How much torque is needed to rotate 75lbs in a 7.5" diameter basket

To calculate torque you need to specify the acceleration of the apparatus and the moment of inertia of the 75 pounds of weight. So we need to know:

1. What rotational velocity do you want to achieve (in RPM)?

2. How quickly do you want the basket to get up to speed (in seconds)?

3. What is the friction this basket needs to overcome as it spins?

4. How is the 75 pounds distributed in the basket? For example - is the weight loaded out at the rim, or is it evenly distributed?

1. I will be building a centrifuge that has a 3hp motor and capable of 3450RPM, at that speed it would generate roughly 1266 g force
2. I will also install a 3ph 3hp max inverter so I can bring it up to speed slowly, time wise does not matter.
3. No friction, the basket will be mounted dirct drive inline to the motor, of course with bearings
4. My intentions would be to distribute the weight evenly, however it won't be balanced, with the design of my unit I plan to mount bearings to add support at the top of the basket. Basket dimensions are 7.5" diameter 12" deep.

Are you able to tell me how much outside force in pounds would my centrifuge generate at 1.266g force, this would help in the choice of bearings I should use for the top support.

Thank you

If the time to spin up to speed doesn't matter, and if you assume no friction, then theoretically the torque needed to get the centrifuge moving is essentially zero. This is because the torque needed is calculated from:


T = I \alpha + T_{friction}


Since T_{friction} = 0 and \alpha (rotational acceleration) is small, you can see that the total torque T is essentailly 0.

Think of this as an analogy: if I asked you how powerful an engine do I need in my car to go 60 MPH, clearly the answer depends on how fast I want to get up to that speed. A Corvette has a bigger engine than a Yugo so it can acelerate faster. If you don't care how long it takes to get up to 60 MPH, and if you believe that friction is 0, you could use a really tiny engine in your car.

The acceleration along the outside rim of the basket at 3450 RPM is indeed about 1266 g. This is an inward acceleration. If your material is evenly distributed along the outside edge of the basket then the centrifuge is nicely balanced and the net force on the center bearing is zero. I think instead you want to consider a worse case, where the 75 pounds is all clumped together and the basket is way out of balance. In this case the force on the center bearing is: 1266g * mass = 1266*75 lb = 9450 pounds.