some pivot... the wrench forms an angle of let's 45 degrees with the east axis... so theta is 45. Let's say the wrench is 0.5m. I exert a force of 10N at the end of the wrench... eastward...

What is the torque about the pivot?

 

[indigojoker]

T=.5*10*sin45

 

[learningphysics]

T=.5*10*sin45

cool. now is \frac{d^2\theta}{dt^2} positive or negative here? taking theta as the angle made with the east axis...

 

[indigojoker]

angle is decreasing so it's negative?

 

[learningphysics]

angle is decreasing so it's negative?

Yes... the angular acceleration is clockwise... in other words the angle the wrench makes with the vertical... call it \alpha... \frac{d^2\alpha}{dt^2} is positive... theta = 90 - alpha... \frac{d^2\theta}{dt^2} = -\frac{d^2\alpha}{dt^2}

You found the torque clockwise... but \frac{d^2\theta}{dt^2} is the counterclockwise angular acceleration...

you could have done the opposite and found the torque counterclockwise... and said T=-.5*10*sin45 (that's the counterclockwise torque)... now in this case you can say T = I\frac{d^2\theta}{dt^2}... but since you said T = +0.5*10*sin45 which is the clockwise torque... you must say T = -I\frac{d^2\theta}{dt^2}

 

[indigojoker]

so say we assumed the counter clockwise notation for the torque for the question for the dipole-dumbbells. so T = I\frac{d^2\theta}{dt^2}

then I would still get:
\frac{d^2\theta}{dt^2} - {\omega}^2\theta = 0

 

[learningphysics]

so say we assumed the counter clockwise notation for the torque for the question for the dipole-dumbbells. so T = I\frac{d^2\theta}{dt^2}

then I would still get:
\frac{d^2\theta}{dt^2} - {\omega}^2\theta = 0

no, because then

T = - 2QEsin(\theta) \frac{A}{2} in order for torque to be counterclocwise.

 

[indigojoker]

ok sure, the wrench is fixed and applying a torque will always rotate it clockwise, however, the dipole-dumbbell rotates both clockwise and counterclockwise. This means that that torque can be negative and positive even if we took a clockwise/counterclockwise notation. Since it's not just going in a circle, why would alpha be negative inthe clockwise notation then?

 

[learningphysics]

ok sure, the wrench is fixed and applying a torque will always rotate it clockwise, however, the dipole-dumbbell rotates both clockwise and counterclockwise. This means that that torque can be negative and positive even if we took a clockwise/counterclockwise notation. Since it's not just going in a circle, why would alpha be negative inthe clockwise notation then?

It's not about whether the torque turns out clockwise or counterclockwise... torque is actually a vector... it has a magnitude and a direction... one way we deal with the direction is by using clockwise... counterclockwise...

It is just like with vectors... suppose I have an object with just one force acting on it is labelled F with an arrow to the right... but F could be a negative number... in which case the force is actually acting towards the left... F is the force acting to the right... but it could be negative...

Now suppose I label the acceleration to the left a... I have a with an arrow pointed to the left... again a itself can be positive or negative... if a is negative then the acceleration is actually to the right... a is the acceleration to the left... but a could be negative...

What is the relationship between F and a here given a mass m?

 

[indigojoker]

ok, could you tell me if this explanation looks right?
http://home.earthlink.net/~suburban-xrisis/phy.pdf

 

[learningphysics]

ok, could you tell me if this explanation looks right?
http://home.earthlink.net/~suburban-xrisis/phy.pdf

Seems like you're mixing up clockwise and counterclockwise...

Do you agree that \frac{d^2\theta}{dt^2} is the counterclockwise angular acceleration? because counterclockwise is the direction of increasing theta...

 

[indigojoker]

oh whoops you're right, wait so if we took the fact that clockwise is positive torque, then wouldn't alpha be positive instead of negative?

 

[indigojoker]

wait, hold on, I think i know why... give me a second to post up another pdf.

 

[indigojoker]

please visit the same link again:
http://home.earthlink.net/~suburban-xrisis/phy.pdf

 

[learningphysics]

please visit the same link again:
http://home.earthlink.net/~suburban-xrisis/phy.pdf

For the second part... instead of calling the angle - theta, I'd rather call the angle as still being theta but where theta happens to be negative...

So in both cases... clockwise torque = -I\frac{d^2\theta}{dt^2}

In both cases counterclockwise torque = I\frac{d^2\theta}{dt^2}

you can pick either convention... makes no difference...

in the second case where theta is negative, \frac{d^2\theta}{dt^2} > 0 in the first case it is less than 0... but the equations remain the same...

 

[indigojoker]

cool. So why would \tau = -Frsin(\theta) when taking the counterclockwise torque notation?

 

[learningphysics]

cool. So why would \tau = -Frsin(\theta) when taking the counterclockwise torque notation?

well... that's how the torque is defined... suppose F is positive, sin(theta) is positive... what is the direction the object is trying to move... clockwise... the torque is a positive value clockwise... hence it is a negative value counterclockwise... and checking the formula -Frsin(theta)... give a negative value... so it works...

F negative, sin(theta) positive... now the object is trying to move counterclockwise... hence the counterclockwise torque should be positive... -Frsin(theta) is positive here... again the formula works...

Examine all combinations of F (positive or negative) and sin(theta) positive and negative... examine which way the torque is... see if the formula -Frsintheta works or not for counterclockwise torque...

If instead you choose F as being the force acting to the left... then everything switches signs... but since you chose F as being to the right,this is how the formulas turn out.

 

[indigojoker]

it seems like the torque really depends on how theta is defined. If theta was along the vertical, then alpha would be positive.

 

[indigojoker]

by the way, this might be overkill but could you check my explanation? I just want to make sure what I have makes sense:
http://home.earthlink.net/~suburban-xrisis/phy.pdf

 

[learningphysics]

by the way, this might be overkill but could you check my explanation? I just want to make sure what I have makes sense:
http://home.earthlink.net/~suburban-xrisis/phy.pdf

I'm sorry indigo... still finding that last section confusing... as I'm sure my explanations are confusing to you...

Why do you have the force on the dipole downwards, then upwards... in both cases, the force should be towards the right in the direction of E.

 

[indigojoker]

sorry, those are the y-componets since the y-componets is the force that the torque considers.