Calculate Torque for Turning Cylinder 90° Clockwise/Counterclockwise

[RanTanPlan]

<< Mentor Note -- thread moved to the schoolwork forums after we found out this is really a homework assignment >>

Hi,
I've got a cylinder which needs to turn 90 deg. clockwise and 90 deg. counterclockwise (seen below from the front and above). How do I calculate the necessary torque on the engine? I can disregard the weight of the bracket holding the cylinder. Any help would be much appriciated!

 

[phinds]

How about you at least list the things that you think you might need to know in order to make such a calculation. For example, does the weight matter? Does it matter how fast you want to turn it? Etc ...

 

[BvU]

Are you in a hurry with this turning, or can it take a few days ? Makes a difference !

Are you aware of the relevant equations ? And the similarity of the equations for rotation with the equations for linear motion (the SUVAT eqns)

This link shows moments of inertia

This homework or work at home ?

Ah, Paul was (a lot ) faster...

 

[RanTanPlan]

Thanks for your replies and links. I have only tried to calculate bending moment. I have no idea how to calculate torque so I am not familiar with the equations. I expect I will need to know the weight of the cylinder, the speed as you mention and maybe moment of inertia of the cylinder.

"This homework or work at home ?" Not sure what you mean?

I guess I need the weight * g * distance to center of shaft. But not sure where to place the forces if that makes sense.

 

[phinds]

"This homework or work at home ?" Not sure what you mean?

Really? It means is this a do-it-yourself project that you are working on or are you doing it for a school assignment.

 

[RanTanPlan]

Ah, it's a school assignment

 

[BvU]

"This homework or work at home ?" Not sure what you mean?

It probably isn't a homework assignment. [edited, ahem...] If it's for a professional application I suggest you search professional assistance, in view of safety and legal stuff.

If you say 'cylinder', we assume a massive cylinder. Or do you mean a hollow cylinder with a bottom (a tin can so to say)? makes a difference !

Again PH was faster and you already replied. Homework: Post in homework and state the FULL problem and relevant equations. You are also required to post your own effort at solution !

 

[RanTanPlan]

It's a massive cylinder.
Oh, I missed there was a homework forum, thanks!

 

[phinds]

It's a massive cylinder.
Oh, I missed there was a homework forum, thanks!

Good to know. Now, you have to make some effort to solve your problem. We're not into giving answers here, we're into helping people figure out how to get answers for themselves.

 

[RanTanPlan]

Sure thing, I realize I asked a too, ah, open question. Well here is my attempt:
Weight of cylinder: 20 kg (evenly distributed load)
Lenght of cylinder: 30 cm

I guess I need a speed of about 15 rpm.

 

[BvU]

I guess I need a speed of about 15 rpm

You mentioned 90 deg clockwise and 90 deg anti-clockwise. With 15 rpm you mean that such a turn should be completed in 1/60 of a second ?
I find that the questions in #3 and #7 haven't got an answer yet.
And I'm prepared to believe 20 * 9.82 = 196, also that 20 kg * 9.82 m/s²= 196 N, but 20 * 9.82 is not 196 N. I don't see what you are calculating with that: what's the 0.1 and what's the 5 cm ?

And, eh...

Now, you have to make some effort to solve your problem.

 

[RanTanPlan]

Right, it has to turn 90 deg. then stop entireIy and later turn the opposite way. I don't understand how 15 rpm becomes 1/60 of a second? I figured it would be 1 sec to make a quarter-turn at 15 rpm?
Sorry, I shouldn't have been stingy with the units:

The 0,1 is just half the length of the cylinder and the 0,05 half the length of that. The black spot is where the shaft is attached. Sorry I'm not doing a very good job explaining what I'm doing, english is not my first language.
I guess my question is if the above is in any way a way to calculate the torque disregarding friction?

 

[RanTanPlan]

Oh, I see I also wrote the cylinder is 30 cm, it is 20 cm.

 

[gneill]

First decide on an angular velocity profile for the desired motion. From that you can find the angular acceleration profile. Torque will depend upon the angular acceleration. If the cylinder is starting from rest at one position, it'll have to accelerate up to some angular velocity and then decelerate back to zero. You haven't mentioned anything about how you will detect when the required position has been reached. Is there a feedback loop of some kind for motion control?

 

[BvU]

don't understand how 15 rpm becomes 1/60 of a second? I figured it would be 1 sec to make a quarter-turn at 15 rpm?

My bad , sorry. 1 second is a lot better. In that one second, it has to accelerate from standstill, and decelerate to standstill again,as @gneill writes, I suppose? Can you do the calculations for a linear motion, with force instead of torque and mass instead of moment of inertia ?

 

[BvU]

I guess my question is if the above is in any way a way to calculate the torque disregarding friction?

Nope. You need moment of inertia - as provided in link in #3

 

[RanTanPlan]

Gneill: Ah okay. I actually don't know about the detection of position...might be motion control.
BvU: Yes exactly. You mean like F=m*a?
"You need moment of inertia - as provided in link in #3" Ok, let's say I have the moment of inertia through the red rotation axis and the accelleration. Is the
torque =moment of inertia*angular acceleration?

 

[gneill]

Gneill: Ah okay. I actually don't know about the detection of position...might be motion control.

You'll need something. Relying on timings alone will find your positioning drifting away from where you expect it to be due to real life things like error tolerances, changes in load as friction changes over time (bearings wear), temperature variations, power supply drift,... Have you accounted for the rotational inertia of the motor?

I once had a wonderful little book by Caxton C. Foster called "Real Time Programming: Neglected Topics" which went over the development of practical control methods when interfacing a computer to real devices like motors. Hands on stuff and code (in BASIC! the book is that old). How to minimize positioning error, how to avoid "hunting" at the target position, etc. If you can find it, pick it up. It's a paperback and probably goes for less than ten dollars (U.S.) for a used copy.

In your diagram your velocity should be angular velocity, not linear. From your velocity profile you should be able to develop the angular acceleration profile. That, along with your moment of inertia and anything you can feed in about frictional torque for your device will get you to the required maximum torque value.

 

[BvU]

Is the torque =moment of inertia*angular acceleration?

Correct: ##\ \ \tau = I\alpha##

 

[RanTanPlan]

Gneill: I see, thanks for the suggestion, might check that out.
Right! but is this what you would call an angular accelleration profile?

BvU: Ok, so I guess I need to find the higest accelleration to multiply with moment of inertia?

 

[gneill]

That's an angular velocity profile. You need to "differentiate" it to find the angular acceleration profile.

 

[RanTanPlan]

Ok, thanks. Oh boy, it's been a long time since I have differentiated anything. So say I differentiate f(x)=ax+b for the part where the accelleration increases and get f'(x)=a because the rate of change in the accelleration is 0. Hmm, guess I'm not sure what I'm looking for..

 

[gneill]

You can "differentiate" directly from your velocity profile diagram. The slope of the line anywhere along the velocity profile is the acceleration at that point.

 

[RanTanPlan]

Ah okay, so I'm looking for the biggest slope..

 

[RanTanPlan]

Anyway, thanks for the help everyone!