0
$\begingroup$

I am a ME student trying to understand this system pictured below. I have a mass sliding on a surface with a coefficient of friction [u], attached by inextensible massless rope to a tension spring of constant [k], and also attached to a flywheel with a moment of inertia [I] which spins as the mass slides x+ and acts as a damper. The forcing function F(t) is of the form Ax^2. The mass is confined to only slide between two stops and the spring is already preloaded under some tension (i will define this as it is stretched by [s]). Essentially a small burst pushes the mass x+ for some distance, damped by friction and the flywheel and then returns to x(0). I would like to assume it is a slightly underdamped system, where it impacts the end stops with some non-zero velocity, and the stops sort of truncate the natural travel. I would imagine that this problem is best solved by using a work-energy balance, but it looks like it fits a second order DE system well. Could someone provide me some guidance on how to solve for the travel function x(t) for this?

I would like to add that I came up with this problem and it is not a homework assignment I just want to cheat on, I drew the diagrams and would just like to get a better understanding of my basic physics.

Below is the block diagram and a graph showing the assumed behavior:

enter image description here

enter image description here

Per the comments below I have added this diagram showing an alternative set up where the rope attaching the mass to the flywheel is wrapped around a bobbin with a smaller diameter than the flywheel. enter image description here

$\endgroup$
0

2 Answers 2

0
$\begingroup$

The Newton equations of motion are

$$m\,\ddot x+k\,(x-x_0)=F(t)+F_c-F_\mu\tag 1$$ $$I\,\ddot \phi+d\,\dot\phi=-F_c\,R\tag 2$$

and the kinematic equations

$$x=R\,\phi\quad\Rightarrow ~,\dot x=R\,\dot\phi~,\ddot x=R\ddot\phi\tag 3$$

you have now three equations for the three unknowns $~\ddot x~,\ddot\phi~$ and the constraint force $~F_c~$

you obtain

$$(I+m\,R^2)\ddot x=\left[ -k \left( x-{x_0} \right) +F \left( t \right) -F_{{\mu}} \right] {R}^{2}-d\,{\dot x} $$

where $~F_\mu=\mu\,m\,g\frac{\dot x}{|\dot x|}~$ and $~F(t)=A\,x^2~$

$\endgroup$
10
  • $\begingroup$ I don't understand the use of x_dot / |x_dot|, is this functioning as a signum function to describe the back and forth action of the mass? I saw one video where they separated the motion into positive and negative paths and solved it like that, but also mentioned the use of a signum to accomplish the same thing. $\endgroup$
    – Austin Fox
    Jan 13 at 18:18
  • $\begingroup$ yes this is a signum function $\endgroup$
    – Eli
    Jan 13 at 18:29
  • $\begingroup$ okay, and just to clarify: d is the flywheel diameter, R is flywheel radius, m is the block mass, and (x-x0) represents the preloaded spring distance from free length? $\endgroup$
    – Austin Fox
    Jan 13 at 18:39
  • $\begingroup$ Yes sorry for that $\endgroup$
    – Eli
    Jan 13 at 19:36
  • $\begingroup$ Sorry to bother you again but my ignorance is getting in my way. What I mean to say in this problem is that there is a burst of force at the beginning resting position which has a force-time function that is -parabolic. but this push only lasts about 0.1 second while the total movement time for the system is about 1 second. I wanted to make sure that the formula you have provided includes this and if not how can it be modified to reflect the brief push I described. $\endgroup$
    – Austin Fox
    Jan 13 at 21:46
-1
$\begingroup$

I would imagine that this problem is best solved by using a work-energy balance, but it looks like it fits a second order DE system well. Could someone provide me some guidance on how to solve for the travel function x(t) for this?

Because there is friction involved kinetic energy is not conserved.

So you need to set up a Newtonian Equation of Motion.

$$F(t)=-k(x-x_0)-mguv-F_{Flywheel}$$ $$ma(t)+k(x-x_0)+mguv+F_{Flywheel}=0\tag{1}$$

For $F_{Flywheel}$ consider the flywheel of radius $R$ and Intertial moment $I$. Motion of the mass requires angular acceleration of the flywheel so that:

$$\tau=I\alpha$$

$$v=\omega R$$

$$\frac{\mathrm{d}v}{\mathrm{d}t}=\alpha R$$

$$F_{Flywheel}R=\tau R$$

$$F_{Flywheel}=\frac{\mathrm{d}v}{\mathrm{d}t}\frac{I}{R}$$

Inserting into $(1)$ we get:

$$Ax^2+k(x-x_0)+mgu\frac{\mathrm{d}x}{\mathrm{d}t}+\frac{\mathrm{d}^2x}{\mathrm{d}t^2}\frac{I}{R}=0\tag{2}$$

This may be solved by means of an integration constant (for second order ODE).

$\endgroup$
2
  • $\begingroup$ Is the (x-xo) term referring to the preload distance, here Xo is the natural resting length and x is the preload distance? $\endgroup$
    – Austin Fox
    Jan 13 at 17:38
  • $\begingroup$ I find the units of the friction force $mguv$ troubling! $mg$ already has units of a force. This means $uv$ should be dimensionless. If $v$ is indeed velocity what does that imply for the friction coefficient $u$?! There's also an equation $v=\dot\omega R$. Shouldn't this be $v=\omega R$? $\endgroup$
    – Newbie
    Jan 13 at 18:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.