Combined Translational & Rotational System Transfer Function

[GreenPrint]

Homework Statement

I'm given the following diagram

And asked to find the transfer function G(s) = \frac{X(s)}{T(s)} and seem to be having some difficulty doing so.

Homework Equations

The Attempt at a Solution

Apparently this is free body diagram

I seem to be having difficulty understanding how this is the free body diagram. To me it would seem as if the the 3 kg m^{2} mass with T(t) applied to would make the 3 kg m^{2} with teeth on it rotate clockwise. The free body diagram above seems to agree with this T(t) is draw clockwise. I don't understand why J_{eq}s^{2}Θ(s) and D_{eq}sΘ(s) are drawn counterclockwise. I don't understand why the force F_{r} acting on the 3 kg m^{2} mass with teeth as a result of the transnational system is equal to F_{2} or why the force is drawn counterclockwise.

For the rectangular mass, I don't really understand what F(s) in the diagram. Apparently it would seem to be the net force on the mass. In which case according to the diagram, the mass is being displaced downwards. Apparently F(s) = (2s^{2}+2s+3)X(s). I'm confused as to why. I know that the force of gravity is pulling down the block by mg. I assume that 2s^{2}X(s). I understand that s^{2}X(s) is the acceleration of the mass. I however don't understand how mg = ma. I understand that the damper is pulling the mass down by the force 2sX(s), so this makes since. I understand that the force of the spring acting on the mass is 3X(s). I'm just confused by the following term s^{2}X(s) and am unsure where it comes from.

Thanks for any help.

 

[BvU]

Hi Green,

Here's a few things a non-expert thinks he recognizes:

As you indicate, T_eq goes clockwise. So the inertial torque and the friction torque go ccw. What you call F_r is F times Radius. The latter happens to be 2 m.

If x is a positive displacement of the 2 kg mass, you neeed a ccw force to achieve that and it is counteracted by the three upward pointing entities in the righthand fbd.

s transforms look mainly at frequency- and transient responses, not so much (not at all?) at steady state; my guess is gravity doesn't come in; the 2s^{2}X(s) in the time domain is inertia: the (reaction) force is F = -ma.
The damper isn't pulling down, it is counteracting: F = -2 v in the time domain.

 

[GreenPrint]

Thanks for your response. Should this go in the advanced physics forum? This question is actually from a control systems class, but I posted it in the physics forums because I'm having a hard time understanding the physics behind this problem, especially the free body diagram.

I was able to determine that T_{eq} goes clockwise by starting at the 3 kg-m^{2}, without teeth, and look at T(t). I took T(t) is pointing in the clockwise direction. As a result the second shaft with gears N_{2} = 20 and N_{3} = 30 would rotate counterclockwise. As a result the shaft connect to the 3 kg-m^{2} with teeth with gear N_{4} = 60 would rotate clockwise. I'm unsure if how I determined this is accurate, but to me it makes since. The T_{eq} as a result of T(t) would seem to be clockwise for the reasons mentioned above.

The first equation I'm trying to solve for is for the rotational part of the system T_{eq}(s) = (J_{eq}s^{2} + D_{eq}s + K_{eq})Θ(s)

To find T_{eq}(s) as a result of T(s) I would do the following T_{eq}(s) = \frac{20}{10}\frac{60}{30}T(s) = (2)(2)T(s) = 4T(s)

To find J_{eq} I reflect the inertia on the 3 kg-m^{2} onto the 3 kg-m^{2} with teeth. I would get (3 kg-m^{2})(\frac{20}{10}\frac{60}{30})^{2} = (3 kg-m^{2})16 = 48 kg-m^{2}. If I add this to the inertia of 3 kg-m^{2} with teeth I would get J_{eq} = 3 kg-m^{2} + 48 kg-m^{2} = 51 kg-m^{2}

To find D_{eq} I would reflect D_{2} onto 3 kg-m^{2} with teeth. I would go about this by doing (\frac{60}{30})^{2}(1 \frac{N-s-m}{rad})= (2)^{2}(1 \frac{N-s-m}{rad}) = 4(1 \frac{N-s-m}{rad}) = 4 \frac{N-s-m}{rad}. I would then have to consider the damper directly connected to the 3 kg-m^{2} with teeth, resulting in D_{eq} = 4 \frac{N-s-m}{rad} + 1 \frac{N-s-m}{rad} = 5 \frac{N-s-m}{rad}.

Since there are no springs directly connected to the rotational system K = 0

Hence my equation T_{eq}(s) = (J_{eq}s^{2} + D_{eq}s)Θ(s) is T_{eq}(s) = ((51 kg-m^{2})s^{2} + (5 \frac{N-s-m}{rad})(s))θ(s). Which all makes since to me. I know have to consider the transnational part of the system.

I however don't understand why on the free body diagram J_{eq}s^{2} and D_{eq}s point in the opposite direction as T_{eq}s^{2} and θ. Perhaps it's something that I was supposed to learn in physics but never did. So I just want to make sure that it's always the case where they are supposed to go in the opposite direction. My guess it's something that has to deal with Newton's Second Law. But this is only one force, and were not trying to find the equal and opposite force, are we? I thought we were just trying to express T_{eq}.

If x(t) which is pointing downwards is taken to be positive, I have a hard time understanding why you would need a counterclockwise force on the on the 3 kg-m^2 inertia mass with teeth. I would assume that you would need some kind of force on this object, but am unsure how you conclude its either clockwise or counterclockwise. How were you able to determine this? I can't seem to visualize a force in either direction, but know that there would need to be one.

Since the 2 kg mass is moving downwards the net force F(s) must be downwards. Apparently for transnational systems I'm trying to solve for F(s) = (J_{eq}s^2 + D_{eq}s + K_{eq})X(s). This doesn't really make any since to me. For transnational systems, why would they have inertial forces acting on them if they are not rotating? Moreover, I'm able to reflect inertial forces, damping forces, and spring forces onto one body in transnational systems? If this is the case, this would simplify the workload in other problems. I'm really curious about this.

For J_{eq}, I don't really understand how or why there even is one but apparently it's 2 kg-m^{2}

For D_{eq} we only have one damper D_{eq} = 2 \frac{N-s}{m}.

For K_{eq} we only have one spring K_{eq} = 3 \frac{N}{m}.

So my equation F(s) = (J_{eq}s^2 + D_{eq}s + K_{eq})X(s) is F(s)=(2 kg-m^{2})s^{2} + (2 \frac{N-s}{m})s + (3 \frac{N}{m}))X(s).

I must now reflect F(s) onto the 3 kg-m^{2} with teeth. To go about doing this I must find F_{eq}(s). Looking at the diagram it's an ideal gear 1:1 so F_{eq}(s) = (\frac{1}{1})^{2}F(s) = F(s)

Because T = r*F, the resulting torque on the 3 kg-m^{2} with teeth is simply 2F(s).

So the net torque on the 3 kg-m^{2} becomes
T_{eq} + 2F(s) = 4T(s)
((51s^{2} + 5s)(s))θ(s) + 2(2s^{2} + 2s + 3)X(s) = 4T(s)
((51s^{2} + 5s)(s))θ(s) + (4s^{2} + 4s + 6)X(s) = 4T(s)

Now because Θ(s) = \frac{X(s)}{2} which I'm not exactly sure why have the feeling that it has something to do with the radius being 2 m, my equation becomes
((51s^{2} + 5s)(s))\frac{X(s)}{2} + (4s^{2} + 4s + 6)X(s) = 4T(s)
(\frac{59}{2}s^{2} + \frac{13}{2}s + 6)X(s) = 4T(s)

Solving for the transfer function \frac{X(s}{T(s)} yield the answer
\frac{X(s)}{T(s)} = \frac{4}{\frac{59}{2}s^{2} + \frac{13}{2}s + 6}
\frac{X(s)}{T(s)} = \frac{8}{59s^{2} + 13s + 12}

So I was able to follow the solutions manual more less and get the answer they had. I am still frustrated because I seem to be still struggling with some of the basic physics aspects that I would like to understand and hope someone can assist me in the process.

 

[BvU]

Hello again,

We make I nice team: I did this as a physicist without much idea of the control engineering applications. Only it was some time ago. And no, it is Introductory level.

Because of the elaborate rack/pinion drawing I kind of skipped the interpretation of N_1,2,3,4, the gearbox, but it's clear to me now. so the D₂ is clear also. And it counteracts T(t) in the time domain.

You did a lot of work and I would need some time to read through. Point is, we seem to speak different languages. I, for one, out of necessity have to start in the time domain and then transform. Preferably step by step, to combine steps later on (and end up with the over-all transfer function). I kind of think that that is also the intention of this exercise.
As far as I can see you skip over a lot of things that are pretty essential. There is no way I can approve of e.g. T_{eq}(s) = 4T(s) because there is so much in between. But I would have no trouble at all with $4\Theta_{eq} = \theta_{in}$. That's what gears do for a living.

Let's define $\theta_1$ input axle, $\theta_2$ intermediate axle, $\theta$ central pinion axle, and
$T_1$ torque on $N_1=10$, $T_2$ torque on $N_2=20$, $T_3$ torque on $N_3=30$, $T_{eq}$ on central pinion.
Furthermore $D_2 = 1$ Nms/rad on the intermediate axle, $D_3 = 1$ Nms/rad on the central pinion axle,
$I1 = 3$ kg m² of the input axle, $I_3 = 3$ kg m² of the central pinion with $R_3 = 2$ m.

There are easy intermediate conversions $\theta_1 = 2\theta_2 =4 \theta$ and $T_2 = 2 \ T1, \ T_{eq} = 2 \ T_3$.

Focusing on and working towards the central pinion we can write a bunch of steps in series (I am brainwashed in the time domain, so I start with that):
$$ T(t) - I_1 \ddot \theta_1 = T_1(t) \quad T_2(t) - D_1 \dot\theta_2 = T_3(t) \quad T_{eq}(t) - I_3 \ddot \theta - D_3 \dot\theta = -R_3\ F(t)$$ This should ultimately give the transfer function for $T(s) \rightarrow T_{eq}(s)\rightarrow F(s)$.
From there to the mass M:
$$F(t) - M\ \ddot x(t) - D_M\ \dot x(t) - k_M\ x(t) = 0$$ Then transform to the s-domain and start writing things out. $F(s) \rightarrow x(s)$ is the easiest part. (Sorry, no time left ).

Some comments: "reflecting zzz onto qqq" seems a dangerous business to me. I wouldn't do it: some things can 't be added up directly; they might be terms in a polynomial that must be multiplied with other polynomials later on. But maybe it is unavoidable here.

To answer some of your questions:
J_{eq}s^{2} and D_{eq}s point in the opposite direction as T_{eq} and θ because they counteract T_{eq}: For the torque left over to move the rack (in the negative x direction !) One has available T - D - J = R F (see above).
Note I have a minus on the righthand side as well: Force on the mass M is up (negative x direction) if torque is positive.

D and J point in the opposite direction as θ because we defined positive $\theta$ and positive T_{eq} as clockwise. (And T_{eq} is the torque on the central pinion in my interpretation).

The sign of the force on M is determined by the choice for the positive x direction: down is positive. Spring does $F = -kx$ and damper does $F = - D \dot x$. Inertia is negative if F on the block is positive: $F_{external} = M\ddot x$ hence $F_{inertia} = - M\ddot x$ all three together: $F_{external} + F_{inertia} + F_{damping} + F_{spring} = 0$ see above.

 

[jaycarl_u]

hello GreenPrint can i ask waht is title of the book you are using in this what is the title who are the authors? can i have the cover picture of the book and the solution manual. thanks for the reply.