What amount of torque is required to spin a 12" dia. Disc weighing 6.25 lb at 1 Rpm? Load is the disc and a complete rotation is 60 seconds . Motor shaft is the axis on horizontal plane , similar too a phono turntable .

You need torque to (a)get it spinning at 1 RPM in the first place, plus (b) to overcome friction of the bearings in the turntable. The answer to (a) depends on how quickly from a standing start you want the turntable to accelerate up to 1 RPM. The basic formula to use is:


T = I \alpha


where 'I' = the moment of inertia of the disc, which is equal to (1/2)mR^2, and \alpha is the acceleration of the disc in units of radians per second squared. Assuming constant acceleration we have


\alpha = \frac{ \Delta \omega}{\Delta t}


where \Delta \omega is the change in rotational velocity, which is from zero rotation to 1 RPM , or \frac {2 \pi} {60} radians per second. \Delta t is the time it takes to reach 1 RPM, which you need to specify. So we have:


T = \frac 1 2 mR^2 (\frac {2 \pi} {60} )(\frac {1 }{\Delta T})


To get it in the correct units we need to divide it all by 32.2 \frac {lbm-ft}{lbf -s^2}.

To work through an example let's assume that you set \Delta t = 1 second:


T = \frac 1 2 (6.25 lbm)(0.5 ft)^2 ( \frac {2 \pi} {60 s})( \frac {1 }{1 s}) (\frac 1 {32.2} \frac {lbf-s^2}{lbm - ft} ) = 0.00254 ft-lb


This is not much at all, and is probably swamped by the friction of the bearings. To determine the torque needed to overcome friction you'll have to do some measurements of your mechanism.

Thank You for showing the formula along with it's completion an answer . WingFling