Write the equations of a mass and spring in different accelerated systems

[Like Tony Stark]

Homework Statement

Consider 1) a platform with a mass and spring rotating with constant angular velocity; 2) a room being accelerated with constant acceleration $a_o$

Relevant Equations

$\Sigma \vec{F}=m\vec{a}$

Hi
I know that the equation of a simple harmonic oscillator is $\ddot x + \omega^2 x=0$. The thing is that I don't know how to get to that equation in the situations given.

In the first situation, I know that
$x) k(x-x_0)=m(\ddot x -x \dot \theta ^2)$
$y) N=m(2 \dot x \dot \theta)$

So $\ddot x -x (\dot \theta ^2 +\frac{k}{m}) +\frac{k}{m} x_0=0$
But how can I eliminate the last term? And what about the $y)$ part?

For the second situation, we have
$x) k(x-x_0)=m(a_o +a_{rel})$
Where $a_o$ is the acceleration of the room and $a_{rel}$ the acceleration of the block relative to the room. So
$\ddot x -\frac{k}{m} x +\frac{k}{m} x_0 +a_o=0$
How should I eliminate the last two terms?

 

[kuruman]

For the rotating frame read this link to see how to do it.
For the frame accelerating linearly, consider the acceleration as fictitious acceleration of gravity.

 

[Like Tony Stark]

Ok, so, for the second situation I thought that the equilibrium position was $d$, so $-kd=ma_o$. Then
$-k(d+x)=m\ddot x + ma_o$
$0=\ddot x +\frac{k}{m} x$
I've got two questions here: 1) is there any difference with the case of $a_o=0$ (a room that's not accelerated)?
2) if there was friction, the equation would be $-g\mu=\ddot x +\frac{k}{m} x$, right? Then, for the first situation, the equilibrium position is $kd=mr\dot \theta^2$, so
$-k(d+x)=m(\ddot r-r\dot \theta^2)$
$0=\ddot r+\frac{k}{m} r$
My question is the same... how would this equation change if the disk wasn't rotating?

 

[haruspex]

Then, for the first situation, the equilibrium position is $kd=mr\dot \theta^2$, so
$-k(d+x)=m(\ddot r-r\dot \theta^2)$
$0=\ddot r+\frac{k}{m} r$

Define d and x. How do they relate kinematically to r?
How do you get the last equation above?

 

[Like Tony Stark]

Define d and x.

Sorry, it was $r$, not $x$. $d=r_{eq}-r_0$ where $r_{eq}$ is the position of equilibrium and $r_0$ the original length

 

[haruspex]

Sorry, it was $r$, not $x$. $d=r_{eq}-r_0$ where $r_{eq}$ is the position of equilibrium and $r_0$ the original length

On the RHS of your equations in post #3 you are taking r as the distance from the centre of the disk, but x seems to be the displacement from the equilibrium position.

 

[Like Tony Stark]

On the RHS of your equations in post #3 you are taking r as the distance from the centre of the disk, but x seems to be the displacement from the equilibrium position.

Yes, but I made a mistake. I've attached an image showing my idea
Consider $d$ as $r$ of equilibrium.

 

[haruspex]

Consider d as r of equilibrium.

THat does not seem to match your diagram.
Tell you what, I'll define the variables and you write the equations:
r is the radius
The below all measured from the base of the spring:
x₀ is relaxed length
x is an arbitrary position of the mass
d is the equilibrium position of the mass

 

[Like Tony Stark]

x₀ is relaxed length
x is an arbitrary position of the mass
d is the equilibrium position of the mass

Thank you
So... $\Delta x_{eq}=d-x_0$. Then $F=-k(\Delta x_{eq} +x)$
Where $\Delta x_{eq}$ is the length of equilibrium

 

[haruspex]

Thank you
So... $\Delta x_{eq}=d-x_0$. Then $F=-k(\Delta x_{eq} +x)$
Where $\Delta x_{eq}$ is the length of equilibrium

I think you mean, where $\Delta x_{eq}$ is the spring extension at equilibrium.
But I don’t understand $F=-k(\Delta x_{eq} +x)$.
Anything you multiply by k should be a spring extension, but x includes x₀, so $\Delta x_{eq} +x$ is not a spring extension.

 

[Like Tony Stark]

but x includes x₀, so $\Delta x_{eq} +x$ is not a spring extension.

Why not? We are adding the deformation that the spring has when the mass is in equilibrium with respect to the platform to a different length measured from that equilibrium position.

 

[haruspex]

Why not? We are adding the deformation that the spring has when the mass is in equilibrium with respect to the platform to a different length measured from that equilibrium position.

In post #8 I defined x as the position of the mass as measured from the base of the spring. If you prefer a different definition, fine, but you need to specify it.

 

[Like Tony Stark]

In post #8 I defined x as the position of the mass as measured from the base of the spring. If you prefer a different definition, fine, but you need to specify it.

Ok, so maybe I made a mess of the variables. The thing is that I got to the equation $0=\ddot r +k/m r$, but it's the same equation that I would get if the platform was stationary. Is this right?

 

[haruspex]

Ok, so maybe I made a mess of the variables. The thing is that I got to the equation $0=\ddot r +k/m r$, but it's the same equation that I would get if the platform was stationary. Is this right?

No.
Please post your working.

 

[Like Tony Stark]

Please post your working.

$-k(d+r)=m(\ddot r-r\dot \theta^2)$
Then, $kd=mr\dot \theta^2$, so
$0=\ddot r+\frac{k}{m} r$

 

[kuruman]

I was following this fine until post #13 where you introduced $r$. What is this variable in terms of the variables that @haruspex defined for you in post #8? It's OK to change symbols but only if you define them using the English language as in "Let $r$ be the $\dots$"

Look at your equation $-k(d+r)=m(\ddot r-r\dot \theta^2)$. On the LHS $d$ is the equilibrium position of the spring, so one would assume that $r$ stands for distance the spring is extended beyond equilibrium. On the RHS you have the term $mr\dot \theta^2$ which a centripetal term. In this term $r$ stands for the additional distance from the center to the mass. You have two different meanings for $r$ in the same equation. That is utterly confusing to me.

I suggest that you do follow one of two courses of action:
1. Path of least resistance: Adopt the definitions in post #8.
2. Do it yourself: Choose your own symbols for the three variables listed in post #8 but make sure they are clearly defined.

 

[Like Tony Stark]

Ok, so using the definitions in post #8, we have
$-k[x-(d-x_0)]=m[\ddot x-(x-r)\theta^2]$
Is that right?

 

[haruspex]

Ok, so using the definitions in post #8, we have
$-k[x-(d-x_0)]=m[\ddot x-(x-r)\theta^2]$
Is that right?

RHS looks ok, except for a missing dot (might as well use $\omega$ instead of $\dot\theta$).
At position x, what is the spring extension?

 

[kuruman]

RHS looks ok, except for a missing dot (might as well use $\omega$ instead of $\dot\theta$).
At position x, what is the spring extension?

If the RHS is correct, I must have misunderstood the meaning of the variables in post #8.
Variable $r$ is defined as "the radius". I would assume that it is the distance from the axis of rotation to the mass. How does that differ from $x$, the "arbitrary position of the mass"? Isn't the base of the spring anchored at the axis of rotation? If so, then,

• when the platform is not rotating and the mass not oscillating, $r=x=x_0$
• when the platform is rotating but the mass is at rest relative to it, $r=x=x_0+\Delta x_{eq}=d$

I do not understand why the (theta dot corrected) centripetal term on the RHS in post #17 is not $-mx\dot \theta^2$.

 

[haruspex]

Isn't the base of the spring anchored at the axis of rotation?

According to the image in post #1, the base of the spring is at the circumference of the disc. The spring extends through the axis of rotation to reach the mass.

 

[kuruman]

According to the image in post #1, the base of the spring is at the circumference of the disc. The spring extends through the axis of rotation to reach the mass.

Ah, yes. Everything has fallen into place now. Thanks.