- #1
SevenToFive
- 56
- 3
We have a 1500lb steel drum that has a 8 foot diameter that we need to rotate to wrap up air filter material. The drum is supported by bearings on both sides and needs to turn at 0.02rpm.
I calculated the torque, using T=Moment of Inertia * Angular Acceleration. For the moment of inertia I used the equation for a hollow cylinder.
Gravity =386.088in/sec^2
For the weight I figured 2500lbs to factor in losses through friction(0.5). Outside diameter with the roll of air filter material would be 109 inches, inside diameter =100 inches. Final rpm = 0.02, Acceleration time is 5 seconds.
The angular acceleration =0.02*2*3.14/60/5 = 0.00042Rad/sec^2
Moment of inertia = 21041.835lb-in-sec^2
T=I*AA = 21041.835lb-in-sec^2 * 0.00042rad/sec^2
T=8.81lb-in
However that value seems very low to me. Any help is greatly appreciated.
I calculated the torque, using T=Moment of Inertia * Angular Acceleration. For the moment of inertia I used the equation for a hollow cylinder.
Gravity =386.088in/sec^2
For the weight I figured 2500lbs to factor in losses through friction(0.5). Outside diameter with the roll of air filter material would be 109 inches, inside diameter =100 inches. Final rpm = 0.02, Acceleration time is 5 seconds.
The angular acceleration =0.02*2*3.14/60/5 = 0.00042Rad/sec^2
Moment of inertia = 21041.835lb-in-sec^2
T=I*AA = 21041.835lb-in-sec^2 * 0.00042rad/sec^2
T=8.81lb-in
However that value seems very low to me. Any help is greatly appreciated.