# Calc Torque Req for Rotating 35lbs @ 25RPM

• lulubell
In summary, Russ is asking for help calculating the torque requirements for a project that is as follows: -Two 5/8" diameter aluminum rods that are 3 feet long and 6" apart from each other need to rotate a maximum load of 35 pounds at a speed of 25 rotations per minute continuously for long periods of time -Can someone please tell him what the torque requirements would be?-Can you draw a diagram?-The description does not state if the rods are balanced or if the axis of rotation is vertical or horizontalAs a summary, the torque required for this scenario is: -Accelerate the system from zero to 25 RPM -Then keep it running at 25 RPM.
lulubell
TL;DR Summary
motor torque needs
I need help calculating the motor torque requirements for a project I'm working on...the senario is as follows...
Two 5/8" diameter aluminum rods that are 3 feet long and 6" apart from each other. These rods need to rotate a maximum load of 35 pounds at a speed of 25 rotations per minute continuously for long periods of time. Can someone please tell me what the torque requirements would be for this scenerio?

Can you draw a diagram? The description doesn't tell us if they are balanced and what the axis of rotation is.

We also need the acceleration.

Last edited:
lulubell said:
Can someone please tell me what the torque requirements would be for this scenerio?
And in addition to Russ' request, can you please tell us your bearing friction/torque numbers and the required angular acceleration of your setup? The torque required to keep a setup rotating at a constant angular velocity with frictionless bearings is...

russ_watters and Tom.G
Welcome to PF.

You have two separate problems here.
1. Accelerate the system from zero to 25 RPM.
2. Then keep it running at 25 RPM.

The analysis will depend on several details.
Is the 35 lb load, a drum that rests on the two shafts, or is it made up of fixed weights attached to the shafts?

Is the 25 RPM that of the rods, or of a drum with a specified diameter, that rests on and is driven by the rotating rods?

Can we assume you will drive it with a particular type of electric motor?
Are the rods both driven by the same one motor?
How are the rods coupled together, by a belt or chain?

Are you tumbling stones? Knowing the application, and having a sketch, would go a long way to answering the question.

Ok...lets see how I do...

First, I've attached a drawing to give you an idea of what I'm talking about, and yes, it is for tumbling stones.

I have six 5/8" bore, sealed, self aligning pillow block bearings, but I'm not sure what you mean by 'bearing friction/torque number', so here is a link to the specs on the bearings... https://www.mcmaster.com/5913K62/

The load can vary a little, but the 35lbs is the max. The load is from two different size molded rubber barrels that are round on the outside, ten sided on the inside, and filled 3/4 of the way full with rocks and media. The 12lb barrels have an 8 3/4" outside diameter, 6 3/4" inside diameter, and are 7 3/4" long and the 6lb barrels are the same diameter but are only 4 1/2" long. And yes, they would sit on and be driven by sprockets on the rotating rods (suggestions on the amount of teeth...?) I say the load varies because I generally run three barrels that could be either of the two weights.

Yes, I have a particular electric motor I would like to use to drive the two rods that the barrels will sit on...
it's an obsolete Bodine gear motor with a 3/8" shaft that I would like to use because it's quieter and cheaper to run, all the specs I was able to get are on the attached drawing. Right now it turns clockwise, but it is reversible if need be. The rods will be chain driven by this motor, all the sprockets I have at this time have 16 and 17 teeth, but if I need to buy different ones to achieve the necessary torque/speed than I will.

Hopefully I covered everything...?!
Thanks to all for all the help!

Last edited by a moderator:
Baluncore
Let's do some initial calculations.

The barrels have an 8-3/4" outside diameter and rest on the 5/8” driver rods. The barrel rests on the smaller diameter rods, so it will rotate slower. The speed of the circumference surface is proportional to the diameter.

If the 5/8” shafts rotate at 25 RPM.
25 RPM * 0.625" / 8.75" = 1.78 RPM
That is one turn of the barrel in 14 seconds, which seems a bit slow to me.

To rotate the drums at 25 RPM, the shafts would need to rotate at;
25 RPM * 8.75" / 0.625" = 350 RPM.
But your 18:1 gear motor output shaft only does 94 RPM.
You would need to gear that up by about 3.75 times faster.

A similar 60 Hz induction motor will run under load at about 1700 RPM. That needs to be reduced to 350 RPM, so you will need;
1700 RPM / 350 RPM = 4.857 gear or chain drive ratio.

Am I correct in assuming the 25 RPM is the drum speed?

The self-aligning bearings should survive if you keep them lubricated.

The main load on the motor will be the tumbling stones.

Consider a virtual drum rotating at 25 RPM, 8.75” in diameter.
The loose stones will ride up the inside of the drum until the top of the heap falls. In effect, the stones are rolling in the bottom half of the drum at twice the drum rotation rate. 35 lb of stone is being lifted, maybe about 5”, twice each 2.4 seconds.

It is easiest to do all computations in SI metric units so power comes out in watts.
35 lb = 15.875 kg.
5” = 0.125 metre.
Acceleration due to gravity g = 9.8 m·s-2.
Force = m·g = 15.875 * 9.8 = 155.6 newton.
Energy per half turn, tumble; E = m·g·h = 155.6 * 0.125 = 19.45 joules.
Power in watts = joules / sec = 19.45 J / 1.2 s = 16.20 watts.
1 HP = 745.7 watt.
16.20 watts = 16.20 / 745.7 = 0.0217 HP = 1/46 HP.
Your motor can deliver 1/30 of a HP, so it could do the job.

If you use the available Bodine gear motor, the gear ratio is unfortunately too far from what is needed.
With 16 tooth sprockets on the 5/8" rollers, the gearing up would need a 16 * 4.857 = 78 tooth sprocket on that gear motor. But 78 tooth is a bit big for a cheap bicycle chain sprocket.

I would fit 12 tooth sprockets on the rollers, since that is the minimum to run quietly.
Then a 12 * 4.857 = 58 tooth sprocket on the gear motor output shaft.

I expect it might be easier to buy another gear motor, so it could be used to direct drive one of the rollers at 350 RPM. That would also eliminate side forces on the gearbox output shaft. The two rollers can then be simply coupled with a loop of chain over two identical sprockets.

An alternative would be to use the Bodine gear motor, but replace the two long shafts with two aluminium tubes; 0.625 * 4.857 = 3” in diameter, then direct drive can be used. You would not need centre support bearings with the thicker tube. You could use a plumber block bearing beyond the end of each tube, but would need to adapt the end of the tube to the smaller axle for the lower cost, smaller and more efficient bearings.

Tom.G
Baluncore said:
Let's do some initial calculations.

The barrels have an 8-3/4" outside diameter and rest on the 5/8” driver rods. The barrel rests on the smaller diameter rods, so it will rotate slower. The speed of the circumference surface is proportional to the diameter.

If the 5/8” shafts rotate at 25 RPM.
25 RPM * 0.625" / 8.75" = 1.78 RPM
That is one turn of the barrel in 14 seconds, which seems a bit slow to me.

To rotate the drums at 25 RPM, the shafts would need to rotate at;
25 RPM * 8.75" / 0.625" = 350 RPM.
But your 18:1 gear motor output shaft only does 94 RPM.
You would need to gear that up by about 3.75 times faster.

A similar 60 Hz induction motor will run under load at about 1700 RPM. That needs to be reduced to 350 RPM, so you will need;
1700 RPM / 350 RPM = 4.857 gear or chain drive ratio.

Am I correct in assuming the 25 RPM is the drum speed?
Yes, the 25 RPM is the drum speed

## What is torque and how is it related to rotating a mass?

Torque is a measure of the rotational force applied to an object. It is related to rotating a mass because it determines how much force is needed to make the object spin around an axis. The formula for torque is τ = r * F * sin(θ), where τ is torque, r is the radius or distance from the axis of rotation, F is the force applied, and θ is the angle between the force and the lever arm.

## How do you calculate the torque required to rotate a 35 lbs mass at 25 RPM?

To calculate the torque required to rotate a mass, you need to know the moment of inertia (I) of the mass and the angular acceleration (α). The formula is τ = I * α. For a constant rotational speed (like 25 RPM), you can use τ = I * ω, where ω is the angular velocity. First, convert 25 RPM to radians per second (ω = 25 * 2π / 60). Then, determine the moment of inertia based on the shape and mass distribution of the object. Multiply I by ω to get the torque.

## What units should be used in the torque calculation?

Torque is typically measured in Newton-meters (Nm) in the International System of Units (SI). When working with pounds (lbs) and inches or feet, you may use pound-feet (lb-ft) or pound-inches (lb-in). Ensure all units are consistent throughout the calculation to avoid errors.

## What is the moment of inertia and how do you find it for a given mass?

The moment of inertia (I) is a measure of an object's resistance to changes in its rotational motion. It depends on the mass distribution relative to the axis of rotation. For simple shapes, there are standard formulas. For example, for a solid cylinder, I = 0.5 * m * r², where m is the mass and r is the radius. For complex shapes, you may need to integrate the mass distribution or use computational methods.

## How does the radius of rotation affect the torque required?

The radius of rotation significantly affects the torque required. A larger radius increases the moment of inertia, which in turn increases the torque needed to achieve the same angular velocity. The relationship is quadratic, meaning that doubling the radius will quadruple the required torque, assuming the mass remains constant.

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