Calculating Motor Torque and RPM

what height and in what time?

THE TILTING PROCESS IS 20 MINUTES.

THE WIDTH OF THE OBJECT TO BE TILTED IS 3800mm AND IT HAS TO BE TILTED BY 180 DEGREES

Your post is not very clear, still on the basis of what I undersatnd

180 Deg in 20minuts = 0.5/20=0.025 RPM

And mass of rod is 12000KG

CG at 3800mm/2=1900mm =190cm

so total torque required is 190*12000=2280000kg-cm

Now lets say you choose a motor of 1400 RPM at rated load

For that you need a gear box ratio of=1400/0.025=56000

so Requirements of torque for motor is about 2280000/56000=40.7kg-cm

On the basis of gear box efficiency you can put a factor on it.

THANK YOU.

THAT HELPED ME. I HAD A DOUBT. I HAD THOUGHT TORQUE = 12000 KG *0.075m

=900 kgm . what is the mistake? PLEASE LET ME KNOW.

It will be much better if you take some time and make a drawing of what you want to do?

Ok

Now you need to specify

1 CTC of motor shaft (gear shaft infact)and 150mm shaft

2 mass of shaft itself

3 mass of plate fixed on motor shaft

4 and also the mass of attachment on other side

SHAFT DIA:150 mm

WEIGHT TO BE TILTED 12000 KGS

CALCULATE RPM OF SHAFT AND TORQUE REQUIRED FOR THE MOTOR TO START.

what is the mistake?

From this initial data of your first post you can only get your answer
" TORQUE = 12000 KG *0.075m = 900 kgm ."

You then added additional info.

THE TILTING PROCESS IS 20 MINUTES.

THE WIDTH OF THE OBJECT TO BE TILTED IS 3800mm AND IT HAS TO BE TILTED BY 180 DEGREES

We now have some extra info and from this the following was derived.

"180 Deg in 20minutes = 0.5/20=0.025 RPM"

"Now lets say you choose a motor of 1400 RPM at rated load"

"For that you need a gear box ratio of=1400/0.025=56000"

Let's assume your calculation "TORQUE = 12000 KG * 0.075m = 900 kgm"... is correct .
You then want a motor / gearbox combo to drive the 150mm shaft @ 0.025 rpm .
The OUTPUT torque of this motor gearbox combo will be 900 kg / m.
The driving motor INPUT torque is however going thru a gearbox and will be 900/56000 = 0.0160 kg-m or 0.16 N/m .

Any help?

From this initial data of your first post you can only get your answer
" TORQUE = 12000 KG *0.075m = 900 kgm ."
Your answer assumes that the total mass of 12000 kg's is acting at a point distance of 0.075m from the shaft centerline ...IS IT?
Or is it a total mass of 12000 kg's but the center of the mass from the shaft centerline is unknown?... in which case you'll need to provide the previous posters with the mass dimensions.

what to do if the weight to be lifted is in L shaped object??????????