# Torque Required to Spin Solid Cylinder

1. Jun 29, 2011

### jafolkerts

I am working on a project and I am having a bit of difficulties with the math. It has been awhile since I took physics in college so please bear with me.

I am attempting to calculate the amount of torque required to spin a solid cylinder resting upon a set of steel rollers. Attached to the cylinder is a flywheel that is 14" in diameter or 0.127 meters in radius.

Relevant Data
Cylinder plus flywheel mass = 2750 kg
Cylinder rolling resistance upon steel rollers (Coefficient of friction ($\mu$)) = 0.0024

At this point, I am merely trying to calculate the amount of effort required to maintain a constant rotation.

Here is my working theory:

First, we need to calculate the force ($F_c{}$) that is required to overcome the rolling resistance of the cylinders on the steel rollers:

$F_c{}$ = $\mu$N
$\mu$ = Coefficient of friction = 0.0024
N = Normal Force = MG (M = mass of cylinder (2750 kg) and G = earth's gravitational acceleration (9.81 m/sec2))
N = (2750 kg)(9.81 m/sec2)
N = 26,977.5 N

$F_c{}$ = (0.0024)(26,977.5 N)
$F_c{}$ = 64.746 N

I assert that it will take at least 64.746 N of force to spin the cylinder. Assuming this is correct, I would now like to calculate the torque required to spin this cylinder by applying this force to the edge of the flywheel using this formula:

Torque = Force x Distance
Torque = (64.746 N)(0.127 m)
Torque = 8.222 Nm

Here's the issue that I am having troubles with: if I increase the flywheel's diameter, intrinsically this should decrease the amount of torque required to spin the cylinder but according to my math/formulas, the torque is also increased. So somewhere I have made an incorrect assumption that I need some help with. Thank you!

2. Jun 29, 2011

### Drakkith

Staff Emeritus
I don't know all the math, but I think the torque applied to the flywheel increases because R increases, but the speed of rotation is reduced. Applying the same force at twice the radius increases the torque by 2x but the speed is reduced by half. (I don't know if the numbers are correct, but you get my point)

I think your mistake is that the amount of torque required to spin the cylinder stays the same, but the force changes since the larger radius gives you more torque per force input.

3. Jun 30, 2011

### Blibbler

Hi - I think you might be tying yourself up in knots as this is a rotational movement and you're using linear mechanics.

F=ma in linear motion

F=Iw in rotational motion where I is the moment of inertia (integral of mass x radius from centre to outer edge of cylinder) and w is angular velocity in radians per second ( it is an acceleration as it involves continuous change of direction for each part of the cylinder - velocity is a vector so a change in either speed or direction is an acceleration).

4. Jun 30, 2011

### rcgldr

That's not the coefficient of friction, it's the rolling resistance as you have stated.

This is correct.

Is the cylinder or the flywheel doing the rolling on the steel rollers?

If it rolls on the cylinder, then the required torque is Fc x Rcylinder. The required force on the flywheel = Fflywheel = Fc x Rcylinder / Rflywheel .

If it rolls on the flywheel, then the torque needed is Fc x Rflywheel, as you have calculated. The torque is greater in this case, but the rate of rotation is slower.

Last edited: Jun 30, 2011
5. Jun 30, 2011

### BobG

I guess so.

It is true that if you need a given amount of torque, you'll have to apply less force to get that torque if you apply that force further away from the center.

But I'm not sure why you think it would take less torque to rotate an object with a larger diameter, since it's not true (having a larger diameter will give it a larger moment of inertia, making it harder to change the rotation rate).

And I'm not quite sure why that's relevant to this situation.

In this case, you have an example of where you're applying a given force (the force of friction) a certain distance away from the center to get a given amount of torque. If you apply that same force (the force of friction) further away from the center, that same force will get you more torque.