Vibrations - Modeling system, equation of motion

[thepikminman]

Vibrations - Modelling system, equation of motion
Hi,

In the first question (question 4) in the attached file, how would you go about modelling the system and finding the equation of motion? All those masses are confusing me, I don't even know where to start.

I don't know whether the angle theta has a significant effect on the displacement of the spring or not, it the masses are fixed or not.

Any ideas??

 

[BvU]

Hello Pmm,

PF Guidelines require a bit more from you before we're allowed to assist.

But I can say a few things:

the angle theta has a significant effect on the displacement of the spring

More the other way around: if the spring contracts or expands, ##\theta## changes.
Masses are fixed on their rods. m1 m2 m3 is a fixed triangle that can rotate around ##O##
And you may assume small oscillations -- so the spring stays vertical

 

[thepikminman]

Hello Pmm,

PF Guidelines require a bit more from you before we're allowed to assist.

But I can say a few things:
More the other way around: if the spring contracts or expands, ##\theta## changes.
Masses are fixed on their rods. m1 m2 m3 is a fixed triangle that can rotate around ##O##
And you may assume small oscillations -- so the spring stays vertical

Thank you!

I drew a FBD, and for the spring side I'm getting F_spring = kx = N. For the angle part I'm getting ω = θt. I tried finding the moments about 0, but don't know the angles to use. I feel like it's something to do with Asin(ωt) but I'm not sure how to get there.

 

[BvU]

For the angle part I'm getting ω = θt.

Please post your work in detail. You don't explain what t in ω = θt is, but it sure looks wrong.
What do you think of figure 5 ? Any conclusions/reasonable assumptions ?
And if that doesn't light the light , what about the wording of parts 4, 5 and 6 ?

 

[thepikminman]

Please post your work in detail. You don't explain what t in ω = θt is, but it sure looks wrong.
What do you think of figure 5 ? Any conclusions/reasonable assumptions ?
And if that doesn't light the light , what about the wording of parts 4, 5 and 6 ?

My bad, t is time, ω is angular velocity in rad/s.

My assumptions from the figure:
1. Damping due to air is negligible.
2. There are no external forces acting on the system.
3. The mass m3 is displaced in the vertical direction only.
4. It is a 1 DOF system (θ is enough to fully describe the position of all parts).

I'm still stuck on where to go from there. In 4,5,6 they talk about angular velocity as a function of time. I know vertical displacement from a rotating object is given by Asinωt, velocity is it's derivative wrt t so, ωAcosωt, acceleration is -ω²Asinωt...I have no clue what A is though. Isn't the amplitude just the length of the circle's radius? So l3 in this case?

And the graph underneath...I don't get it, it's a plot of θ vs θ isn't it? ω_nt = θ, and so is angular displacement right??

 

[BvU]

4 is the important one. ##\theta## is your main character in this whole play. The whole thing is at equilibrium when ##\theta = 0 ## as you can see from fig. 5. Out of equilibrium there is a restoring force (magnitude ?) and there is some inertia, so there will be oscillations, just like with a mass on a spring. Since we are talking angles, we should speak the language: torque, angular acceleration, and so on.
It's time for some equations, and sure enough the exercise starts off with that. Any suggestion ?

 

[thepikminman]

4 is the important one. ##\theta## is your main character in this whole play. The whole thing is at equilibrium when ##\theta = 0 ## as you can see from fig. 5. Out of equilibrium there is a restoring force (magnitude ?) and there is some inertia, so there will be oscillations, just like with a mass on a spring. Since we are talking angles, we should speak the language: torque, angular acceleration, and so on.
It's time for some equations, and sure enough the exercise starts off with that. Any suggestion ?

I know that ∑M₀= I₀θ'' ... but I don't know how to find I. I tried using I = ml² and got ΣM₀ = -θ'' (m₁l₁² + m₂l₂² + m₃l₃²) + kl₃sinθ but I can't see where that gets me.

Can you tell me, what am I looking to even arrive at with these types of problems? I know it's the equation of motion, but how do I know when I've found it, what am I looking to express?

And my lecture notes keep using ω_n and f_n both for the natural frequency interchangeably. Which is correct? I thought ω_n was angular velocity?

 

[BvU]

Yep, ##\omega = 2\pi f##.

what am I looking to even arrive at

You are working towards somehing that (see the figure) looks like a sine function of time. Like with a mass on a spring. The restoring force is opposite to and proportional with the distance from equilibrium: ma = -kxHere you have mass(es) and a spring, but this time it is not an ##x## or ##y## but something with an angle. So you have to translate F = ma to angular lingo.

 

[thepikminman]

Yep, ##\omega = 2\pi f##.
You are working towards somehing that (see the figure) looks like a sine function of time. Like with a mass on a spring. The restoring force is opposite to and proportional with the distance from equilibrium: ma = -kxHere you have mass(es) and a spring, but this time it is not an ##x## or ##y## but something with an angle. So you have to translate F = ma to angular lingo.

That's what I did in my last post with " ΣM₀ = -θ'' (m₁l₁² + m₂l₂² + m₃l₃²) + kl₃sinθ". Is that the correct answer?

 

[BvU]

Sorry I overlooked that.
Sort of, yes: There is only one torque: ##kl_3\sin\theta ## ( except I that miss a sign) .
So you get ##\ I\alpha = \sum\tau\ ## .

And for small ##\theta## you may approach ##\sin\theta \approx \theta##

 

[thepikminman]

Sorry I overlooked that.
Sort of, yes: There is only one torque: ##kl_3\sin\theta ## ( except I that miss a sign) .
So you get ##\ I\alpha = \sum\tau\ ## .

And for small ##\theta## you may approach ##\sin\theta \approx \theta##

Thank you!

What do you mean by only one torque though?
And what is α in Iα= Σtorques?

 

[BvU]

##\alpha = \dot \omega = \ddot \theta##, the angular acceleration.

What do you mean by only one torque though?

Very good observation !
And I realize I am making a mistake: the center of mass of m1, m2, m3 is not at ##O##.
But from fig 5 equilibrium is at ##\theta = 0 ## and the motion is a sine, so:
My best estimate is that the exercise composer intended you to ignore the torque contribution due to gravity working on the center of mass: the angle of l₂ is not given and you would need it.

Would @haruspex agree in this ?

 

[thepikminman]

##\alpha = \dot \omega = \ddot \theta##, the angular acceleration.
Very good observation !
And I realize I am making a mistake: the center of mass of m1, m2, m3 is not at ##O##.
But from fig 5 equilibrium is at ##\theta = 0 ## and the motion is a sine, so:
My best estimate is that the exercise composer intended you to ignore the torque contribution due to gravity working on the center of mass: the angle of l₂ is not given and you would need it.

Would @haruspex agree in this ?

I'm confused now :P I thought each mass contributes to the torque about point 0? That lecturer is a sneaky one so i wouldn't be surprised if he's trying to confused the class.

 

[BvU]

each mass contributes to the torque about point 0

Very good observation (again! There is something to be said for sneaky lecturers ...) .

True. I took into account that l₁ and l₃ are almost horizontal. So their torques are constant (to within ##\theta^2## ) and the contribution goes into the equilibrium position. l₂ looks pretty vertical, so the contribution from m₂ varies proportional to ##\theta## and that is the same order of magnitude as that of the spring. The orientation of l₂ is not given. With a vertical orientation you would get a torque of 10 kg x 10 m/s² 0.27 m x ##\sin \theta## in a direction opposite to that of the spring -- effectively reducing the 120 N/m to 93 N/m.

If I were you I'd ignore it (if you were supposed to do something with that the exercise would become rather complicated and the orientation of l₂ would have been given ) and only consider the contributions of the masses to the moment of inertia.

I don't mean to confuse you. Had overlooked the ##m_i\, g\; l_i \cos\phi_i## torques -- but I suspect the exercise composer had too !

 

[thepikminman]

Very good observation (again! There is something to be said for sneaky lecturers ...) .

True. I took into account that l₁ and l₃ are almost horizontal. So their torques are constant (to within ##\theta^2## ) and the contribution goes into the equilibrium position. l₂ looks pretty vertical, so the contribution from m₂ varies proportional to ##\theta## and that is the same order of magnitude as that of the spring. The orientation of l₂ is not given. With a vertical orientation you would get a torque of 10 kg x 10 m/s² 0.27 m x ##\sin \theta## in a direction opposite to that of the spring -- effectively reducing the 120 N/m to 93 N/m.

If I were you I'd ignore it (if you were supposed to do something with that the exercise would become rather complicated and the orientation of l₂ would have been given ) and only consider the contributions of the masses to the moment of inertia.

I don't mean to confuse you. Had overlooked the ##m_i\, g\; l_i \cos\phi_i## torques -- but I suspect the exercise composer had too !

Ok...I think i kind of get it...

For part b though, it asks for the natural frequency of the system (not frequencies), if its just the usual sqrt(k/m)*(1/2π), what do I use for the mass value if it's the natural frequency of the entire system?

Is the natural frequency of this system even sqrt(k/m)*(1/2π)?

 

[haruspex]

Without seeing the data table it is hard to be sure, but the lack of angle constants implies to me that l₂ is supposed to be at right angles to the others. And I gather the equilibrium position is with the main rod horizontal. Unclear whether that is the relaxed spring position, but I don't think that matters.

That all being so, deltas to the gravitational torques of m₁ and m₃ are second order quantities and can be ignored. That from m₂ is first order.

So we get ##\ddot\theta\Sigma m_il_i^2+(-gm_2l_2+kl_3)\theta=0##, yes?

 

[thepikminman]

Without seeing the data table it is hard to be sure, but the lack of angle constants implies to me that l₂ is supposed to be at right angles to the others. And I gather the equilibrium position is with the main rod horizontal. Unclear whether that is the relaxed spring position, but I don't think that matters.

That all being so, deltas to the gravitational torques of m₁ and m₃ are second order quantities and can be ignored. That from m₂ is first order.

So we get ##\ddot\theta\Sigma m_il_i^2+(-gm_2l_2+kl_3)\theta=0##, yes?

I get the ##\ddot\theta\Sigma m_il_i^2## part and the kl₃ part, but why does m2 effect the moment about 0 and not m1 and m3? I'd have thought if anything it would be the opposite, m1 and m3 effect it but m2 doesn't because it's virtually vertical...

Why not θ''Σm_il_i²+(-Σgm_il_i+kl₃)θ=0??

Also, is the natural frequency just √(k/m), and the value of m to use just m1+m2+m3?

 

[haruspex]

I get the ##\ddot\theta\Sigma m_il_i^2## part and the kl₃ part, but why does m2 effect the moment about 0 and not m1 and m3? I'd have thought if anything it would be the opposite, m1 and m3 effect it but m2 doesn't because it's virtually vertical...

We are concerned with the small change to the moment that results from the small angular displacement. If a mass m is on rod length l at angle α+θ measured anticlockwise from positive x axis, then the gravitational moment is mgl cos(α+θ). The difference from the equilibrium position is mgl(cos(α+θ)-cos(α)). For small θ that approximates -mgl sin(α)θ. For m₁ and m₃, sin(α)=0; for m₂ sin(α)=1.

You can think of m₂ as an upside-down pendulum.

 

[thepikminman]

We are concerned with the small change to the moment that results from the small angular displacement. If a mass m is on rod length l at angle α+θ measured anticlockwise from positive x axis, then the gravitational moment is mgl cos(α+θ). The difference from the equilibrium position is mgl(cos(α+θ)-cos(α)). For small θ that approximates -mgl sin(α)θ. For m₁ and m₃, sin(α)=0; for m₂ sin(α)=1.

You can think of m₂ as an upside-down pendulum.

thanks I get it now!

To find the formula for natural frequency; from the equation of motion, I see that there is a θ'' and a θ but no θ'. I tried solving the characteristic equation by letting the θ' coefficient = 0 but that didn't work. Am I on the right track? And for the natural frequency value do you use (m₁+m₂+m₃) for the mass?

 

[haruspex]

there is a θ'' and a θ but no θ'.

Right, but that is how it should be for undamped SHM.

do you use (m1+m2+m3) for the mass?

Use that for the mass where? You cannot plug this into an off-the-shelf equation for a mass on a spring. Work with the differential equation.

 

[thepikminman]

Right, but that is how it should be for undamped SHM.
Work with the differential equation.

I'm trying to solve it but the answer is getting crazy... Is this not undamped SHM?

 

[haruspex]

I'm trying to solve it but the answer is getting crazy... Is this not undamped SHM?

It is. Please post your working.

 

[thepikminman]

It is. Please post your working.

Ok, it's quite long:

To save time i'll say m₁l₁² + m₂l₂² + m₃l₃² = I_n

So the equation of motion is θ''I_n -(m₂gl₂ + l₃k)θ = 0 then the characteristic equation is I_ns² + 0s - (gl₂ + l₃k) = 0. Finding the roots I used -I_n ±√(0² + 4(l_n)((gl₂ + l₃k) / 2I_n. This is where it gets too complicated. Is this the right method?

 

[haruspex]

θ''In -(m2gl2 + l3k)θ = 0

It goes crazy because you have a sign wrong.

 

[thepikminman]

It goes crazy because you have a sign wrong.

I thought it should be minus because -(m₂gl₂ + l₃k) is going to opposite direction to the moments from the masses??

But I don't see how that will make it any less crazy...any clue? I've been working on this single problem 8 hours a day, literally for just this question, for the past two days

 

[haruspex]

I thought it should be minus because -(m₂gl₂ + l₃k) is going to opposite direction to the moments from the masses??

But I don't see how that will make it any less crazy... something? anything? a clue? hint? nudge in the right direction? I've been working on this single problem 8 hours a day, literally for just this question, for the past two days and have my exam on friday..

##\theta## is positive anticlockwise, so its derivatives are too. Does a positive displacement increase or decrease the anticlockwise moment from m₂ ? Does it increase or decrease the anticlockwise moment from the spring?

Once you have recognised a DE as SHM, the easiest way is to plug in a generic solution Acos(ωt+φ) and solve.

 

[thepikminman]

##\theta## is positive anticlockwise, so its derivatives are too. Does a positive displacement increase or decrease the anticlockwise moment from m₂ ? Does it increase or decrease the anticlockwise moment from the spring?

Once you have recognised a DE as SHM, the easiest way is to plug in a generic solution Acos(ωt+φ) and solve.

Thank you.

I tried that, but for values of A and φ where do I get the condition values for x at t = 0 and x' at t = 0?

 

[haruspex]

Thank you.

I tried that, but for values of A and φ where do I get the condition values for x at t = 0 and x' at t = 0?

Figure 5? (But you can do better than trying to read the slope at time zero.)

 

[thepikminman]

Figure 5? (But you can do better than trying to read the slope at time zero.)

But that gives the angular displacement, and it doesn't seem exact... can I take the reading of 0.02 radians and use x = l3*sin(0.02) to get x? Even if I can, I can't see how I can get the velocity.

 

[haruspex]

But that gives the angular displacement, and it doesn't seem exact... can I take the reading of 0.02 radians and use x = l3*sin(0.02) to get x? Even if I can, I can't see how I can get the velocity.

You don't have to use velocity. What other parameter is in the generic equation?

 

[thepikminman]

Figure 5? (But you can do better than trying to read the slope at time zero.)

Ok, I tried it a different way, can you tell me if this is right:

for SHM, θ(t) = θ^sin(ω_nt) (where θ^ = amplitude).

⇒ θ'' = -ω_n²θ

I_totalθ'' = -kθ

⇒I_total-ω_n²θ = -kθ

⇒ω_n = √k/I_total

?

 

[haruspex]

Ok, I tried it a different way, can you tell me if this is right:

for SHM, θ(t) = θ^sin(ω_nt) (where θ^ = amplitude).

⇒ θ'' = -ω_n²θ

I_totalθ'' = -kθ

⇒I_total-ω_n²θ = -kθ

⇒ω_n = √k/I_total

?

Let's get the DE right first. Can you use my hints in post 26 to correct your sign error in post 23?

 

[thepikminman]

##\theta## is positive anticlockwise, so its derivatives are too. Does a positive displacement increase or decrease the anticlockwise moment from m₂ ? Does it increase or decrease the anticlockwise moment from the spring?

Once you have recognised a DE as SHM, the easiest way is to plug in a generic solution Acos(ωt+φ) and solve.

A positive (angular) displacement would increase the anticlockwise moment, and decrease the anticlockwise moment from the spring (but increase the spring force)

Ok so the DE should be: θ''I_n + 0θ' + (m₂gl₂ + kl₃) θ= 0

But then from the -b formula, how can I get the sqaure root of 4*[m₁l₁² + m₂l₂² + m₃l₃²]*(m₂gl₂ + kl₃)? I've no idea

 

[haruspex]

A positive (angular) displacement would increase the anticlockwise moment, and decrease the anticlockwise moment from the spring

So why do you have the same sign for each in the equation?

how can I get the sqaure root of 4*[m1l12 + m2l22 + m3l32]*(m2gl2 + kl3)?

Just deduce ω from the graph.

 

[thepikminman]

So why do you have the same sign for each in the equation?

Just deduce ω from the graph.

Haha I had the minus in my first answer and you said it was wrong ? That's why I'm confused

And for deducing ω from the graph, I see that θ = 0.02 at ωt=0, but how do i get ω from that?