Hello, I'm stuck on what should be a fairly simple problem. Trying to calculate the size of motor required to rotate steel pipe at 0.33RPM around its axis, on 2 sets of rollers, 4 wheels total (hard rubber). Pipe 48" OD Mild Steel pipe 0.5WT 24m in Length Rollers 4 x 12" Rubber wheels (40" Center to Center) on bearings This is where is stand. Torque required to accelerate the system (0.00 RPM to 0.33 RPM in 1 sec) Ta=Iα I=(∏/32)ρL(D1^4-D2^4) 1.219 D1 – External diameter of the disc [m] 1.194 D2 – Internal diameter of the disc [m] 7,850.000 ρ – Density of the load [kg/m3] 9,069.050 m – Weight of the load [kg] 24.000 L – Length of the load [m] I=3,300.689 α=(w1-w0)/t 0.000 ω0 – Initial velocity of the motor [rad/s] 0.035 ω1 – Final velocity of the motor [rad/s] or 0.33 RPM 1.000 t – Time for velocity change α=0.035 Ta=114.064 N*m Now the next part i am not too sure about, Torque required to overcome rolling resistance : Rolling resistance coefficient - f = 0.0077m Hard Rubber on Steel http://www.roymech.co.uk/Useful_Tables/Tribology/co_of_frict.htm F = f x W/R Fr=(0.0077m*9069.050kg*9.81 kg/s^2)/0.610m Trr= 1123.031 N *.610m = 685.049 N*m Total Torque required to rotate the pipe at 0.33rpms in 1 sec. T=Ta+Trr=799 N*m or 589 Lb*ft This seems low to me and this is only the torque at the pipe, once geared that torque will be significantly lower. will I know i'm missing the bearing friction torque required but that cant be that much. What else should I be calculating?, if anything. I have attached a drawing of the system, once I find the torque required for rotate the pipe I should have the required input torque dividing that Torque by the gear ratios. One more thing, Bonus question. Some of these types of pipe have a seam welded down the middle along the length of the pipe, it projects from the pipe a few mm sort of like a speed bump on the pipe. Is there a way to calculate the torque required to over come the rollers once the seams comes in contact with the roller wheels? Thanks in advance.