## Torque required to rotate a hollow cylinder

First let me start off by saying I was a liberal arts major so I need some help with some formulas I've researched.
I'm rotating a hollow cylinder that is 8ft in dia. x 8 ft long. The cylinder weighs 3750 lbs and will be filled with 20,093 lbs of watered sewage sludge. Total weight = 23,843lbs. The cylinder sits and rotates on 4 free spinning rollers. The water drains off over a 6 hour period and the remaining dry material weighs approx 6000 - 7500 lbs. I assumed that a drive package with enough torque to turn the cylinder with the full load of watered sludge would also turn the cylinder with the dry material, even though the material has changed and the dry material has an angle of repose that the drive package will have to overcome when turning the cylinder. Any insight or suggestions on that assumption would be helpful.

I understand that Torque=Moment of Inertia * angular acceleration

Moment of Inertia of hollow cylinder = Mass * r^2

First question: Can mass be expressed in lbs or does it have to be Newtons? As I understand it, there is a difference in mass and weight and I'm not sure how this affects the calculations.

Second question: I've seen different ways of calculating angular acceleration. I understand it to be:

Ang. acceleration in rad/sec^2 = change in velocity/ change in time * 2(pi)
I may be incorrect and this is where I need help.
The tank rotates at 0.04 rpm. (yes it's really that slow) and the acceleration time is from 0-0.04 in 5 seconds.

Once I determine the torque required I can determine hp and size a drive package appropriately. Thanks in advance!
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 Two observations. You are making the problem more difficult than it needs to be while at the same time omitting lots of critical information required to correctly design the drive system. You would probably have to do some testing to fully define the design requirements. But if this were my job assignment, I'd ignore all that and ask myself under what conditions would my torque requirements be the highest? Perhaps that might be with all the sludge clumped up 3 feet off center, and at that time in the cycle the sludge had dried to a weight of 10 kips. Torque required would therefore be 30,000 ft-lbs. Under all other operating conditions the drive requirements might be much less, but this load would define the system.

 Quote by ary520 First question: Can mass be expressed in lbs or does it have to be Newtons? As I understand it, there is a difference in mass and weight and I'm not sure how this affects the calculations.
Mass isn't expressed in Newtons. Newtons are mass * acceleration

Newtons are a measure of force.

Pounds are a little tricky to deal with when considering mass because it is a vague term. Typically you would use pounds as a force, not as a mass. But with everyday objects you can think of a pound of something the way you would think of a kilogram of something, even though one is a force and the other is a mass.

I think the best thing to do is convert everything to metric first, do the calculations, then convert back if necessary.

## Torque required to rotate a hollow cylinder

I love the English system, except for the mass weight confusion. I absolutely refuse to deal with slugs. So I also work these problems in metric, and convert back if I have to.

 Quote by Pkruse I love the English system, except for the mass weight confusion. I absolutely refuse to deal with slugs. So I also work these problems in metric, and convert back if I have to.
Don't get me wrong... I love the English system too. There's nuance and beauty and story in it. Metric feels cold too calculated.
 We seem to have similar opinions on that.
 Yeah, I mean you just can't order a 41cm pizza. It's gotta be 16".
 It's all about "Peak Torque" as Pkruse alludes. Consider everything you can think of: T-peak = T-acceleration + T-friction + T-offsetload + T-blah,blah,blah Then, the lazy (smart?) engineer's way to solve this is double or triple T-peak for a factor of safety, then select your gearmotor & drive system for that. Torque is cheap. Breakdowns, repairs, and lost production is outrageously expensive. IMHO, you should run the numbers in Imperial System if you are dealing with "pounds", "feet", and "inches". Be rigorous, learn lbf, lbm, slugs, ft-lbf, and the conversion constant g0, what they mean and how to work with those numbers. Once you get the hang of it, it ain't so bad. And finally if you are a liberal arts major and are trying to do engineering without proper training, well, ....ummm....how to say this?....ummmm....aaahhhh....
 And finally if you are a liberal arts major and are trying to do engineering without proper training, well, ....ummm....how to say this?....ummmm....aaahhhh....[/QUOTE] Actually I work with a lot of people with no engineering background who deisgn and build machinery every day. It is possible. You don't have to be a narcissistic engineer to do this stuff.
 So writes the person who can spell "narcissistic" but can't spell "design" properly. Or figure out how to make the text-quoting function work.

 Quote by tygerdawg so writes the person who can spell "narcissistic" but can't spell "design" properly. Or figure out how to make the text-quoting function work.
lol, tiger dog.