Mass Moment of Inertia Calculation

[jdkuhndog]

I am trying to determine the HP for a little motor to rotate a hook. There is a gear around the hook which meshes with the little gearmotor I am using - there is a 6:1 ratio between the output gear on the gearmotor and the gear around the hook. The output gear on the gearmotor rotates at 12.2 RPM.

The hook will have a load hanging on it that is a cylinder in the vertical position. It is 2' in diameter and weighs 35000#.

This is what I have done:

Mass moment of Inertia (Im)= (m*r^2)/2
Im = (35000# * (1 ft)^2)/2 = 17500 #ft

RPM of hook = 12.2 / 6 = 2.03RPM

w = (2.03 * 2 * pi)/60 = 0.213 RAD/sec

a = w/t = (0.213)/5 sec = 0.0426 RAD/sec^2 (assume 5 seconds to accelerate load to full speed)

T = (I*a)/g = 17500 * 0.0426 / 32.2 = 23.4 ft#

T at gearmotor = 23.4/6 = 3.9 ft# = 46.8 in#

HP = 3.9 * 12.2/5250 = 0.009 HP (I have selected an 1/8 HP motor).


Is this done correctly ??? I didn't break things out into slugs, so I wasn't sure if this was correct.

Thanks for any input.

 

[hacksaw]

how do you plan to stop the rotation?

 

[jdkuhndog]

I assumed (always a bad thing to do) that if it took such a small Torque/HP to accelerate the Inertia, then an equally small Torque would be required to decelerate it...therefore I will assume simply taking power off of the gearmotor will be sufficient to bring this load to a stop.

Am I wrong ?

Thanks.

 

[hacksaw]

17500 ft-lbs is a lot of torque to dissapate...

equivalent of a footbal player trying to stop an elephant...

 

[jdkuhndog]

Sorry - I had the units wrong in my example.

The mass moment of Inertia is 17500 # ft^2.

The torque is only 23.4 ft# as shown.

Thanks !

 

[hacksaw]

what is the friction load (dynamic and break-away torques)?

 

[jdkuhndog]

Friction Load and Break-Away torques are negligible.