# How can we calculate the required torque to rotate 1200 kg?

1. Mar 28, 2015

### Rajastc

I want to calculate torque to rotate 1200 kg of mass which is attached in 1.5" dia shaft. Shaft is supported by 2 bearings at its end. Neglect bearing frication. Please help.

2. Mar 28, 2015

### Staff: Mentor

Without friction every non-zero torque will let it start rotating - the acceleration depends on the torque of course.

3. Mar 28, 2015

### Jobrag

Once you've got it spinning not very much, I think that you need to ask a different question, explain what you're trying to do first.

4. Mar 28, 2015

### Rajastc

Thanks mfb & Jobrag, Actually i need to design maintenance cart, where i need to support 1200 kg of assembly. Need to rotate 180 degree for first set of maintenance and then rotate again another 180 degree to complete the remaining work.

Planning to use spur gear mechanism and hand wheel to rotate. so i want to calculate torque to rotate 1200 kg.

5. Mar 28, 2015

### Staff: Mentor

This will be determined by the friction your system has. And by time constraints, assuming you don't want to wait an hour to rotate it.

6. Mar 28, 2015

### Jobrag

This would be easier with a pint and a sheet of paper. How symmetrical is the piece of equipment, how are you supporting the the bits on the shaft, can you post a photograph of the equipment.

7. Mar 28, 2015

### Rajastc

#### Attached Files:

• ###### Picture1.jpg
File size:
23.1 KB
Views:
640
8. Mar 28, 2015

### Jobrag

If the bit of equipment is as symmetrical as your sketch shows the torque required to rotate it will be negligible.

9. Mar 28, 2015

### Rajastc

if the mechanism is symmetrical, respective load, torque will be negligible. And consider no bearing friction. Right?

But the formula for torque is
T= mass x g x Radius x coefficient of frication.

10. Mar 28, 2015

### Jobrag

In the set up you show the relevant r will be close to zero. Thought experiment with the set up you have will the piece of equipment tend to stay where it is or will the heavy side swing down?

11. Mar 28, 2015

### tygerdawg

What is missing in this discussion is the mass moment of inertia of the object to be rotated. Sure, mass = 1200kg. But you also must rotationally accelerate & decelerate that mass, and that requires torque. That is calculated with a rotational mass moment of inertia formula. Either you get the MMI from a CAD model with mass properties or calculate it yourself. Many sources for MMI of primitive shapes (like cylinders, cubes, etc.) that will provide estimates.

For inertia, Torque = MMI x AngularAcceleration.

AngularAcceleration is approximately (ChangeInAngularVelocity) / (ChangeInTime). So you must specify your motion parameters much better in order to gain accuracy.

Finally, turning the object from rest requires PeakTorque = (TorqueToAccelerate) + (TorqueForFriction) + (TorqueToKeepTurning) + (TorqueForAnythingElse) + etc.

12. Mar 28, 2015

### Staff: Mentor

That's what I said earlier: if friction is negligile, even very small torque values will be sufficient.

13. Mar 29, 2015

### Rajastc

Thanks tygerdawg , mfb & Jobrag..I got fair idea to proceed further.

The formula for torque:
1. Torque to accelerate= MMI x Acceleration.
2. Torque for frication= T=μ*m*g*R
where :
T : the required torque
μ : the friction coefficient
m : the weight of the cylinder
g : earth gravity
R : radius at contact bearings/cylinder
Can you confirm the above torque for frication formula?

3. Torque to keep turning :
In my case , no need continuous rotation. but for understanding, can you give some more information?

14. Mar 29, 2015

### Staff: Mentor

Looks good.
Torque to keep it moving is just friction, but with the coefficient of kinetic friction instead of static friction.

15. Mar 30, 2015

### Kishan Majethia

To initiate rotation we will need to consider co-efficient of static friction and "torque required to keep it turning" will require to consider co-efficient of kinetic friction. so
Torque required to keep turning = Torque to accelerate= MMI x Acceleration + Torque for frication= T=μkinetic*m*g*R

16. Mar 31, 2015

### Rajastc

Hi All, Thank for the great help. I got good understanding to proceed further.