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I need to find the torque that will rotate a stainless steel drum that is filled with small steel balls. The total volume of this drum is 210 cubic feet, and there will be 38 cubic feet of material inside at any given point in time. It is a horizontal cylinder that will rotate through its central axis. What is the torque required to start turning the drum.

Mass of Drum: 13,500 lb
Mass of Material inside: 13,700 lb
Diameter of Drum: 38"
Length of Drum: 297"

Drum to Rotate at 10 rpm

Friction in the bearings and the drive is negligible.

I know I have to consider that the material will move up the side of the drum as it rotates, but I'm not sure how to go about doing this.

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start with the following approximation: at the moment the drum starts turning, assume the balls and whatnot resting in the bottom of the drum do not move around but are for that moment stuck together so they will rotate as a single mass around the axis of the rolling drum. This distribution of mass will have a well-defined moment of inertia you can calculate, which we call (I). Next, determine how fast you wish the drum to accelerate, this number is (alpha). the starting torque is then torque (tau) = (I) x (alpha).

alternatively, you can go to the website of someone who manufactures ball mills (which is what you described) and see what they recommend for the torque and RPM of the motor drive of a ball mill of a certain shape and mass.

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  • $\begingroup$ And this should be a maximum after 90 degrees of rotation... $\endgroup$
    – Solar Mike
    Oct 26 '17 at 9:59

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