Calculating Torque for a 21,000 lb Merry-Go-Round at 4 RPMs: A Design Challenge

[adcoleman4]

Hi everyone,

I'm charged with designing a merry-go-round for a class project. It is to be driven by an electric motor and I chose to power it with a gear train. I need to pick a specific horsepower motor to drive the pinion and subsequent gears after it, finally leading to an inner ring gear that turns the whole structure.

My question is this, if the whole merry-go-round weighs about 21,000 pounds, how do I determine the amount of torque the inner ring gear must be applied to turn it at a speed of 4 RPMs?

I've got a good start but am not confident with my answer. I calculated the inertia of three separate bodies that comprise the merry-go-round and found that to be about 108,000 lbf-ft^2. At 4 RPM, I calculated the angular acceleration of the 36'-diameter merry-go-round to be 0.00833 rad/sec^2. I've found some equations online that say torque is equated by this equation:

τ = (I/g)*α and I got about 27.98 lbf-ft. That seems low for a 21,000 lb merry-go-round.

Can anyone on here tell me if I am correct or where I possibly went wrong?

And also could anyone give me any insight as to configure my gear train between the inner ring gear and the motor to accurately size the motor?

Thanks a lot!

 

[Travis_King]

First. In order to get a clearer picture of the system requirements, you must think of the process the other way around (i.e. not from motor to driven wheel). You must first start with the big wheel, then work your way backwards out to the motor, taking into consideration losses and efficiency along the way and then adding a factor of safety. If that's how you are already thinking of it, great!

Some math errors, not sure where you went wrong, but:
21,000 lb converts to 652 slugs.
I = .5 * 652 * 18^2 = 105652 slug-ft^2

T=I*a = 105652 slug-ft^2 *.00833 rad/s^2 = 881 lbf-ft

Try it in metric and you'll get the same thing!