Torque Required to Rotate a Hollow Cylinder

This will give you the basics...

Hollow Cylinder

Equation:

and

Open Calculator

Where:

T =	Required Torque, lb-ft
WK 2 =	Inertia of load to be accelerated lb-ft 2 (See moment of inertia calculations)
=	Change of speed, rpm
t =	Time to accelerate the load, seconds
W =	Weight of object, lb
R1 =	Outside Radius of cylinder, ft
R2 =	Inside Radius of cylinder, ft

http://www.engineersedge.com/motors/hollow_cylinder_axis_torque_force_equation.htm

Yes, but..........................

The diameter of the cylinder is immaterial.

What really matters is the diameter of the drive, and driven sprockets, which OP has not given us.

Also complicating this is the rolling resistance of the drum against the idler rollers that support the weight.

The diameter of the drive and driven sprockets are more relative to the speed of the drum which I have given as 0.04 rpm.

The diameter of the cylinder is very relevant to the moment of inertia, and the capacity of material it will hold.

Sorry.

Good luck.

I shall try to explain how I see your problem and its solution.

The cylinder is filled with the sludge and the outside is empty of any thing. When it starts to rotate the internal wall will slide on the sludge which is still at rest and accelerate progressively due to shear strain, same for the bottom. From layer to layer the sludge depending on its viscosity (apparent) will accelerate till it turns all to same speed as cylinder. At same time the weight will load the rollers and a rolling and sliding (if the bearings are journals) resistance will also appear. The 5 seconds depend on the shear resistance the sludge will offer to the turning cylinder. If the dimensions are di= internal diameter/h= height/ρ1= specific mass of barrel material/ρ2= specific mass of sludge (wet)/τ=shear strain at wall between cylinder and sludge/Mr= resistance moment of rollers. Your moment will be M= J*(dω/dt)+(π*di*h*di/2+π/16*di^3)*τ+Mr. The problem is more complex since the shear depends on the relative sliding speed between layers. If you work in metric units the problem of lbs and slugs is easier to solve!

Just to comment on slugs - slug is a unit of mass = 32 lb. 1 lbf accelerates 1 slug at 1 ft/sec². Analogous to bewton and kg in SI system. If you want to use formulas without any g factors, need to use lbf for force and slug for mass. If you want force in lbf and mass in lbm you need to put g = 32 ft/sec² in at the appropriate place (as in earlier issues of Perry, Chemical Engineers' Handbook)

Metric units would be less confusing to use for your calculations. Here is a link to Online Conversions. Slugs and other quaint English units are out of step with the rest of the world and will probably disappear when the few remaining holdouts retire.

Your objective, as I understand it, is to select a motor that will bring a 7500 lb. (3402 kg) 8 ft diameter cylinder, 80% full of "a watery sludge mixture," from rest to 0.04 rpm in no more than 5 seconds. The cylinder rotates horizontally, supported by rollers. So there will be a free surface in the cylinder, and between the sludge mixture and the cylinder wall will be an air space. The moment of inertia calculation for the cylinder charged with sludge mixture will therefore be complicated because of its non-uniform mass distribution about the axis of rotation.

The "wet sludge" (50-60lb/cu.ft) is only a third of the "watery sludge mixture" and that other 2/3 needs to be taken into account for the moment of inertia calculation.

Do you have to fill the cylinder before you rotate it? Startup torque could be reduced a lot if you could rotate an empty cylinder and then fill it once you have loaded it with angular momentum. The motor would then sustain the rotation, and would not need to start a huge load from rest. You could also rig some falling weight to assist starting from rest, which is when your motor will be most challenged.

The cylinder itself weighs 7500lbs and is filled approx 80% full of the watery sludge mix so the total weight is around 20,106 lbs. The cylinder can be started before it's filled. The water in the cylinder quickly drains off and only the solids remain, which leaves it about 1/3 full of a wet sludge. When the cylinder is first filled it really doesn't rotate the load, it seems to spin around it since it is mostly water. As the water drains off, the weight decreases but it then starts to turn a solid load that adheres to the side of the tank and has an angle of repose of about 45%.

Betwen the sludge mixture and the cylinder wall are filter tiles that allow water to drain off.

I agree that calculating the moment of inertia of the sludge mixture is the real difficulty here. The rotation is slow so the angle of repose of the sludge is approx. 45%. I haven't found any examples online where this is demonstrated, and if it is I believe I will have a hard time determining the necessary information. For example, the shear resistance of the sludge (as one poster previously mentioned) is critical.

I havent gotten into the rolling resistance of the rollers but I know that can be figured out.

I appreciate everyone's input.

BIGG, everything I've researched says that is the right way to approach it. Calculating the inertia of the empty tank is no problem. I'm having difficulty getting started with determining the torque required to continue turning the cylinder with the sludge at a 35-45 deg angle of repose.

BIGG you mentioned attacking it as a circular segment and I've done some research on that approach. What I don't understand is how determining the center of gravity of the load will help find the torque required to turn it. I cant find anything that explains this. The cylinder is about 1/3 full, and with the angle of repose I can determine what my segment would be, I just don't know how to eventually get a torque value from that.

Thanks for the input. I think I'm on the right track to solving this but my lack of physics education is killing me.

I've delayed long enough. The distance, x, of the CG to the center of a semi-circle is 0.424 * r, where r is the radius. Imagine the sludge to be the white half of the cylinder below (end view). Then the weight of that much sludge would be acting at a distance x from the center of rotation. Isn't that the definition of a torque?

If you want to try 1/3 full at 45deg angle, x for a circular segment is (2r³sin³a)/(3A) where A=r²(2a-sin²a)/2 and a is half the included angle. Then you'll have to use a little vector geometry to get the horizontal component of r for the moment length.

Being the rude SOB I am, I have to ask: how did someone with a 'lack of physics education' become responsible for solving this problem?

Dear ary520,

If all mathematical calculations and other methods already stated are not suffice, you may try with a very crude method to determine the starting torque as follow:

Pl. fixup a suitable torque wrench on the shaft and try to rotate the shaft; note down the reading in the torque wrench when the shaft is starting to move.

Thanks,

Manindra