Acceleration at the tip of a rotating flexible rod

In summary: The gui allows me to input a target angle.The target angle is the angle the arm is pointing towards.I have a setup which has a flexible rod that is rotated by a motor. The motor has encoder unit on it which gives the position in quad counts. This we can easily convert to degrees or radians. I have made many tests and the readings are perfect. So I used the plot of the position values and found the angular acceleration. This is convereted to m/s^2 by multiplying with the length of the rod. Though its a flexible rod and this is not valid, this would give me an idea.At the same time, at the tip of the flexbile rod an
  • #1
Ben9622111222
31
0
Hello,I have a setup which has a flexible rod that is rotated by a motor. The motor has encoder unit on it which gives the position in quad counts. This we can easily convert to degrees or radians. I have made many tests and the readings are perfect. So I used the plot of the position values and found the angular acceleration. This is convereted to m/s^2 by multiplying with the length of the rod. Though its a flexible rod and this is not valid, this would give me an idea.

At the same time, at the tip of the flexbile rod an accelerometer sensor is fixed and the acceleration at the tip is noted in m/s^2. These values are almost 1.5 to 1.8 times higher than the encoder calculated value. I have made many tests on the sensor as well and its is also working perfectly fine.

The question is that, if this is actually expected? Can tip acceleration be higher than at the axis? Can I trust my values?

If not, could anyone suggest or tip the reason for the higher values... Kindly reply.
 
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  • #2
Hi Ben,

Angular acceleration times arm length gives tangential acceleration. There is also a radial acceleration component in the rotating frame.
What does the accelerometer sensor measure ? Does it have three axes or does it work otherwise ?
 
  • #3
The sensor has 3 axis, but its mounted such that z axis observed. See the attached pictures for clear understanding please.The sensor is LIS344ALH from ST Microelectronics.
 

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  • #4
Well, that is peculiar. I remember an old thread; perhaps you can find something there...
 
  • #5
Thank you. I just read through the thread. It is a similar setup but he can trust his readings because he just have one set of them... the ones from the sensor. In my case, there are 2 sets of readings, both for the same motion, but at different positions and different sensors. I have to trust both values as well as both have been checked and recheked according to specifications and functionality..

The question I have is this... Is there by any chance, a possibility that the acceleration at the tip can be higher than that at the axis.

The situation here has a flexible rod.
 
  • #6
Ben9622111222 said:
Is there by any chance, a possibility that the acceleration at the tip can be higher than that at the axis.
Hard to say without a good idea of what's actually happening. But yes, if the motor accelerates and the rod flexes, you can add the contributions to the acceleration. But the flexing leads to oscillations, I suspect, so that's an effect that should average out at some point. Is there a recording of the acceleration versus time that shows these oscillations ?
Can you check that constant angular speed gives a (close to) zero acceleration reading ?
 
  • #7
You can see a Oscilloscope plot of the movement. At higher inputs you can see the plot that has prominent oscillations. Yes, if i give a very small input, the systtem rotates very slowly, or i would say, in a constant ang. velocity and the acceleration value is so small, close to zero. The funny thing is, at these speeds the encoder and the sensor values match.

(Inputs means I have a gui where i can give in values from 0 to infinity. The higher the value, lesser the time to reach final position. But there is a control loop regulating the motor current, so above a certain input, the time to reach final position remains constant.)
 

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  • #8
Picture is more than a thousand words. But, just to make sure: what we see is a plot of what versus what ? Sensor output (*) ? And what is the sequence (*) ?
I see even the 91322 with the small amplitude has an oscillation superposed - so why the encoder and sensor should match there and not for the other picture is then strange.

The 'fft' in the file name and the naming of the axes are rather confusing to me. Also, your gui allows you to input what ? The target angle ? And then it issues a burst of power of the right amplitude and a fixed duration ?

(*) seems logical from the context - you describe only one sensor - and if your input is a target angle, then the contraption first accelerates and halfway it decelerates.
 
  • #9
Ohh... sorry... my mistake.
The bold plot is the sensor output in volts vs time in milli seconds and u can see an FFT generated for that (bottom of the screen, thin line). Ignore that FFT plot.

As to the pictures above, none are plots for the situation for a very low gui input. Its so hard to get them on oscilloscope because its close to zero and hence wouldnot trigger the scope to get a reading. So no image of that.

Ya, I can input the target angle, which I always give as 90 degrees. See picture for this. If you know EPOS I can skip an explanation. Otherwise, the easiest explanation is that the profile acceleration and profile decceleration are the inputs, and higher the value ( I started with 100 and 100 and in steps went till 10000 and 10000), the higher values of acceleration are plotted on the oscilloscope. The pictures in #7, one was 1000 and 1000 for acc. and decc. respectively and other is for 2000 and 2000. You can see the voltage is more and the oscillations are more prominent in one figure than the other.
 

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  • #10
BvU said:
(*) seems logical from the context - you describe only one sensor - and if your input is a target angle, then the contraption first accelerates and halfway it decelerates.
Yes, that's true, its a triangular velocity curve movement. An internet serach found picture is attached.
 

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  • #11
Hehe, I'm flattered you should think I might know Epos (I don't. But it looks like a nice tool. It even has an auto-tuning facility !)

Epos.jpg has a box that seems to say "sinusoidal". Would that be for ##\ddot \theta##, or just an approximation for ##\theta## itself ?

The fft is interesting too: you do see the oscillations (probably - the left dashed line) and the main noise component (right dashed line) sitting there.

Trying to link that to the time scale fails, so I must be doing something wrong: I see about 3.5 oscillations/division which would be 330 Hz, not the 33 or 34.

Coming back to your original problem (the factor 1.5 to 1.8): I 'm afraid I can't help you very much with that. Perhaps you can experiment with damping or stiffening the rod ? Another approach might be to filter out the oscillation frequency somehow, either digitally or with a few R and C ?
 
  • #12
Yes 3 oscillations per division. Each divison is 100 ms. So 33.33 ms is one oscillation. Inverse is 30 Hz...Rough calculation. Correct time stamps give 35 Hz exactly. Thanks for the help anyway. Will try out something.
 
  • #13
One (perhaps 2) more question: with the 1000 acc picture, the -1 division acceleration is reached in ##\approx.## 50 ms, and in the 2000 picture much faster. Is that one of the parameters you chose, or was it Epos that did that ? Also: the deceleration phase doesn't end as abruptly, so there's something going on too ?
 
  • #14
The time is controlled by the epos algoritham. I have nothing to do with that.

Yes, obviously decceleration would not end abruptly. Its a flexible rod, so due to sudden halt, there is free vibrations at the tip. Also the regulator plays a part in making the stopping a bit smoother. both contribute to this shape of the plot.
 
  • #15
Ben9622111222 said:
At the same time, at the tip of the flexbile rod an accelerometer sensor is fixed and the acceleration at the tip is noted in m/s^2. These values are almost 1.5 to 1.8 times higher than the encoder calculated value.
Your picture says that that this is in the vertical plane? How are you accounting for the 1g offset in the the accelerometer reading?
 
  • #16
A.T. said:
How are you accounting for the 1g offset in the the accelerometer reading?

Suppose the rod is standing parallel to the Earth's surface. Now the sensor reads 1,73 Volts from ground. For the accelartion I take this value as the base. Suppose the acceleration value read is 450 mV from ground, then the value is 1730-450 = 1280 mV. 1280/200 = 64 m/s^2. where 200 is the sensitivity of sensor.

When the rotation is complete, the rod is perpendicular to the Earth's surface. The sensor reads 1,53 volts now. ( beacsue sensitivity is 200 mV/g). hence 200 less than other value. To calculate the decceleration I use 1,53.

Please see the pictures in #7. You can notice the offset in intial and final positions. The scale of oscilloscope plot is 1:2.
 
  • #17
Ben9622111222 said:
Hello,

I have a setup which has a flexible rod that is rotated by a motor. The motor has encoder unit on it which gives the position in quad counts. This we can easily convert to degrees or radians. I have made many tests and the readings are perfect. So I used the plot of the position values and found the angular acceleration. This is convereted to m/s^2 by multiplying with the length of the rod. Though its a flexible rod and this is not valid, this would give me an idea.

At the same time, at the tip of the flexbile rod an accelerometer sensor is fixed and the acceleration at the tip is noted in m/s^2. These values are almost 1.5 to 1.8 times higher than the encoder calculated value. I have made many tests on the sensor as well and its is also working perfectly fine.

Interesting setup. You have 2 measurements of acceleration- one at the base of the rod, measured in the 'lab frame' of reference, and another at the tip, measured in the rotating frame of reference. If you transform one to the other under the assumption that the rod is a rigid beam, the residual difference will be due to the rod flexing- and will be periodic, but I can imagine the specifics can become quite complicated since there's competition between the resonant oscillation frequency of the rod (a function of rod length and 'bending modulus') and the rotation frequency of the base- this would be the driving frequency for the rod oscillation.

This is an excellent undergrad problem to work on...
 
  • #18
Andy Resnick said:
I can imagine the specifics can become quite complicated since there's competition between the resonant oscillation frequency of the rod (a function of rod length and 'bending modulus') and the rotation frequency of the base- this would be the driving frequency for the rod oscillation.

This is a point I have been working for some days now. I have made some observations and one among that is:

The time to achieve maximum acceleration is a multiple of the resonant frequency. Please see picture attached in #10. Such a situation is ideal. In reality acceleration curve will have a slope to reach max. Can this be a cause for the high value? Like a exitation effect?

I had thought about it. but the problem is that, incase of very high speed acceleration, the time to achieeve constant acc. or max. acc is 9 ms and the resonact frequecy is 28,6 ms. The above point doesn't stand here. So what can be the reason here. Can it be, that such sudden change lead to more flexing and tip vibration and affect more.
 
  • #19
Ben9622111222 said:
<snip> resonact frequecy is 28,6 ms. <snip>.

I don't understand what you mean- do you mean the period of beam oscillation at it's lowest resonance? How was this determined? Resonant frequencies of cantilevered beams can be calculated fairly easily *if* you know the Young's modulus of the material.

How did you determine a resonant period of 28.6 ms?
 
  • #20
Andy Resnick said:
How did you determine a resonant period of 28.6 ms?

You can see the pictures in #7. The time between 2 consecutive peaks is the natural frequency. It was found as 35 Hz which is 28,6 ms. Also an FFT analysis would give the same.

Another method was also used. A function generator was connected to the motor. And the frequency was increased by sweeping from 0 to 2000Hz. I could find upto 12 modes. All three methods gave the same result of 35 Hz as first mode.
 
  • #21
Ben9622111222 said:
You can see the pictures in #7. The time between 2 consecutive peaks is the natural frequency. It was found as 35 Hz which is 28,6 ms. Also an FFT analysis would give the same.

Another method was also used. A function generator was connected to the motor. And the frequency was increased by sweeping from 0 to 2000Hz. I could find upto 12 modes. All three methods gave the same result of 35 Hz as first mode.

Great- now see how well that agrees with a model for a cantilevered beam of length L:

http://iitg.vlab.co.in/?sub=62&brch=175&sim=1080&cnt=1
 
  • #22
There's a few questions I've got for you -

In "FFT full curve (1).jpg", this is for one full rotation, right? (not swinging back and forth?)

Have you considered that there might be a time delay between your two signals?
Supposing the motor applies a torque at the center, the rod flexes, and then the tip begins to accelerate. You could simply try shifting one set of time data by a small amount to see if it helps resolve your 1.5 to 1.8 factor.

In your "FFT full curve (1).jpg" images the overall sine wave shape looks like the effect of rotating in the vertical plane while measuring force radially. Have you double checked that your sensor is measuring in the right direction? I'd say turn your apparatus so that it rotates in the horizontal plane, and see if you get the similar results.
 
  • #23
Andy Resnick said:
Great- now see how well that agrees with a model for a cantilevered beam of length L:

http://iitg.vlab.co.in/?sub=62&brch=175&sim=1080&cnt=1

The ''figure'' shows the setup. While testing the beam showed vibration charachteristics of free-free beam and not of cantilever beam, that is, at 35 Hz which is the first mode, the beam vibrated with 2 nodes.

My end objective is to develop a Mathematical model to describe the beam rotation. This will, i hope help predict the flexing nature, tip displacement, residual vibration after stopping etc.

So, I have now developed a state-space equation, which I am not sure if its right or wrong, but anyways, the equation is also attached with this thread. Finding this equation was pretty easy. Now to solve this equation, I need 4 terms.
1.Damping ratio
2.Natural frequency
3.beam boudary conditions
4.Angular acceleration.

I have found 1 and 2 . Finding 3 is the link you have shared. Unfortunately the system is too complex to analyse that way(Overhanging section and point mass etc).(Atleast for me). If I can use 1,2 and 4 and simply put a trial value for 3, I will get a result. This result is the nature of the tip movement which I already have with me using the accelerometer sensor. I can then compare the two and find the value of 3.

So essentuialy I have 3 unknowns. Acceleration , boundary condition and the correctness of the Matlab model.

The problem is that, I can't make sure which acceleration value is correct. The encoder or the sensor or neither. I have to make sure the acceleration value is correct. After this I can find the trial value for 3 that gives me comparable simulation results with the sensor values. Then I have all 4 values for an specific movement.

Except acceleration all other values are constant for all movements. So now I can make other movement speeds and compare with sensor values. This will ultimately prove right or wrong the correctness of the state space equation.

I hope you get it.
 

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  • #24
elegysix said:
In your "FFT full curve (1).jpg" images the overall sine wave shape looks like the effect of rotating in the vertical plane while measuring force radially. Have you double checked that your sensor is measuring in the right direction? I'd say turn your apparatus so that it rotates in the horizontal plane, and see if you get the similar results.

Yes, I have done the measurements in horizontal plane as well. The values are the same. And the movement is just one direction. 90 degrees, stop.
elegysix said:
Have you considered that there might be a time delay between your two signals?
Supposing the motor applies a torque at the center, the rod flexes, and then the tip begins to accelerate. You could simply try shifting one set of time data by a small amount to see if it helps resolve your 1.5 to 1.8 factor.

The thing is I am not looking for the acceleration at a time t. I am looking for the maximum value of acceleration. It can be that the 2 methods read in different time stamps, but the maximum value need to be in same right?
 
  • #25
Ben9622111222 said:
So I used the plot of the position values and found the angular acceleration.
How exactly did you do that? Differentiating measured noisy values twice will just increase the noise. Fitting a smooth curve will reduce peak accelerations.
 
  • #26
A.T. said:
How exactly did you do that? Differentiating measured noisy values twice will just increase the noise. Fitting a smooth curve will reduce peak accelerations.

For a position curve which is plotted for constant acc. equation is s=s0+v0t+0,5at^2. Please see the attched screenshot.I find a regression equation of the same form.
 

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  • #27
Ben9622111222 said:
For a position curve which is plotted for constant acc. equation is s=s0+v0t+0,5at^2. Please see the attched screenshot.I find a regression equation of the same form.
This assumes that the acceleration is constant during that period. It doesn't give you the peak acceleration, just the average acceleration during that period.

But that aside, it might be possible that for a flexible rod, that the peak angular acceleration of the tip is higher than peak of the hub. If the tip initially lags behind due to inertia (less acceleration than the hub), then it must catch up at some point (higher acceleration than the hub), in order to traverse the same angle in total at some later point.
 
  • #28
A.T. said:
This assumes that the acceleration is constant during that period. It doesn't give you the peak acceleration, just the average acceleration during that period.

Yes, that is true, but note that the current is in saturation here... which obviously means that is the max. acceleration possible.

A.T. said:
But that aside, it might be possible that for a flexible rod, that the peak angular acceleration of the tip is higher than peak of the hub. If the tip initially lags behind due to inertia (less acceleration than the hub), then it must catch up at some point (higher acceleration than the hub), in order to traverse the same angle in total at some later point.

This topic is exactly what I want to be answered. This was a good suggestion to explain the difference in values, thank you. If I can make sure that there is a valid explanation or proof that the values are true, and its is so due to so and so reasons I can go ahead or worst case a confirmation that something is wrong, then I can try to rectify it.

Right now I am stuck without knowing to make a decision.
 
  • #29
Ben9622111222 said:
Yes, that is true, but note that the current is in saturation here... which obviously means that is the max. acceleration possible.
It means constant current, and assuming constant voltage a constant power input, right? Does this imply constant output torque?

Ben9622111222 said:
Right now I am stuck without knowing to make a decision.
You need a 3rd measurement. For example, capture the kinematics with a stroboscopic light and a camera on long exposure, or a high speed video camera.
 
  • #30
I have figured out the problem. In order to understand the tip movement the peaks of the sensor readings should be used. This is direct indication of the tip deflection or oscillations to be precise.

This peaks cannot be compared directly with the encoder value. Why? Because, at the encoder postion there is no oscillation effect, its all dampened. So the average of the oscillation peaks has to compared with the encoder value. And now values match.

Now everything is understood and clear. Thanks all for the time and valuable comments.
 

1. What is acceleration at the tip of a rotating flexible rod?

Acceleration at the tip of a rotating flexible rod refers to the rate of change of velocity at the end of a rod that is rotating around a fixed point. It is a measure of how quickly the tip of the rod is changing direction as it rotates.

2. How is acceleration at the tip of a rotating flexible rod calculated?

The acceleration at the tip of a rotating flexible rod can be calculated using the formula a = ω^2r, where a is the acceleration, ω is the angular velocity, and r is the radius of the rod.

3. What factors affect the acceleration at the tip of a rotating flexible rod?

The acceleration at the tip of a rotating flexible rod is affected by the angular velocity, the length and flexibility of the rod, and the distance from the fixed point of rotation.

4. How does the acceleration at the tip of a rotating flexible rod change with time?

The acceleration at the tip of a rotating flexible rod remains constant if the angular velocity and length of the rod are constant. However, if the angular velocity changes or the rod becomes more or less flexible, the acceleration at the tip will also change over time.

5. What are some real-world applications of understanding acceleration at the tip of a rotating flexible rod?

Understanding acceleration at the tip of a rotating flexible rod is important in fields such as engineering and physics. It can be used to analyze the behavior of rotating objects, such as propellers or helicopter blades, and to design structures that can withstand rotational forces. It is also relevant in sports, such as gymnastics and figure skating, where the flexibility and rotation of the body play a crucial role in performance.

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