Combined Translational & Rotational System Transfer Function

In summary, the 3 kg mass with teeth on it rotated counterclockwise due to the transnational system. The force of gravity was pulling it down, but the damper was countering that. The other term in the equation, s^{2}X(s), is the acceleration of the mass.
  • #1
GreenPrint
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0

Homework Statement



I'm given the following diagram

AvChBTn.png


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

Homework Equations


The Attempt at a Solution



Apparently this is free body diagram

uqgDxRO.png


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

For the rectangular mass, I don't really understand what [itex]F(s)[/itex] 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 [itex]F(s) = (2s^{2}+2s+3)X(s)[/itex]. I'm confused as to why. I know that the force of gravity is pulling down the block by [itex]mg[/itex]. I assume that [itex]2s^{2}X(s)[/itex]. I understand that [itex]s^{2}X(s)[/itex] is the acceleration of the mass. I however don't understand how [itex]mg = ma[/itex]. I understand that the damper is pulling the mass down by the force [itex]2sX(s)[/itex], so this makes since. I understand that the force of the spring acting on the mass is [itex]3X(s)[/itex]. I'm just confused by the following term [itex]s^{2}X(s)[/itex] and am unsure where it comes from.

Thanks for any help.
 
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  • #2
Hi Green,

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

As you indicate, Teq goes clockwise. So the inertial torque and the friction torque go ccw. What you call Fr 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 [itex]2s^{2}X(s)[/itex] 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.
 
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  • #3
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 [itex]T_{eq}[/itex] goes clockwise by starting at the [itex]3 kg-m^{2}[/itex], without teeth, and look at [itex]T(t)[/itex]. I took [itex]T(t)[/itex] is pointing in the clockwise direction. As a result the second shaft with gears [itex]N_{2} = 20[/itex] and [itex]N_{3} = 30[/itex] would rotate counterclockwise. As a result the shaft connect to the [itex]3 kg-m^{2}[/itex] with teeth with gear [itex]N_{4} = 60[/itex] would rotate clockwise. I'm unsure if how I determined this is accurate, but to me it makes since. The [itex]T_{eq}[/itex] as a result of [itex]T(t)[/itex] 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 [itex]T_{eq}(s) = (J_{eq}s^{2} + D_{eq}s + K_{eq})Θ(s)[/itex]

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

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

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

Since there are no springs directly connected to the rotational system [itex]K = 0[/itex]

Hence my equation [itex]T_{eq}(s) = (J_{eq}s^{2} + D_{eq}s)Θ(s)[/itex] is [itex]T_{eq}(s) = ((51 kg-m^{2})s^{2} + (5 \frac{N-s-m}{rad})(s))θ(s)[/itex]. 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 [itex]J_{eq}s^{2}[/itex] and [itex]D_{eq}s[/itex] point in the opposite direction as [itex]T_{eq}s^{2}[/itex] and [itex]θ[/itex]. 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 [itex]T_{eq}[/itex].

If [itex]x(t)[/itex] 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 [itex]3 kg-m^2[/itex] 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 [itex]2 kg[/itex] mass is moving downwards the net force [itex]F(s)[/itex] must be downwards. Apparently for transnational systems I'm trying to solve for [itex]F(s) = (J_{eq}s^2 + D_{eq}s + K_{eq})X(s)[/itex]. 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 [itex]J_{eq}[/itex], I don't really understand how or why there even is one but apparently it's [itex]2 kg-m^{2}[/itex]

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

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

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

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

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

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

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

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

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.
 
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  • #4
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 N1,2,3,4, 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. [itex]T_{eq}(s) = 4T(s)[/itex] 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 :smile:).

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:
[itex]J_{eq}s^{2}[/itex] and [itex]D_{eq}s[/itex] point in the opposite direction as [itex]T_{eq}[/itex] and [itex]θ[/itex] because they counteract [itex]T_{eq}[/itex]: 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 [itex]θ[/itex] because we defined positive ##\theta## and positive [itex]T_{eq}[/itex] as clockwise. (And [itex]T_{eq}[/itex] 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.
 
  • #5
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.
 

Related to Combined Translational & Rotational System Transfer Function

1. What is a Combined Translational & Rotational System Transfer Function?

A Combined Translational & Rotational System Transfer Function is a mathematical representation of the relationship between the input and output signals in a system that combines both translational and rotational motion. It describes how the system will respond to different inputs and helps to analyze and design the system for optimal performance.

2. How is the Combined Translational & Rotational System Transfer Function different from a regular transfer function?

The Combined Translational & Rotational System Transfer Function is different from a regular transfer function in that it takes into account both translational and rotational motion, whereas a regular transfer function only considers one type of motion. This makes it more suitable for analyzing systems that involve both types of motion, such as a rotating machine with a linear actuator.

3. What factors are included in the Combined Translational & Rotational System Transfer Function?

The Combined Translational & Rotational System Transfer Function includes factors such as the mass, inertia, damping, and stiffness of the system, as well as the input force or torque and the output displacement or angle. These factors together determine how the system will respond to different inputs.

4. How is the Combined Translational & Rotational System Transfer Function used in practical applications?

The Combined Translational & Rotational System Transfer Function is used in practical applications to design and analyze systems that involve both translational and rotational motion, such as robots, vehicles, and machinery. It helps engineers to optimize the performance of these systems and ensure they function as intended.

5. Can the Combined Translational & Rotational System Transfer Function be used for systems with multiple inputs and outputs?

Yes, the Combined Translational & Rotational System Transfer Function can be used for systems with multiple inputs and outputs. In this case, the transfer function will be a matrix that relates all the input signals to all the output signals. This allows for more complex systems to be analyzed and designed using the transfer function approach.

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